PTA Program Design Ladder Competition (Questions 1~180)

The article directories

• • 1, Hello World (5 分)
  • 2. Print the hourglass (20 points)
  • 3. Single-digit statistics (15 points)
  • 4. Calculate Celsius temperature (5 marks)
  • 5. Seat Number (15 points)
  • 6. Continuous factor (20 points)
  • 7, A-B (20 marks)
  • 8. Calculate the index (5 marks)
  • 9, Calculate the factorial and (10 marks)
  • 10, Simple questions (5 分)
  • 11. Square with Obama (15 points)
  • 12. Checking ID cards (15 points)
  • 13. How many are there (15 points)
  • 14. Big Ben (10 minutes)
  • Who will pour first (15 points)
  • 16. Too handsome to have any friends (20 points)
  • 17. Say important words three times (5 marks)
  • 18. Odd and Even split (10 points)
  • 19. Output GPLT (20 points)
  • 20, The day after tomorrow (5 marks)
  • 21, A+B (15 分)
  • 22, I Love GPLT (5 分)
  • 23. Rent (20 points)
  • 24. Are you too fat (5 points)
  • 25. A gang of one (15 points)
  • 26. Are you too fat? (10 points)
  • 27, Left pad (20 points)
  • 28. Year of Birth (15 points)
  • 29. Thumbs up (20 points)
  • 30. Valentine’s Day (15 points)
  • 31, A times B (5 marks)
  • 32, A divided by B (10 marks)
  • 33. New World (5 points)
  • 34. Ancient Style layout (20 points)
  • 35. Best Height difference between couples (10 points)
  • 36. Looking for 250 (10 points)
  • 37. Date formatting (5 marks)
  • Reading Room (20 points)
  • 39. Sure win (15 points)
  • 40, the universe is invincible hello (5 points)
  • 41. Single (20 points)
  • 42. Pretending to sleep (10 points)
  • 43, MATRIX A times B (15 marks)
  • 44, the NTH string (15 points)
  • 45. Discount (5 points)
  • 46. 2018 We need to win (5 points)
  • 47. Electron Wang (10 points)
  • 48. Who is the winner (10 points)
  • 49. De-weighting linked lists (25 points)
  • 50. Hall of Fame and Vouchers (25 points)
  • 51. Make small change (30 cents)
  • 52. (3n+1) Guess (15 points)
  • 53, Write this number (20 marks)
  • I want to pass! (20 points)
  • 55, Change the format of output integer (15 marks)
  • 56. Prime Pair Conjecture (20 points)
  • 57. Problem with loop right shift of array elements (20 points)
  • 58. Being sarcastic (20 points)
  • 59, The Derivative of a Unary Polynomial (25 marks)
  • 60, A+B and C (15 分)
  • 61, Prime numbers (20 points)
  • 62, Theory of Moral ability (25 points)
  • A divided by B (20 marks)
  • 64. Hammer, Paper, Scissors (20 points)
  • 65. Moon cake (25 points)
  • 66. Single digit statistics (15 points)
  • 67、 D进制的A+B (20 分)
  • 68, The minimum number of groups (20 points)
  • 69, Check id card (15 points)
  • 70. Code with Obama (15 points)
  • 71. Stroke (15 points)
  • 72, The sum of the number of combinations (15 marks)
  • 73, True or false (15 points)
  • 74. Image Filtering (15 points)
  • 75. Wifi Password (15 points)
  • 76. Checking passwords (15 points)
  • 77. I won’t tell you (15 points)
  • 78, How many different values are there (20 points)
  • 79, Dangerous goods packing (25 points)
  • 80, N- Defensive number (15 points)
  • 81, Programming in C is fun! (5 points)
  • 82. Taxi Fare (15 points)
  • 83, Finding the Partial Sum of power Series Expansion (20 marks)
  • 84, What is a computer? (5 points)
  • 85. Assign x squared to Y (5 marks)
  • 86. Calculate the running time of the train (15 minutes)
  • 87, Calculate the salary (15 points)
  • 88, Day K candle picture (15 points)
  • 89. Are you too fat (10 points)
  • 90. Find the smallest string (15 points)
  • 91. Bubble sorting of Strings (20 points)
  • 92, The organization of the week information (10 points)
  • 93. Even and Odd (10 points)
  • 94. Pirates’ Spoils (25 points)
  • 95. Speeding judgment (10 points)
  • 96, JMU-DS – Order table interval element deletion (15 marks)
  • 97, Find the specified character (15 marks)
  • 98. Four Operations on complex numbers (15 points)
  • 99. Output triangle character array (15 points)
  • 100. Find the longest string (15 points)
  • 101, The classification of integers (20 points)
  • 102. Length of Service (20 points)
  • 103, 7-10 Array looping left (20 points)
  • Find the reciprocal KTH term of a chained linear list (20 marks)
  • 105, Square Circle to the right (20 points)
  • 106, The minimum number of groups (20 points)
  • 107. Monkeys eat peaches (15 points)
  • 108, String uppercase conversion (15 points)
  • 109. Calculation of piecewise Functions [3] (10 marks)
  • 110, Find integer (10 marks)
  • 111. Tourism Planning (25 points)
  • Number of Black Holes (20 points)
  • 113. The Hare and the Tortoise (20 points)
  • 114, Spiral square (20 points)
  • 115. Delete substrings from a string (20 points)
  • 116, A-B (20 points)
  • 117, Guess the number (20 points)
  • 118, 40059 Four Operations (15 points)
  • Who is the tallest in the dormitory (20 points)
  • 120. Stepped electricity price (15 points)
  • 121. Speedway penalties (15 points)
  • 122, catch the mouse ah ~ lost or earned? (20 points)
  • 123. Find a ball with a scale (10 points)
  • 124, Find the set of integers that meet the given conditions (15 marks)
  • 125. Count 24 points (25 points) with playing cards
  • 126, Approximate PI (15 marks)
  • 127. A Simple Calculator for Two Numbers (10 points)
  • 128, Introduction to Arithmetic: Addition, Subtraction, Multiplication and Division (10 marks)
  • 129, The NTH string (15 points)
  • 130. Calculating the value of a symbolic function (10 marks)
  • 12-24 hours (15 points)
  • 132. Grade Change (15 points)
  • Finding the number of combinations (15 marks)
  • 134. Output full array (20 points)
  • 135. Magic Coupon (25 points)
  • 136. Find the minimum value (20 points)
  • 137. Input and display of address book (10 points)
  • 138. Calculate how far an object falls (5 points)
  • 139. Time conversion (15 marks)
  • 140. Han Xin Points soldiers (10 points)
  • 141, BCD decryption (10 points)
  • 142. Number of traps (15 points)
  • 143. Simplified Insertion sort (15 points)
  • 144. Comparison of Rational Numbers (10 marks)
  • 145, Calculating the wages of employees (15 points)
  • 146. The Monkey chooses the King (20 points)
  • 147. Three digits in reverse order (10 points)
  • 148. Scoring Rules (5 points)
  • 149. Judges score (5 points)
  • 150, Sum of Special A Series (20 marks)
  • 151. Change coins (20 cents)
  • 152. Drop Ball (20 points)
  • 153, Output student scores (20 points)
  • 154, Find the number of digits and the sum of the digits (15 marks)
  • 155, The simplest fraction (15 marks)
  • 156. I am the Flag Raiser (10 points)
  • 157. RMB exchange (15 marks)
  • 158. Median of two ordered sequences (25 points)
  • 159. Find the student with the highest total score (15 points)
  • 160. Finding the Partial Sum of Simple Staggered Sequences with Given Precision (15 marks)
  • 161. Length of Words (15 marks)
  • 162. Table Output (5 marks)
  • 163. Packing Problems (20 points)
  • Find the first N terms of staggered sequence and (15 marks)
  • 165. Finding the mean square Deviation of Set Data (15 marks)
  • 166. Currency conversion (20 cents)
  • 167. Finding the Local Maximum of the Matrix (15 marks)
  • 168, Check the price of fruit (15 points)
  • 169. Mixed type data Formatting input (5 points)
  • 170, BCD decryption (10 marks)
  • 171. String Pattern Matching (25 points)
  • 179, Integers (5 分)
  • 173. PTA makes me refreshed (5 marks)
  • 174. Temperature conversion (5 marks)
  • 175. Output diamond pattern (5 marks)
  • 176. Fish or meat (10 points)

It’s not easy to make. Please give me a thumbs up if you like. I’m Wang Rui.

1, Hello World (5 分)

There’s no input for this super simple problem.

All you have to do is print the famous phrase “Hello World!” in one line. That’s it.

Example: None Example output: Hello World!

public class Main {
	public static void main(String[] args) {
		System.out.println("Hello World!"); }}Copy the code

2. Print the hourglass (20 points)

import java.util.Scanner;
public class Main {
	public static void main(String[] args) {
		int i,j,k;		// for loop
		int n;			// There are several positive integers
		int b=1;		// Indicates the maximum number of rows
		int h=1;		// Count the rows
		int sum=1;		// At least 1 star will be printed
		int g=0;		// Determines the element to print Spaces
		int index;		// Remember the number of stars printed in the bottom half of the line
		String Str;		/ / symbol
		char c;
		Scanner dx = new Scanner(System.in);
		n = dx.nextInt();
		Str = dx.nextLine();	// Enter a string
		c = Str.charAt(1);		// Use the string-to-character method to get the character
		if(n<7) {
			System.out.println(c);
			System.out.println(n-1);
			System.exit(0);		// Request to terminate the JVM
		}
		else {
			while(sum<=n) {
				b = b+2;
				sum = sum + b*2;
				h++;
			}
		}
		h=h-1;		// Since K starts at 1, there will be one more line. Subtracting 1 is the true number of lines in the top and bottom half
		index = h;	// Protect the lower half of the line
		sum = sum - 2*b;	// Subtract the maximum number of rows on both sides, that is, the number of overcounts
		sum = n -sum;
		for(i=h; i>=1; i--)// Print the upper part [including the middle symbol]
		{
			for(j=1; j<=g; j++) System.out.print("");
			for(k=1; k<=2*i-1; k++) System.out.print(c); System.out.println(); g++; }for(i=2; i<=index; i++){for(j=1; j<=(g-2); j++){ System.out.print("");
			}
			for(k=1; k<=2*i-1;k++){
				System.out.print(c);
			}
			System.out.println();
			g--;
		}
		System.out.println(sum);
	}
}
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3. Single-digit statistics (15 points)

L1-003 Unit digit statistics (15 minutes) Given a k-digit integer N= D k−1 10 K −1 +… + D 1 10 1 + D 0 (0≤ D I ≤9, I =0…, K −1, D k−1 >0), write a program to count the number of occurrences of each different unit digit. For example, given N=100311, there are two zeros, three ones, and one three.

Input format: Each input contains 1 test case, that is, a positive integer N with no more than 1000 digits.

Output format: For each different one digit in N, output the digit D and the number of times M in N in the format of D:M. The output is required in ascending order of D.

Example Input: 100311 Example output: 0:2 1:33 3:1

import java.util.Scanner;

public class Main {
	
	public static void main(String[] args) {
		
		int shu[] = new int[10];
		Scanner sc = new Scanner(System.in);
		String a = sc.nextLine();
		int len = a.length();
		for(int i = 0 ; i < len ; i++) {
			shu[a.charAt(i)-48] + +; }for(int i = 0 ; i < 10 ; i++) {
			if(shu[i]! =0)
				System.out.println(i+":"+shu[i]); }}}Copy the code

4. Calculate Celsius temperature (5 marks)

import java.util.Scanner;
public class Main{
	public static void main(String[] args) {
			Scanner scanner = new Scanner(System.in);
			int f = scanner.nextInt();
			System.out.println("Celsius = " + (5*(f-32) /9)); }}Copy the code

5. Seat Number (15 points)

Each PAT candidate will be assigned two seat numbers when taking the exam, one is a test seat and one is an exam seat. Under normal circumstances, the examinee first gets the test seat number when entering the test machine. After taking a seat and entering the test machine state, the system will display the examinee’s test seat number. During the test, the examinee needs to change to the test seat. But some candidates are late, the test machine has been over, they can only take the test machine seat number to ask you for help, from the background to find out their test seat number.

Input format: Enter the first line to give a positive integer N (≤1000), followed by N lines, each line gives a candidate’s information: admission ticket number Test seat number Test seat number. The admission ticket number consists of 16 digits and the seats are numbered from 1 to N. Ensure that each person’s admission ticket number is different and that at no time will two people be assigned to the same seat.

After the candidate information, a positive integer M (≤N) is given. The following line contains M test seat numbers to be queried, separated by Spaces.

Output format: Output the test ticket number and test seat number of each candidate in a line separated by a space.

Example: 4 3310120150912233 2 4 3310120150912119 4 1 3310120150912126 1 3 3310120150912002 3 23 4 Example: 3310120150912119 1 3310120150912002 2

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
public class Main{
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n=Integer.parseInt(br.readLine());
		HashMap map = new HashMap();
		while(--n>=0){
			String s=br.readLine();
			String s2[]=s.split("");
			map.put(s2[1],s2[0] +""+s2[2]);
		}
		int m=Integer.parseInt(br.readLine());
		String s3[]=br.readLine().split("");
		int j=0;
		while(--m>=0){ System.out.println(map.get(s3[j])); j++; }}}Copy the code

6. Continuous factor (20 points)

import java.util.Scanner;
public class Main {
	public static void main(String[] args) {
		Scanner s=new Scanner(System.in);
		long n=s.nextInt();
		long start=0,len=0;
		long sum=0;
		for(int i=2; i<Math.sqrt(n); i++) { sum=1;
			for(intj=i; sum*i<=n; j++) { sum*=j;if(n%sum==0&&j-i+1>len) {
					start=i;
					len=j-i+1; }}}if(start==0) {
			start=n;
			len=1;
		}
		System.out.println(len);
		for(int i=0; i<len-1; i++) { System.out.print(start+i+"*");
		}
		System.out.print(start+len-1); }}Copy the code

7, A-B (20 marks)

They ask you to calculate A−B. The trouble is that both A and B are strings — that is, A−B is formed by removing all the characters contained in B from string A.

Input format: Input gives strings A and B in sequence on 2 lines. Both strings must be no longer than 10 or 4, and each string must consist of visible ASCII characters and whitespace characters, ending with a newline character.

Output format: Print the result string of A−B in one line.

Example: I love GPLT! It’s a fun game! Aeiou Example output: I LV GPLT! It ‘s fn gm!

#include<stdio.h>
#include<string.h>
int main(a)
{
	int i=0,j=0;			// For loops and array subscripts
	char str1[10010];		// Store two strings A and B
	char str2[10010];		// Used to store the string letters to be deleted
	int length;				// To store the length of a string
	gets(str1);
	gets(str2);
	length = strlen(str2);	// Just remember the length of the str2 string
	while(str1[i]! ='\ 0')	// start with str1
	{
		for(j=0; j<length; j++)// Put this letter in str2 and see if there is a corresponding letter, if there is a corresponding letter
		{					// Then the for loop will jump out prematurely, and j cannot be equal to length,
			if(str1[i]==str2[j])// Remember the key words in advance! So we can be sure from this that if the for loop executes
				break;			// At the end of the day, there is no corresponding letter in str2,
		}						// We need to display it.
		if(j==length)
			printf("%c",str1[i]);
		i++;
	}
	printf("\n");
	return 0;
}
	
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8. Calculate the index (5 marks)

#include<stdio.h>
#include<math.h>
int main(a)
{
	int n;
	int result;
	scanf("%d",&n);
	if(n<1 || n>10)		return 0;
	result=pow(2,n);
	printf("2^%d = %d\n",n,result);
	return 0;
}
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9, Calculate the factorial and (10 marks)

For a given positive integer N, you need to calculate S=1! + 2! + 3! +… +N! .

Input format: Input is given a positive integer N up to 10 in a line.

Output format: Outputs the value of S in a single line.

Input example: 3 Output example: 9 Author: Chen Yue Unit: Zhejiang University Time limit: 400 MS Memory limit: 64 MB

#include<stdio.h>
int main(a)
{
	int N;
	int i;	// for loop
	int sum=0,mix=1;	/ / sum
	scanf("%d",&N);
	if(N<1 || N>10)		return 0;
	for(i=1; i<=N; i++) { mix=i*mix; sum=sum+mix; }printf("%d\n",sum);
	return 0;
}
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10, Simple questions (5 分)

#include<stdio.h>
int main(a)
{
	printf("This is a simple problem.\n");
	return 0;
}
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11. Square with Obama (15 points)

US President Barack Obama has become the first US President to write computer code, not only calling for everyone to learn to code, but even leading by example. In late 2014, to mark the official launch of Computer Science Education Week, Obama wrote simple computer code: Draw a square on a screen. Now you can draw with him!

Input format: Enter a line that gives the square edge length N (3≤N≤21) and some character C that makes up the square edge, separated by one space.

Output format: Prints the square drawn by the given character C. But notice that the spacing between rows is larger than the spacing between columns, so to make the result look more square, we actually output the number of rows as 50% of the number of columns (rounded).

Input example: 10 A Output example: AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA aaAAAAAAAAAA

#include<stdio.h>
int main(a)
{
	int N;
	char c;
	int i,j;
	scanf("%d %c",&N,&c);
	if(N%2= =0) {for(i=0; i<N/2; i++){for(j=0; j<N; j++){printf("%c",c);
			}
				printf("\n"); }}else{
		for(i=0; i<N/2+1; i++){// Since the decimal point is always 0.5 anyway, add 1
			for(j=0; j<N; j++){printf("%c",c);
			}
			printf("\n"); }}return 0;
}
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12. Checking ID cards (15 points)

#include <stdio.h>  
#include <stdlib.h>  
   
int charToInt(char c)
{  
     return (int)(c-'0');  
 }  
 int main(int argc, char *argv[]) 
 {  
     int n,i,j,k=0,s,  
     a[]={7.9.10.5.8.4.2.1.6.3.7.9.10.5.8.4.2};  
     char cur,b[11] = {'1'.'0'.'X'.'9'.'8'.'7'.'6'.'5'.'4'.'3'.'2'},  
     c[20];  
     scanf("%d",&n);  
     for(i = 0; i < n; i++) { s=0;  
         scanf("%s",&c);  
         for(j= 0; j<17; j++) {if(c[j]>='0'&&c[j]<='9')
			 {  
                 s+=charToInt(c[j])*a[j];  
			 }
			 else
			 {  
                 printf("%s",c);  
                 if(i<n- 1)
				 {  
                     printf("\n");  
                 }  
                 s=- 1;  
                 break; }}if(b[s%11]==c[j]&&s! =- 1)
		 {  
             k++;  
         }
		 else if(b[s%11]! =c[j]&&s! =- 1)
		 {  
             printf("%s",c);  
             if(i<n- 1)
			 {  
                 printf("\n"); }}}if(k==n)
	 {  
         printf("All passed");  
     }  
     return 0;  
}  
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13. How many are there (15 points)

An integer is defined as the ratio of the number of twos in it to the number of digits. If the number is negative, the degree increases by 0.5 times; If it’s still an even number, it doubles again. For example, the number 13142223336 is an 11-digit number with three 2’s, and it is both negative and even, so its degree of second offense is calculated as 3/11×1.5×2×100%, which is about 81.82%. Calculate how two a given integer is.

Input format: The first line of input gives an integer N with no more than 50 digits.

Output format: Output the degree of N in a line, with two decimal points reserved.

Thanks to Mr. Duan Xiaoyun of Anyang Normal University and Mr. Li Fulong of Software Engineering Class 5 for the supplementary test data!

#include<stdio.h>
#include<string.h>
int main(a)
{
	char N[50];
	int i;
	float count=0,sum=0; 
	scanf("%s",&N);
	int x=strlen(N);		// Calculates the length of the string
	for(i=0; i<x; i++){// Count how many twos there are
		if(N[i]=='2')
			count++;		
	}
	if(N[0] = =The '-') {/ / negative
			sum=(count/(x- 1)) *1.5*100;
	}else{						/ / positive
		sum=count/x*100;
	}
		if(N[x- 1] %2= =0)			/ / even
			sum*=2;
	printf("%.2f%%",sum);
	return 0;
}
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14. Big Ben (10 minutes)

#include<stdio.h>
main()
{
	int m, n,h,i,k;
	scanf("%d:%d", &m, &n);
	if (m >= 13 && m < 24)
	{
		h = m - 12;
		if(n ! =0)
			k = h + 1;
		else if (n == 0)
			k = h;

		for (i = 0; i < k; i++)
		{
			printf("Dang"); }}else
		printf("Only %02d:%02d. Too early to Dang.",m,n);
}
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Who will pour first (15 points)

L1-019 Who pours first (15 points) is an interesting part of the ancient Chinese wine culture. The method that two people draw fist on wine table is: each mouth shout out a number, at the same time with hand draw out a number. If the sum of their numbers is exactly the same, the loser loses, and the loser is fined a drink. If two players win or two players lose, the next round continues until there is a single winner.

The following is the record of a and B’s drinking capacity (the maximum number of glasses they can drink without falling) and the record of pulling fists. Please judge which one of them will fall first.

Input format: Input the first line to give the amount of alcohol (not more than 100 non-negative integer) of both party A and party B, separated by Spaces. The next line gives a positive integer N (≤100), followed by N lines, each giving the record of a single stroke in the format:

A shout a row b shout B row where shout refers to the number shouted, row refers to the number drawn, both are positive integers not more than 100 (both hands row together).

Output format: Output the first person to fall in the first line: A for A, B for B. The second line tells you how many drinks the person who didn’t pour had. They guarantee that one of them will fall. Note that the program is terminated until someone falls down, and the following data need not be processed.

Example: 1 16 8 10 9 12 5 10 5 10 3 8 5 12 12 18 1 13 4 16 12 15 15 1 1 16 Example: A 1

#include<stdio.h>
int main(a)

{

	int N,A,B;		/ / A, B

	int i;

	int countA=0,countB=0;

	int Ahan[100],Ahua[100],Bhan[101],Bhua[100];

	scanf("%d%d",&A,&B);

	scanf("%d",&N);

	for(i=0; i<N; i++)scanf("%d%d%d%d",&Ahan[i],&Ahua[i],&Bhan[i],&Bhua[i]);

	for(i=0; i<N; i++){if((Ahua[i]==(Ahan[i]+Bhan[i]))&&(Bhua[i]==(Bhan[i]+Bhan[i])))

				continue;

		if(Ahua[i]==(Ahan[i]+Bhan[i])){

			countA++;				// The number of cups increases by 1

			A--;					// Drink less alcohol

			if(A<0) {printf("A\n%d\n",countB);		//count does not print that

				break; }}else
 
			if(Bhua[i]==(Ahan[i]+Bhan[i])){

				countB++;

				B--;

				if(B<0) {printf("B\n%d\n",countA);

					break; }}}return 0;

}
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16. Too handsome to have any friends (20 points)

#include <stdio.h>
#include <string.h>
int love[100010];		// Store the ID space
int main(a)
{
	int n;			// There are several circles of friends
	int k;			// There are several people in each circle of friends
	int m;			// Number of people to be queried
	int id;			/ / id number
	int newid;		// Id to be queried
	int i,j;		// For loops and array subscripts
	int flag=0;		// It is used to determine if the output result is too handsome to have friends
	scanf("%d",&n);
	    
        for(i=0; i<n; i++) {scanf("%d",&k);
            for(j=0; j<k; j++) {scanf("%d",&id);
                if(k == 1)break;// Only your own circle of friends
                love[id] = 1;   // Everyone's ID must be different, so let's set the value of all known friends to 1}}scanf("%d",&m);
        for(i=0; i<m; i++) {scanf("%d",&newid);
            if(! love[newid])// If the condition is not true or false, the non-false is true property, it already exists
            {               // There is no need for the output. To take advantage of this feature, we just need to put the output that does not exist in advance
                if(++flag > 1) printf("");
                printf("%05d",newid);
                love[newid] = 1; }}if(flag == 0) printf("No one is handsome");
        printf("\n");
    return 0;
}
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17. Say important words three times (5 marks)

L1-021 Important Words Say three times (5 marks) No input for this super simple question.

All you have to do is say the very important words — “I’m gonna WIN!” Print it three times in a row.

Note that each row is on one line, and there should be no extra characters except carriage returns on each line.

Example: None Example: I’m gonna WIN! I ‘m gonna WIN! I ‘m gonna WIN!

#include <stdio.h>
void f(a)
{
	printf("I'm gonna WIN! \n");
}
int main(a)
{
	f();
	f();
	f();
	return 0;
}
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18. Odd and Even split (10 points)

#include <stdio.h>
int main(a)
{
	int N;				// How many numbers to enter
	int i;				// for loop
	int a[1000];		// The given maximum array space stores the input data in case the subscript is out of bounds
	int num1=0,num2=0;	// use it to calculate odd/even numbers
	scanf("%d",&N);
	if(N<=0 || N>1000) return 0;	// Natural numbers greater than 0 are positive integers
	for(i=0; i<N; i++)scanf("%d",&a[i]);
	for(i=0; i<N; i++) {if(a[i]%2! =0)
			num1++;
		else
			num2++;
	}
	printf("%d %d\n",num1,num2);
	return 0;
}
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19. Output GPLT (20 points)

L1-023 Output GPLT (20 marks) Given a string of up to 10000 English letters. Please reorder the characters, press GPLTGPLT… This order is printed, ignoring other characters. Of course, the number of the four characters (case insensitive) is not necessarily the same. If a character has been printed, the remaining characters are printed in GPLT order until all characters have been printed.

Input format: A line containing a non-empty string of up to 10000 English letters.

Output format: Output the sorted string in a line as required by the topic. They make sure the output is not empty.

Input the sample: pcTclnGloRgLrtLhgljkLhGFauPewSKgt output sample: GPLTGPLTGLTGLGLL

#include<stdio.h>
#include<string.h>
int main(a)
{
	char str[10010];			// Memory space for storing input strings
	int g=0,p=0,l=0,t=0;		// These four variables are used to count the number of occurrences in the string
	int i,j;					// For loops and array subscripts
	int n;						// Get the length of the string entered
	gets(str);					// Provide input function
	n = strlen(str);			// Get the length of STR string and assign it to n
	for(i=0; i<n; i++)Start with the first letter and look for g, g, T, T, L, L,p, p
	{							// If there is one, the number of occurrences will be counted in the for loop, because each corresponding sequence will be +1
		if(str[i]=='g' || str[i]=='G')
			g++;
		else if(str[i]=='p' || str[i]=='P')
			p++;
		else if(str[i]=='l' || str[i]=='L')
			l++;
		else if(str[i]=='t' || str[i]=='T')
			t++;
	}
	while(n)					// For each letter that corresponds to the output, subtract 1
	{							// No need to print until they all return to the initial value of 0, then n=0, judged false end of the loop
		if(g! =0)
		{	printf("G");	g--;	}
		if(p! =0)
		{	printf("P");	p--;	}
		if(l! =0)
		{	printf("L");	l--;	}
		if(t! =0)
		{	printf("T");	t--;	}
		if(g==0 && p==0 && l==0 && t==0)
			n=0;
	}
	printf("\n");
	return 0;
}
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20, The day after tomorrow (5 marks)

#include<stdio.h>
int main(a)
{
	int d;	/ / input
	scanf("%d",&d);
	if(d<1 || d>7)	return 0;
	if(d<=5)
		printf("%d",d+2);
	else
		printf("%d",d7 -+2);
	return 0;
}
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21, A+B (15 分)

L1-025 positive integers A+B (15 marks) the objective of the problem is simply to find the sum of two positive integers A and B, where A and B are in the interval [1,1000]. Slightly troublesome, the input is not guaranteed to be two positive integers.

Input format: Input gives A and B on one line, separated by Spaces. The problem is that A and B don’t have to be positive integers. Sometimes they can be numbers out of range, negative numbers, real numbers with A decimal point, or even A bunch of garbled code.

Note: We consider the first space in the input to be the separation of A and B. They make sure there is at least one space and that B is not an empty string.

Output format: If the input is indeed two positive integers, the output is in the format A + B = and. If an input is not desirable, output at the appropriate location? And obviously the sum is also right? .

Input Example 1:123 456 Output Example 1:123 + 456 = 579 Input example 2:22. 18 Output Example 2:? + 18 =? Example 3: -100 blabla bla… 33 Example 3:? +? =?

import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Main {
	static String a,b;
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		a = scan.next();
		b = scan.next();
		boolean a1 = isNumeric(a);
		boolean a2 = isNumeric(b);
		if(a1&&a2) {			// If both numbers are numbers
		int number1 = Integer.valueOf(a);
		int number2 = Integer.valueOf(b);
		System.out.println(number1 + "+" + number2 + "=" + (number1+number2));
		}
		else if(a1==true&&a2==false) {//a is a number b is not
			int number1 = Integer.valueOf(a);
			System.out.println(number1 + "+" + "?" + "=" + "?");
		}else if(a1==false&&a2==true) {		//b is a number a is not
			int number1 = Integer.valueOf(b);
			System.out.println("?" + "+" + number1 + "=" + "?");
		}else {								/ / is not
			System.out.println("?" + "+" + "?" + "=" + "?"); }}public static boolean isNumeric(String str ){		// Check if it is a number
         Pattern pattern = Pattern.compile("[0-9] *");
         Matcher isNum = pattern.matcher(str);
         if( !isNum.matches() ){
             return false;
         }
         return true; }}Copy the code

22, I Love GPLT (5 分)

#include<stdio.h>
#include<string.h>
int main(a)
{
	char a[]="I Love GPLT";
	int m=strlen(a);
	for(int i=0; i<m; i++)printf("%c\n",a[i]);
}
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23. Rent (20 points)

L1-027 Rental (20 points) Here’s a picture that used to go viral on Sina Weibo:

For a time online a cry for help, urgent ask this how to break. Index [0]=2 arR [2]=1, index[1]=0 arR [0]=8, index[2]=3 arR [3]=0, and so on… It’s easy to get the phone number 18013820100.

This requires you to write a program that generates this code for any phone number — in fact, just the first two lines.

Input format: Input gives an 11-digit mobile phone number in one line.

Output format: Generate the first two lines of code for the input number, where the numbers in the ARR must be given in descending order.

Example: int[] arr = new int[]{8,3,2,1,0}; Int [] index = new int[]{3,0,4,3,1,0,2,4,3,4,4};

import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
 
public class Main {
 
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		ArrayList<Integer> alist = new ArrayList<Integer>();
		String s = sc.nextLine();// Get the phone number
		// Store the phone number in the alist array
		for (int i = 0; i < s.length(); i++) {
			if(! alist.contains(Integer.valueOf(s.charAt(i) +""))) {
				alist.add(Integer.valueOf(s.charAt(i) + "")); }}// Sort (from smallest to largest)
		Collections.sort(alist);
		/ / reverse
		Collections.reverse(alist);
		System.out.print("int[] arr = new int[]{");
		/ / output
		for (int i = 0; i < alist.size(); i++) {
			if (i == 0) {
				System.out.print(alist.get(i));
			} else {
				System.out.print("," + alist.get(i));
			}
		}
		System.out.println("};);
		System.out.print("int[] index = new int[]{");
		
		// Outputs the subscript of the phone number sequence
		for (int i = 0; i < s.length(); i++) {
			if (i == 0) {
				System.out.print(alist.indexOf(Integer.valueOf(s.charAt(i) + "")));
			} else {
				System.out.print("," + alist.indexOf(Integer.valueOf(s.charAt(i) + "")));
			}
		}
		System.out.println("};); }}Copy the code

24. Are you too fat (5 points)

#include <stdio.h>
int main(a)
{
	float H;		// Enter a positive integer to indicate someone's height
	scanf("%f",&H);
	if(H<=100 || H>300)		return 0;
	According to the formula :(kg) standard weight = height -100*0.9
	//1 kg =2 kg
	printf("%.1f\n",((H- 100.) *0.9) *2);		// The result is a decimal number
	return 0;
}
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25. A gang of one (15 points)

L1-030 a group (15 marks) A group study group is a common learning organization in primary and secondary schools, where teachers put the top students in groups with the bottom students. In this case, please write a program to help the teacher to automatically complete the assignment, that is, after getting the rank of the class students, in the current group of students, the most ranked students and the most ranked students of the opposite sex into a group.

Input format: The first line of input gives a positive even number N (≤50), which is the number of students in the class. After that, N lines are given each student’s gender (0 for girl, 1 for boy) and name (a non-empty string of up to eight English letters) in descending order, separated by one space. The ratio of boys to girls in this class is 1:1, and there is no joint ranking.

Output format: Each line outputs a group of two students’ names separated by one space. The students with higher ranking are in front and the students with lower ranking are behind. The group outputs are ranked in descending order of the previous students.

Example: 8 0 Amy 1 Tom 1 Bill 0 Cindy 0 Maya 1 John 1 Jack 0 Linda Example: Amy Jack Tom Linda Bill Maya Cindy John

#include<stdio.h>
int main(a)
{
	int N;				// Indicates the total size of the class
	int sex[50];		/ / gender
	char name[50] [15];	/ / name
	int i,j,k;			// for loop
	scanf("%d",&N);
	if(N<=0 || N>50 ||N%2! =0)	return 0;	// An even number, between 1 and 50
	for(i=0; i<N; i++)scanf("%d %s",&sex[i],&name[i][0]);
	for(i=0; i<N/2; i++)// Control the number of rows, since there are 8 people, so maximum 4 rows
	{
		for(j=i; j==i; j++)// Control searches from the highest ranking down to the highest ranking
		{
			for(k=N- 1; k>=N/2; k--)// Control the search from low to high, because there are 8 people, so the bottom 4 correspond to the top 4, only 4 times
			{
				if(sex[j]! =sex[k]&&sex[j]<=1&&sex[k]<=1)
				{
					sex[j]=10;		// Because the boys are 1 and the girls are 0, we can exclude the selected ones by using the logic judgment above
					sex[k]=10;		// Because the boys are 1 and the girls are 0, we can exclude the selected ones by using the logic judgment above
					printf("%s %s\n",name[i],name[k]);
					break;
				}
			}
		}
	}
}
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26. Are you too fat? (10 points)

#include<stdio.h>
#include<math.h>
int main(a)
{
	int n;				// There are several people
	float height;		/ / height
	float weight;		// True weight
	float biao;			// Standard weight
	int i;				// For loops and array subscripts
	scanf("%d",&n);
	for(i=0; i<n; i++) {scanf("%f %f",&height,&weight);
		biao = (height- 100.) *0.9*2;	// Get the standard weight in kg
		if(fabs(weight-biao)<biao*0.1)
			printf("You are wan mei! \n");
		else if(biao>weight)
			printf("You are tai shou le! \n");
		else
			printf("You are tai pang le! \n");
	}
	return 0;
}
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27, Left pad (20 points)

According to the news on Sina Weibo, there was a developer who was dissatisfied with NPM (Node Package Manager) and withdrew his open source code, including a module called left-pad, This is the module that crashed javascript React/Babel. What kind of module is this? It’s just padding something in front of the string to a certain length. For example, if you populate the string GPLT with a length of 10, the result of the left-pad call should be *****GPLT. The Node community has had an emergency release of an alternative to left-pad, which has been heavily mocked. Now you will implement this module.

Input format: The first line of input gives a positive integer N (≤10 4) and a character, separated by a space, which is the length of the resulting string and the character used for filling. The second line gives the original non-empty string, ending with carriage return.

Output format: Output the result string in one line.

Example 1:15 _I love GPLT Example 1: ____I love GPLT Example 2:4 * This is a sample for cut Example 2: CUT

#include<stdio.h>
#include<string.h>
int main(a)
{
	int  i;				// For loops and array subscripts
	int  n;				// Fill in the length of the string
	char c;				// Fill the string with characters
	char a[60000];		// Apply for string input space
	int lenght;			// The length of the string
	scanf("%d %c",&n,&c);
	getchar();
	gets(a);
	lenght=strlen(a);	// Calculates the length of the string
	if(lenght<n)		// When the input string < fills the string length
	{
		for(i=0; i<n-lenght; i++)printf("%c",c);
		puts(a);
	}
	else				/ / otherwise
	{
		for(i=lenght-n; i<lenght; i++)printf("%c",a[i]);
	}
	return 0;
}
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28. Year of Birth (15 points)

#include<stdio.h>
#include<string.h>
int main(a){
	int cnt=0,same=0;
	int a[10];
	//memset(a,0,sizeof(a));
	int year,t,d;
	int i;
	scanf("%d%d",&year,&d);
	t=year;
	while(1) {memset(a,0.sizeof(a));
		for(i=0; i<4; i++){ a[t%10] = 1;
			t/=10;
		}
		same=0;
		for(i=0; i<10; i++) same+=a[i];if(same==d) break;
		cnt++;
		year++;
		t=year;
	}
	printf("%d %04d",cnt,year);
	return 0;
}
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29. Thumbs up (20 points)

L1-034 Likes (20 points) There is a “like” function on Weibo where you can show your support by giving a “like” to a blog post you like. Every blog post has tags that define its identity, and the type of blog you like indirectly defines your identity. Write a program that analyzes a person’s characteristics by counting their likes.

Input format: Input in the first line gives a positive integer N (≤1000), which is the number of blog posts that the user likes. In the following N lines, each line gives a feature description of a blog that has been liked by the blog in the format of “K F 1… F K”, where 1≤K≤10, and F I (I =1…,K) is the number of feature labels. We number all feature labels from 1 to 1000. Numbers are separated by Spaces.

Output format: count the feature tag that most frequently appears in all the liked blog posts, output its number and occurrence times in a line, with 1 space between the numbers. If there is a tie, the one with the largest number is printed.

Example: 4 3 889 233 2 5 100 3 233 2 73 4 3 73 889 2 2 233 123 Example: 233 3

#include<stdio.h>
#define N 1001
int main(a)
{	
	int n;					// The number of likes
	int i,j;				// For loops and array subscripts
	int k;					// Indicates that several attribute tags are required
	int app[N]={0};			// Store label number
	int id;					/ / id number
	int index1=0;			// The maximum number of occurrences of the tag
	int index2=0;			// mark the maximum number of occurrences
	scanf("%d",&n);
	for(i=0; i<n; i++) {scanf("%d",&k);
		for(j=0; j<k; j++) {scanf("%d",&id);
			app[id]++;		// This is equivalent to a[3]++ for every value entered,
							// Increment by 1 for that coordinate}}for(i=0; i<N; i++)// Check from the smallest number according to the for loop
	{
		if(app[i]>=index1)	// If a[I] contains a value greater than the maximum number of times previously marked,
		{					// Then index1 should mark the number of occurrences
			index1=app[i];	// Because the loop is from small to large, so if there is exactly equal and ID
			index2=i;		If the number is still larger than before, the if statement is also executed
		}					// This avoids the problem of comparing number sizes in the same number of times
	}
	printf("%d %d\n",index2,index1);
	return 0;
}
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30. Valentine’s Day (15 points)

#include<math.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(a)
{
	char str1[10010] [11];	// Save space for strings
	int i;					// For loops and array subscripts
	int n=0;				// Decide to output the result, according to the list of friends into three cases
	int t;			
	for(i=0; i<1000; i++) { gets(str1[i]);if(strcmp(str1[i],".") = =0)	/ / "." Represents the end of input, using a string comparison function
		{	t=i;break;	}
		n++;				// Each time you have a friend, n increases by itself, using n as the output standard
	}
	if(n>=14)
		printf("%s and %s are inviting you to dinner... \n",str1[1],str1[13]);
	else if(n<2)
		printf("Momo... No one is for you ... \n");
	else if(n<14 && n>=2)
		printf("%s is the only one for you... \n",str1[1]);
	return 0;
}
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31, A times B (5 marks)

L1-036 A times B (5 marks) Here’s A problem you can solve in less than 10 seconds: Given two integers A and B with absolute values of 100 or less, print the value of A times B.

Input format: The first line contains two integers A and B (−100≤A,B≤100), separated by Spaces.

Output format: Print the value of A times B in one line.

Example Value: -8 13 Example value: -104

#include<stdio.h>
int main(a)
{
	int A,B;
	scanf("%d %d",&A,&B);
	if(A<- 100. || A>100 || B<- 100. || B>100)	return 0;
	printf("%d\n",A*B);
	return 0;
}
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32, A divided by B (10 marks)

#include<stdio.h>
int main(a)
{
    int a,b;
	float s;
    scanf("%d%d",&a,&b);
    s=1.0*a/b;			// The best way to compute a decimal like this is by using the
    if(b>0)				//*0.1 to keep two decimal places, because this does not affect the value
        printf("%d/%d=%.2f",a,b,s);	// If both are initially set to double, the rounding may not be handled properly when reserved
    if(b==0)
        printf("%d/%d=Error",a,b);
    if(b<0)
        printf("%d/(%d)=%.2f",a,b,s);
    return 0;
}
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33. New World (5 points)

L1-038 New World (5 marks) No input for this super easy question.

All you need to do is print the programmer’s king James phrase “Hello World” on the first line and the updated “Hello New World” on the second line.

Example value: Hello World Hello New World

#include<stdio.h>
int main(a)
{
	printf("Hello World\n");
	printf("Hello New World\n");
	return 0;
}
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34. Ancient Style layout (20 points)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(void)
{
	int i,j,n;
	scanf("%d",&n);
	char s[10010];
	getchar();
	gets(s);
	int t = strlen(s);
	int k = t/n;
	if(t%n! =0)
	k++; 
	for(i=0; i<n; i++) {for(j=n*k-n+i; j>=0; j=j-n) {printf("%c",s[j]);
		}
		printf("\n");
	}
	return 0;
}
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35. Best Height difference between couples (10 points)

Using data from multiple couples, experts found that the optimal height difference between couples follows a formula :(the woman’s height) ×1.09 = (the man’s height). If so, the height difference between the two of you, whether holding hands, hugging, or kissing, is the most harmonious difference.

Here you write a program, for any user to calculate the best height of his/her lovers.

Input format: The first line is a positive integer N (≤10), indicating the number of users to be queried. Then N lines, each line gives the gender and height of the user to query according to the format of “gender height”, where “gender” is “F” for female, “M” for male; “Height” is a real number between [1.0, 3.0].

Output format: For each query, calculate the best height of the couple for the user in one line, reserving 2 decimal places.

Example Input: 2 M 1.75F 1.8 Example output: 1.61 1.96

#include<stdio.h>
int main(a)
{
	int N;				
	char sex[10];		
	float	height[10];	
	int i;			
	scanf("%d",&N);
	if(N<1 || N>10)	return 0;
	for(i=0; i<N; i++)scanf("%s%f",&sex[i],&height[i]);
	for(i=0; i<N; i++) {if(sex[i] == 'M')
			printf("%.2f\n",height[i]/1.09);
		else
			printf("%.2f\n",height[i]*1.09);
	}
		return 0;
}
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36. Looking for 250 (10 points)

#include<stdio.h>
int main(a)
{
	int i;			
	int a[10000];	
	for(i=0; i<10000; i++) {scanf("%d",&a[i]);
		if(a[i]==250)
		{
			printf("%d\n",i+1);
			break; }}return 0;
}
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37. Date formatting (5 marks)

L1-042 Date formatting (5 marks) Different countries in the world have different dates. For example, Americans tend to write “month-day-year”, while Chinese people tend to write “year-month-day”. Please write a program below, automatically read in the United States format of the date to rewrite the Chinese custom date.

Input format: Enter the month, day, and year in the MM-DD-YYYY format. They guarantee that the dates given are legal from New Year’s Day 1900 to the present.

Output format: The year, month, and day are displayed in the format YYYY-MM-DD.

Example: 03-15-2017 Example: 2017-03-15

#include<stdio.h>
int main(a)
{
	int month;	
	int day;	
	int year;	
	scanf("%d-%d-%d",&month,&day,&year);
	if(year>=1900)
	printf("%d-%02d-%02d\n",year,month,day);
	return 0;
}
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Reading Room (20 points)

#include <stdio.h>

int main(a)
{
    int N;
    scanf("%d", &N);

    while (N--) {
        int id;
        int books[1001] [2] = {0};
        char key;
        int hh, mm; 

        while (scanf("%d %c %d:%d", &id, &key, &hh, &mm) && id) {

            // If you want to borrow
            if (key == 'S') {
                books[id][0] = - 1;              // Set the borrow flag
                books[id][1] = hh * 60 + mm;    // Count the time in minutes
            } else if (books[id][0] = =- 1) {
                // Return the book, and the previous record of borrowing

                // The total number of books borrowed increases by 1
                books[0] [0] + +;// Calculate the borrowing time
                int total = hh * 60 + mm - books[id][1];

                // Add total borrowing time
                books[0] [1] += total;       

                // Restore the loan mark of the book to allow borrowing next time
                books[id][0] + +; }// if-else

        } // while id ! = 0

        int ave_m = 0;
        if (books[0] [0] > 0) {
            double m = books[0] [1] * 1.0 / books[0] [0];
            // Round to the bottom
            ave_m = (int)(m + 0.5);
        }

        printf("%d %d\n", books[0] [0], ave_m);

    } // while (N-- > 0)

    return 0;
 }
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39. Sure win (15 points)

L1-044 Secure Win (15 points) You should be able to play “paper, hammer, scissors” : two people give hand signals at the same time, and the rules are shown below:

Now ask you to write a steady win not lose the program, according to the other party’s move, give the corresponding win. But! To keep your opponent from losing too badly, you need to draw every K times.

Input format: Input begins with a positive integer K (≤10), the number of draw intervals, on the first line. Then each line gives the opponent a move: ChuiZi for hammer, JianDao for scissors, Bu for paper. End indicates the End of input. This line should not be treated as a move.

Output format: for each input move, according to the requirements of the output win or draw move. Each move takes one line.

Example: 2 ChuiZi JianDao Bu JianDao Bu ChuiZi ChuiZi End Example: Bu ChuiZi Bu ChuiZi JianDao ChuiZi Bu

#include<math.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(a)
{
	int i;			// for loop
	int k;			// Indicates the number of intervals between draws
	int index;		// mark the number of draws
	char name[20];	// Enter the name of the offer
	scanf("%d",&k);
	index=k;		// mark the interval times
	while(1)
	{
		scanf("%s",name);
		if(strcmp(name,"End") = =0)
			break;
		else
		{
			if(index==0)
			{
				if(strcmp(name,"ChuiZi") = =0)
					printf("ChuiZi\n");
				if(strcmp(name,"JianDao") = =0)
					printf("JianDao\n");
				if(strcmp(name,"Bu") = =0)
					printf("Bu\n");
				index = k;
			}
			else
			{
				if(strcmp(name,"ChuiZi") = =0)
					printf("Bu\n");
				if(strcmp(name,"JianDao") = =0)
					printf("ChuiZi\n");
				if(strcmp(name,"Bu") = =0)
					printf("JianDao\n");
				index--;			// Each time you win, the interval is 1 less. When =0, draw}}}return 0;
}
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40, the universe is invincible hello (5 points)

#include<stdio.h>
int main(void)
{
	char s[8];
	gets(s);
	printf("Hello ");
	puts(s);
}
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41. Single (20 points)

L1-046 Single (20 points) The so-called “bachelor” here, I don’t mean single, I mean all the numbers made up of ones, such as 1, 11, 111, 1111, etc. Legend has it that any single person can be divisible by an odd number that does not end in 5. For example, 111111 is divisible by 13. Now, your program reads an integer x, which must be odd and does not end in 5. Then, after calculation, two numbers are printed: the first number, S, indicates that x times S is a single, and the second number, N, is the number of digits of the single. Of course, there are not unique solutions, and they want you to output the smallest solution.

Tip: One obvious solution is to gradually increase the number of singles until x is divisible. The difficulty is that s can be a very large number — for example, if the program inputs 31, it outputs 3584229390681 and 15, because 31 times 3584229390681 yields 111111111111111, which is a total of 15 ones.

Input format: Input gives a positive odd number x (<1000) on a line that does not end in 5.

Output format: Print the corresponding minimum s and n on a single line, separated by 1 space.

Example: 31 Example: 3584229390681 15

#include<stdio.h>  
int main (a)
 {  
    int n, len=0,p=0,now=1;  
    char ans[1000];  
    scanf("%d",&n);  
    while (!0)
{  
   len++;                 / / 1,11,111,111... Try to see if it's divisible
        if(p||now/n)          P ==0; p==0; p==0
        ans[p++]='0'+now/n;   // convert to character
        now%=n;               // The remainder is the key to analog division
        if(now==0) 
{                   // The remainder is 0
            ans[p]='\ 0'; // String terminator is added to a string created by yourself step by step
            printf("%s %d\n", ans, len);  
            break;  
        }  
        now=now*10+1; // Multiply by 10 is the key to analog division
    }  
    return 0;  
}
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42. Pretending to sleep (10 points)

#include<stdio.h>
int main(void)
{
	int N;	// Enter a positive integer
	scanf("%d",&N);
	char name[8] [4];	// Name // Be sure to set one more digit here
	int huxi[8];		// Respiration rate
	int maibo[8];		/ / the pulse
	int i;				// For loops and array subscripts
	for(i=0; i<N; i++)// Provide input function
		scanf("%s%d%d",&name[i][0],&huxi[i],&maibo[i]);
	for(i=0; i<N; i++)if(huxi[i]<15 || huxi[i]>20 || maibo[i]<50 || maibo[i]>70)
			puts(name[i]);
}
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43, MATRIX A times B (15 marks)

Given two matrices A and B, you are asked to compute their product matrix AB. Note that only matrices that match in size can be multiplied. That is, if A has R, A rows and C, A columns, and B has R, B rows and C, B columns, then the two matrices can only be multiplied if C, A and R, B are equal.

Input format: The input gives two matrices A and B. For each matrix, the number of rows R and the number of columns C are given in one row, followed by R rows, each row is given C integers, separated by one space, with no extra space at the beginning and end of the row. The input ensures that R and C of both matrices are positive and that the absolute values of all integers do not exceed 100.

Output format: If the sizes of the two input matrices match, the product matrix AB is output in the format of the input. Otherwise, Error: Ca! = Rb, where Ca is the number of columns of A and Rb is the number of rows of B.

Example 1:2 3 1 2 3 4 5 63 4 7 8 9 0-1-2 3-4 5 6 7 8 Example 1:24 20 22 24 16 53 58 63 28 Example 2: 3 2 38 26 43-5 0 17 3 2-11 57 99 68 81 72 Example 2: Error: 2! = 3

#include<stdio.h>
#define N 100
#define M 100
int main(a)
{
	int a[N][M]={0},b[N][M]={0},c[N][M]={0};
	int i,j,k;				// For loops and array subscripts
	int ra,ca,rb,cb,rc,cc;	Row and column of rectangle A, row and column of rectangle B, and row and column of rectangle C
	int n;					//
	scanf("%d %d",&ra,&ca);
	for(i=0; i<ra; i++)/ / rectangle A
		for(j=0; j<ca; j++)scanf("%d",&a[i][j]);
	scanf("%d %d",&rb,&cb);

	for(i=0; i<rb; i++)B / / rectangle
		for(j=0; j<cb; j++)scanf("%d",&b[i][j]);
	if(ca == rb)
	{	// Notice that since we are multiplying two rectangles, we can be sure that these two rectangles are different looking,
		// When the same condition is determined by us, the other two conditions will determine the row and column criteria of rectangle C, so
		// Here we should combine the rows of rectangle A and the columns of rectangle B as the shape of rectangle C to participate in the calculation

		rc = ra;	// The rows of rectangle A are the rows of rectangle C
		cc = cb;	// The columns of rectangle B are used as columns of rectangle C
		n = ca;		// Here is the factor that determines the k cycle
		printf("%d %d\n",rc,cc);
		for(i=0; i<rc; i++)// Control the number of rows
		{
			for(j=0; j<cc; j++)// Control column number
			{
				for(k=0; k<n; k++)// Multiply the elements in the first row of rectangle A by the elements in the first column of rectangle B
				{					// And so on
					c[i][j]+=a[i][k]*b[k][j];
				}
				if(j! =cc- 1)
					printf("%d ",c[i][j]);
				else
					printf("%d",c[i][j]);
			}
			printf("\n");			// After each column of rectangle C is printed, the line is wrapped}}else	// Output this message when the data matches exactly
		printf("Error: %d ! = %d\n",ca,rb);
	return 0;
}
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44, the NTH string (15 points)

#include<stdio.h>
#include<math.h>
#define L 6

int main(a)
{
	char ch[26] = {'z'.'y'.'x'.'w'.'v'.'u'.'t'.'s'.'r'.'q'.'p'.'o'.'n'.'m'
	             ,'l'.'k'.'j'.'i'.'h'.'g'.'f'.'e'.'d'.'c'.'b'.'a'};
	char arr[L];
	int n,l,a,i;         //l is the length of the sequence, n is the position of the sequence in reverse order, and a is the position of a character of the desired string in the ch array
	scanf("%d %d",&l,&n);
	n=n- 1;               // The last character is correct after subtracting 1
	for(i=0; i<l; i++) { a=n/pow(26,l-i- 1);
		arr[i]=ch[a];
		n=n-a*pow(26,l-i- 1);
	}
	for(i=0; i<l; i++) {printf("%c",arr[i]);
	}
	return 0;
 }
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45. Discount (5 points)

L1-051 Discount (5 cents) When you go to the mall to find a sale, calculating the price after the sale can be quite a mental exercise. For example, if the original price is ￥988 and marked 30% off, the discount price should be ￥988 x 70% = ￥691.60. Write a program to calculate the discount price for the customer.

Input format: Input the original price (positive integer not more than 10,000 yuan) and discount (integer within [1, 9]) of the product in one line, separated by Spaces.

Output format: Output the discount price of the product in one line, with 2 decimal places reserved.

Example Value: 988 7 Example value: 691.60

#include<stdio.h>
int main(void)
{
	double before_money;			/ / the original price
	double after_money;			/ / after the discount
	double zhe;					/ / discount
	scanf("%lf %lf",&before_money,&zhe);
	after_money=before_money*zhe/10;
	printf("%.2lf\n",after_money);
}
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46. 2018 We need to win (5 points)

#include<stdio.h>
int main(void)
{
	printf("2018\n");
	printf("wo3 men2 yao4 ying2 ! \n");
	return 0;
}
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47. Electron Wang (10 points)

Dogs are said to have the IQ of a four-year-old human child and are even more intelligent when it comes to adding numbers. For example, if you put two piles of balls on the ground, with one ball and two balls respectively, the clever wang will use “Wang! Wang! Woof!” That means 1 plus 2 is 3.

This requires you to create a simulation program for the tamagotchi, calculate the sum based on the number of two piles of balls recognized by the electronic eye, and give the answer using the bark of the dog.

Input format: Input gives two positive integers A and B in the [1, 9] range on A line, separated by Spaces.

Output format: output A + B Wang! .

Example Value: 2 1 Example value: Wang! Wang! Wan

#include<stdio.h>
int main(void)
{
	int A;	
	int B;
	scanf("%d %d",&A,&B);
	for(int i=1; i<=A+B; i++)printf("Wang!");
	return 0;
}
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48. Who is the winner (10 points)

#include<stdio.h>
int main(void)
{
	int Pa;			// A's number of votes
	int Pb;			// B's number of votes
	int num;		// Judges vote
	int a=0,b=0;	// Count the votes of judges A and B
	scanf("%d %d",&Pa,&Pb);
	for(int i=0; i<3; i++) {scanf("%d",&num);
			if(num==0)
				a++;
			else
				b++;
	}
	if(Pa>Pb&&a! =0)
		printf("The winner is a: %d + %d\n",Pa,a);
	else if(Pa<Pb&&b! =0)
		printf("The winner is b: %d + %d\n",Pb,b);
	else if(Pa<Pb&&a==3)
		printf("The winner is a: %d + %d\n",Pa,a);
	else if(Pa>Pb&&b==3)
		printf("The winner is b: %d + %d\n",Pb,b);
}
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49. De-weighting linked lists (25 points)

Given a linked list L with integer key values, you need to delete all key values that have the same absolute value. So for every key K, only the first node whose absolute value is K is preserved. Meanwhile, all deleted nodes must be saved on another linked list. For example, given L is 21→-15→-15→-7→15, you need to output the de-duplicated list 21→-15→-7, and the deleted list -15→15.

Input format: Input on the first line gives the address of the first node of L and a positive integer N (≤10 5, the total number of nodes). The address of a node is a non-negative 5-bit integer, and the NULL address is represented by −1.

Then N lines, each describing a node in the following format:

The address key is the next node where the address is the address of that node, the key is an integer with absolute value up to 10 or 4, and the next node is the address of the next node.

Output format: output the de-duplicated list first, and then output the deleted list. Each node is in a row and output in the format of input.

Example: 00100 5 99999-7 87654 23854-15 00000 87654 15-1 00000-15 99999 00100 21 23854 Example: 00100 21 23854 23854-15 99999 99999-7-1 00000-15 87654 87654 15-1

#include <stdio.h>
#include <string.h>
#include <math.h>
#define N 10*10*10*10*10+1
struct Node
{
    int jian,xia;		//jian > xia > next node
}node[N];
int main(a)
{
	int l,n;		//l→ first address, n→ several addresses, input several rounds
    int first;		// The starting address of each round
	int i,j;		// For loops and array subscripts
	int tmp[10001] = {0};
	// With the loop split list, check if the subscript appears, if so, assign the value to the deleted list
	int one[N],two[N];
	// One → represents the deleted list, and two → represents the deleted list
	int o=0,t=0;
	int num;		//num gets the key value of each address
	scanf("%d%d",&l,&n);
        for(i=0; i<n; i++) {scanf("%d",&first);
            scanf("%d%d",&node[first].jian,&node[first].xia);
        }
        for(i=l; node[i].xia ! =- 1;)
        {
            int num=abs(node[i].jian);
            if(tmp[num]==0)
            {
                tmp[num]=1;
                one[o++]=i;		// Storage address
            }
            else
            {
                two[t++]=i;
            }
            i=node[i].xia;
        }
		num=abs(node[i].jian);// Take note of the last subscript
        if(tmp[num]==0) { one[o++]=i; }else{ two[t++]=i; }printf("%05d %d ",one[0],node[one[0]].jian);
            for(j=1; j<o; j++) {printf("%05d\n",one[j]);
                printf("%05d %d ",one[j],node[one[j]].jian);
            }
            printf("%d\n".- 1);
            if(t! =0)		// Do not delete the linked list
            {
                printf("%05d %d ",two[0],node[two[0]].jian);
                for(j=1; j<t; j++) {printf("%05d\n",two[j]);
                    printf("%05d %d ",two[j],node[two[j]].jian);
                }
                printf("%d\n".- 1);
            }
			return 0;
}
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50. Hall of Fame and Vouchers (25 points)

Example Input: 10 80 5 [email protected] 78 [email protected] 87 [email protected] 65 [email protected] 96 [email protected] 39 [email protected] 87 [email protected] 80 [email protected] 88 [email protected] 80 [email protected] 70 Example output: 360 1 [email protected] 96 2 [email protected] 88 3 [email protected] 87 3 [email protected] 87 5 [email protected] 80 5 [email protected] 80

Hall of Fame and Vouchers (25 points)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct stu{
int id;/ / place
char edu[25];/ / account
int score;/ / result
};
// Sort algorithm?
int cmp(const void * a,const void * b)
{ 
	struct stu aa = *(struct stu *)a;
	struct stu bb = *(struct stu *)b;
	if(aa.score == bb.score)
	return strcmp(aa.edu,bb.edu); // When the score is the same, we compare the account.
	else
	return aa.score > bb.score ? - 1 : 1;
	}
	int main(a)
	{
	int N,G,K;
	struct stu stud[10005].
	while(scanf("%d%d%d",&N,&G,&K) ! = EOF){int i;
	int ans = 0;
	
	for(i = 0 ; i < N ; i++){
	scanf("%s %d",stud[i].edu,&stud[i].score);
	if(stud[i].score >= 60 && stud[i].score < G){
	ans += 20;
	}
	if(stud[i].score >= G && stud[i].score <= 100){
	ans += 50; }}printf("%d\n",ans);// Export vouchers?
	
	qsort(stud,N,sizeof(stud[0]),cmp);
	int num = 1;
	stud[0].id = 1;
	
	for(i = 1 ; i < N ; i++) // The ranking is handled after the order is sorted.
	{
	num++;
	if(stud[i].score == stud[i- 1].score){
	stud[i].id = stud[i- 1].id;
	}else{ stud[i].id = num; }}for(i = 0 ; i < K ; i++)
	printf("%d %s %d\n",stud[i].id,stud[i].edu,stud[i].score);
	int k = K;
	while(stud[K- 1].score == stud[k].score){// There are no students ranked less than or equal to K after the output of K students.
	printf("%d %s %d\n",stud[k].id,stud[k].edu,stud[k].score); k++; }}return 0;
}
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51. Make small change (30 cents)

Han Meimei likes to go shopping everywhere in the universe. Now she visits a Martian store and discovers that it has a special rule: you can pay with coins from any planet, but no change, and of course no debt. Han Meimei has 10 or 14 coins from all over the world at hand and needs you to help her calculate whether it is possible to come up with the exact amount to pay.

Input format: The first line of input gives two positive integers: N (≤10 4) is the total number of coins, and M (≤10 2) is the amount that Han Meimei should pay. The second line gives the positive integer face values of N coins. Numbers are separated by Spaces.

Output format: Output the face value of the coin in a line V 1 ≤V 2 ≤… V 1 +V 2 +… K = M + V. Digits are separated by one space. There must be no extra space at the beginning and end of each line. If the solution is not unique, the smallest sequence is output. If there is No Solution, output No Solution.

Note: we say that the sequence {A [1], A [2], and…} than {B [1], [2] B,…} “small”, refers to the existence k p 1 make [I] A [I] = B for all I < k was established, and A [k] < B [k].

Example 1:8 9 5 9 8 7 2 3 4 1 Example 1:1 3 5 Example 2:4 8 7 2 4 3 Example 2: No Solution

Collect small change (30 cents)

#include<stdio.h>  
#include<algorithm>  
#include<vector>  
using namespace std;  
int values[10001],dp[101],choose[10001] [101];  
bool cmp(int a,int b)  
{  
    return a>b;  
}  
int main(a)  
{  
    int i,j,n,m,k,t;  
    scanf("%d %d",&n,&m);  
    for(i=1; i<=n; i++) {scanf("%d",&values[i]);  
    }  
    sort(values+1,values+n+1,cmp);  
    for(i=1; i<=n; i++) {for(j=m; j>=values[i]; j--) {if(dp[j]<=dp[j-values[i]]+values[i])  
            {  
                choose[i][j]=1; dp[j]=dp[j-values[i]]+values[i]; }}}if(dp[m]! =m) {printf("No Solution\n");  
        return 0;  
    }  
    int index=n,sum=m;  
    vector<int> arr;  
    while(sum>0)  
    {  
        if(choose[index][sum]==1)  
        {  
  
           arr.push_back(values[index]);  
           sum-=values[index];  
        }  
        index--;  
    }  
    for(i=0; i<arr.size(); i++) {if(i==0)  
        {  
            printf("%d",arr[i]);  
        }  
        else  
        {  
            printf(" %d",arr[i]); }}printf("\n");  
    return 0;  
}
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52. (3n+1) Guess (15 points)

#include<stdio.h>
int main(a)
{
	int i,j;
	int n;
	int k=0;
	scanf("%d",&n);
	if(n==0) {printf("0");return 0;
	}
	for(i=0;; i++) {if(n%2= =0)
			n/=2;
		else
		{
			n= (3*n+1) /2;
		}
		k++;
		if(n==1)
		break;
	}
	printf("%d",k);
}
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53, Write this number (20 marks)

Write down the number (20 points). Read in a positive integer n and count the sum of its digits. Write down each digit of the sum in Pinyin.

Input format: Each test input contains 1 test case, which gives the value of the natural number N. I’m going to make sure that n is less than 10, 100.

Output format: output each digit of the sum of the digits of N in a line, there is 1 space between the pinyin digits, but there is no space after the last pinyin digits in a line.

The input sample: 1234567890987654321123456789 output sample: yi SAN wu

Write the number (20 points).

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main(a)
{
	int k=0;
	int sum=0,i=0,j;
	char n[1001];
	gets(n);
	int len = strlen(n);
	for(i=0; i<len; i++) { sum+=n[i]-'0';
	}
	//printf("%d\n",sum); // Do I get 135?
	int dayin[10];
	//printf("%d\n",lenght);
	i=0;
	while(sum! =0)
	{
		dayin[i++]=sum%10;
		sum/=10;
	}
	for(j=i- 1; j>=0; j--) {if(j! =0) {switch(dayin[j])
		{
		case 1:printf("yi ");break;
		case 2:printf("er ");break;
		case 3:printf("san ");break;
		case 4:printf("si ");break;
		case 5:printf("wu ");break;
		case 6:printf("liu ");break;
		case 7:printf("qi ");break;
		case 8:printf("ba ");break;
		case 9:printf("jiu ");break;
		case 0:printf("ling ");break; }}else
		{
		switch(dayin[j])
		{
		case 1:printf("yi");break;
		case 2:printf("er");break;
		case 3:printf("san");break;
		case 4:printf("si");break;
		case 5:printf("wu");break;
		case 6:printf("liu");break;
		case 7:printf("qi");break;
		case 8:printf("ba");break;
		case 9:printf("jiu");break;
		case 0:printf("ling");break; }}}return 0;
}
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I want to pass! (20 points)

#include<stdio.h>
int main(a)
{
    int n;          
    scanf("%d",&n);             
    while(n--){
        char str[110];
        int np=0,nt=0,other=0,lp,lt;
        scanf("%s",str);
        int len=strlen(str);
        for(int i=0; i<len; i++){// Count the number of P,T, and other letters and the positions of P and T
            if(str[i]=='P'){
                np++;
                lp=i;
            }else if(str[i]=='T'){
                nt++;
                lt=i;
            }else if(str[i]! ='A') 
                other++;
        }
        if((np! =1)||(nt! =1)||(other! =0)||(lt-lp<=1)) {// The number of P and T must be one, with no other letters, and at least one A between P and T
            printf("NO\n");
            continue;
        }
        int x=lp,y=lt-lp- 1,z=len-lt- 1;              
        if(x*y==z)
        printf("YES\n");
        else printf("NO\n");
    }
    return 0;

} 
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55, Change the format of output integer (15 marks)

Let’s use the letter B for “hundred”, S for “ten”, and 12… N to represent the non-zero one digit number n (<10), reformatted to output any positive integer with no more than three digits. For example, 234 should be printed as BBSSS1234 because it has two hundreds, three tens, and a four in the ones place.

Input format: Each test input contains 1 test case, given as a positive integer n (<1000).

Output format: The output of each test case is one line, and n is output in the specified format.

Example 1:234 Example 1: BBSSS1234 Example 2:23 Example 2: SS123

Output integer (15 marks)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main(a)
{
	int n;
	int i=0,j,num=1;
	int k[100] = {0},k1=0;
	scanf("%d",&n);
	while(n! =0)
	{
		k[k1++]=n%10;
		n/=10;
	}
	int ci=k1;
	for(i=k1- 1; i>=0; i--) {if(ci==3)
		{
			while(k[i]--){
				printf("B"); }}else if(ci==2)
		{
			while(k[i]--){
				printf("S"); }}else
		{
			while(k[i]--){
				printf("%d",num);
				num++;
			}
		}	
		ci--;	
	}
	return 0;
}
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56. Prime Pair Conjecture (20 points)

#include<stdio.h>
#include<math.h>
int main(a)
{
	int i,j;
	int r=0;
	int n;
	int a[2900],k=0;
	scanf("%d",&n);
	for(i=5; i<=n; i+=2)
	{
		for(j=3; j*j<=i; j+=2) {if(i%j==0)
			break;
		}
		if(j==i)
		a[k++]=i;	// Save all primes from 2 to n
	}
	for(i=0; i<k; i++) {if(a[i+1]-a[i]==2)
		r++;	
	}
	printf("%d\n",r);
	
	
	
}
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57. Problem with loop right shift of array elements (20 points)

If there are N (>0) integers in array A, move each integer to the right by M (≥0) without allowing the use of another array. That is, the data in A is transformed from (A 0 A 1… A N−1) to (A N−M… A N−1 A 0 A 1… A N−M−1) (the last M numbers are moved to the first M positions in A cycle). If you need to consider moving data as little as possible, how do you design the way to move it?

Input format: Each input contains one test case. Input N (1≤N≤100) and M (≥0) in line 1. Line 2 enters N integers separated by Spaces.

Output format: Output the sequence of integers that have been moved to the right by M bits in a row, separated by Spaces. There must be no extra Spaces at the end of the sequence.

Example Value: 6 2 1 2 3 4 5 6 Example value: 5 6 1 2 3 4

1008 Loop right shift of array elements (20 marks)

#include<stdio.h>
int main(a)
{
	int kr[100],k=0;
	int n,m;
	int i,j;
	scanf("%d%d",&n,&m);
	m%=n;		// if n
	for(i=0; i<n; i++)// Print the beginning of the array
	scanf("%d",&kr[i]);
	for(i=n-m; i<n; i++)printf("%d ",kr[i]);
	for(i=0; i<n-m; i++)//
	{
		if(i! =n-m- 1)
		printf("%d ",kr[i]);
		else
		printf("%d",kr[i]);
	}
	
	/* Second output: very cool and classic */
	/* for (int i = 0; i < N - M - 1; i++) printf("%d ", ary[i]); printf("%d", ary[N - M - 1]); * /
	
	
}
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58. Being sarcastic (20 points)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main(a)
{
	int i=0,j;
	char c;
	char str[100] [100];
	int dan=80;	// Each group of words is entered in positive order, but the subscript is controlled here, so the words are automatically sorted from the end forward
	while((c=getchar())! ='\n')
	{
		if(c==' ')
		{
			str[dan][i]='\ 0';
			dan--;
			i=0;
			continue;
		}
		else
		{
			str[dan][i]=c;
			i++;
		}
	}
	str[dan][i]='\ 0';	// The line breaks out of the loop, so remember to add the terminator to the last word
	for(i=dan; i<80; i++)// The input is in reverse order
	printf("%s ",str[i]);
	printf("%s",str[i]);	// The last word entered has no Spaces
	
}
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59, The Derivative of a Unary Polynomial (25 marks)

Derivation of unary Polynomials (25 marks)

Design functions to find derivatives of unary polynomials. (Note: the first derivative of x n (n is an integer) is nx n−1.)

Input format: Input polynomial non-zero coefficients and exponents (absolute values are integers not more than 1000) in exponential descending mode. Numbers are separated by Spaces.

Output format: Output the coefficients and exponents of the non-zero terms of the derivative polynomial in the same format as the input. Digits are separated by Spaces, but there can be no extra space at the end. Note that the “zero polynomial” has both an exponent and a coefficient of 0, but is represented as 0, 0.

Example Value: 3 4-5 2 6 1-2 0 Example value: 12 3-10 1 6 0

Derivation of 1010 Unary Polynomials (25 marks)

#include<stdio.h>
int main(a)
{
	int k=0;
	int xi,zhi;
	while(scanf("%d %d",&xi,&zhi)! =EOF) {if(zhi)	// There is no case where the exponent is 0, so no output is required
		{
			if(k! =0)
			putchar(' ');
			k++;
			printf("%d %d",xi*zhi,zhi- 1);	// The coefficient is multiplied by the exponent, exponent -1}}if(k==0)	// If the derivative is 0, output "0 0"
	printf("0 0\n");
	return 0;
 } 
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60, A+B and C (15 分)

#include<stdio.h>
int main(a)
{
  // The input data is very large, so we should define it as long to ensure that the data is not lost
	int i;
	int k=1;
	long sum=0;
	int n;
	long a,b,c;
	scanf("%d",&n);
	for(i=0; i<n; i++) {scanf("%ld%ld%ld",&a,&b,&c);
		sum=a+b;
			if(sum>c)
			printf("Case #%d: true\n",k);
			else
			printf("Case #%d: false\n",k); k++; }}Copy the code

61, Prime numbers (20 points)

Let PI be the ith prime number. Given two positive integers M≤N≤10 4, please print all primes from P M to P N.

Input format: Input gives M and N on a single line, separated by Spaces.

Output format: All primes from P M to P N are displayed in one line. Each 10 digits are separated by Spaces. There must be no extra space at the end of the line.

Example Value: 5 27 Example value: 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103

#include<stdio.h>
#include<math.h>
int scnprim(int n);
int main(a)
{	
	int i,m,n,count=0;
	scanf("%d %d",&m,&n);
 
	for(i=2;; i++) {if(scnprim(i))
		{
			count++;
			if(m<=count&&n>=count)
			{	if((count-m+1) %10= =0)
				printf("%d\n",i);
				else if(count! =n)printf("%d ",i);
				else
				printf("%d",i); }}if(n<count)
		{
			break; }}return 0;
}
int scnprim(int n)// Filter prime numbers
{
	int sqrtc,j;
	sqrtc=(int)sqrt(n);
	for(j=2; j<=sqrtc; j++) {if(n%j==0)
			return 0;
 
	}
	return n;
}
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62, Theory of Moral ability (25 points)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct Student
{
	int name;	
	int de;
	int cai;
	int sum;
}love1[100100],love2[100100],love3[100100],love4[100100],love5[100100];	// Love1 is the total student
int comp(const void* a,const void* b)
{
	struct Student *aa = (struct Student *)a;
	struct Student *bb = (struct Student *)b;
	if(aa->sum ! = bb->sum)return ((bb->sum) - (aa->sum));
	else if(aa->de ! = bb->de)return ((bb->de) - (aa->de));
	else
	return ((aa->name) - (bb->name));
}
int main(a)
{
	int one=0,two=0,three=0,four=0,five=0;
	int i,j;
	int N,L,H;
	scanf("%d%d%d",&N,&L,&H);
	for(i=0; i<N; i++) {scanf("%d%d%d",&love1[i].name,&love1[i].de,&love1[i].cai);
		love1[i].sum = love1[i].de+love1[i].cai;
	}
	for(i=0; i<N; i++) {if(love1[i].de>=L && love1[i].cai>=L)
		{
			one++;
			if(love1[i].de>=H && love1[i].cai>=H)
			{	love2[two] = love1[i];		two++;		}
			else if(love1[i].cai<H && love1[i].de>=H)
			{	love3[three] = love1[i];	three++;	}
			else if(love1[i].de<H && love1[i].cai<H && love1[i].de>=love1[i].cai)
			{	love4[four] = love1[i];		four++;		}
			else{ love5[five] = love1[i]; five++; }}}printf("%d\n",one);
	qsort(love2,two,sizeof(love2[0]),comp);
	qsort(love3,three,sizeof(love3[0]),comp);
	qsort(love4,four,sizeof(love4[0]),comp);
	qsort(love5,five,sizeof(love5[0]),comp);
	
	for(i=0; i<two; i++)printf("%d %d %d\n",love2[i].name,love2[i].de,love2[i].cai);
	for(i=0; i<three; i++)printf("%d %d %d\n",love3[i].name,love3[i].de,love3[i].cai);
	for(i=0; i<four; i++)printf("%d %d %d\n",love4[i].name,love4[i].de,love4[i].cai);
	for(i=0; i<five; i++)printf("%d %d %d\n",love5[i].name,love5[i].de,love5[i].cai);
}
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A divided by B (20 marks)

In this case, calculate A/B, where A is A positive integer with no more than 1000 digits and B is A positive integer with 1 digit. You need to output the quotient Q and remainder R so that A=B×Q+R is true.

Input format: Enter A and B in A row, separated by 1 space.

Output format: Output Q and R in one line, separated by 1 space.

Example Value: 123456789050987654321 7 Example value: 17636684150141093474 3

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main(a)
{
	int i,j;
	int R,B,Q;
	// remainder, divisor, quotient
	char s[1000010];	/ / dividend
	scanf("%s%d",&s,&B);
	int len = strlen(s);
	Q = (s[0] -'0')/B;
	R = (s[0] -'0')%B;
	if(Q! =0||len==1)      // The divisor is larger than the quotient
		printf("%d",Q); 
	for(i=1; i<len; i++) { Q = (R*10 + s[i]-'0') / B;
		R = (R*10 + s[i]-'0') % B;
		printf("%d",Q); 
	}
	printf(" %d\n",R);
}
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64. Hammer, Paper, Scissors (20 points)

#include<stdio.h>
int main(a)
{
	int n;
	int i;
	int a=0,b=0,c=0,d=0,e=0,f=0;
	int one=0,two=0,three=0,four=0,five=0,six=0;
	char js,ys;		// Represent a's best move, represent B's best move
	int ping=0;
	char j,y;	// Represent a, B's move
	scanf("%d",&n);
	getchar();
	for(i=0; i<n; i++) {scanf("%c %c",&j,&y);
			 if(j=='C'&&y=='J')
		{ a++;		one++;		}
		else if(j=='J'&&y=='B')
		{	b++;	two++;		}
		else if(j=='B'&&y=='C')
		{	c++;	three++;	}
		else if(j=='B'&&y=='B')
		{		ping++;			}
		else if(j=='J'&&y=='J')
		{		ping++;			}
		else if(j=='C'&&y=='C')
		{		ping++;			}
			 if(y=='C'&&j=='J')
		{ 	d++;	four++;		}
		else if(y=='J'&&j=='B')
		{	e++;	five++;		}
		else if(y=='B'&&j=='C')
		{	f++;	six++;		}
		getchar();
	}
	
	// First judge a's most successful move
	if(one>two&&one>three)
		js='C';
	else if(two>one&&two>three)
		js='J';
	else if(three>one&&three>two)
		js='B';
	else if(two==one&&two>three)
		js='C';
	else if(two==three&&three>one)
		js='B';
	else if(one==three&&two>one)
		js='J';
	else
		js='B';
		
	// Determine b's most successful move
	if(four>five&&four>six)
		ys='C';
	else if(five>four&&five>six)
		ys='J';
	else if(six>four&&six>five)
		ys='B';
	else if(five==four&&five>three)
		ys='C';
	else if(five==six&&six>four)
		ys='B';
	else if(four==six&&five>four)
		ys='J';
	else
		ys='B';
		
		
	printf("%d %d %d\n",a+b+c,ping,(n-(a+b+c+ping)));
	printf("%d %d %d\n",d+e+f,ping,(n-(d+e+f+ping)));
	printf("%c %c\n",js,ys);
	
	
	return 0;
	
	
}
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65. Moon cake (25 points)

1020 Moon Cake (25 points)

Mooncakes are a traditional food eaten by Chinese people during the Mid-Autumn Festival. There are many mooncakes with different flavors in different regions. Now given all kinds of mooncake inventory, total price, and the maximum market demand, please calculate the maximum profit you can get.

Note: part of stock is allowed to be withdrawn when selling. An example is given as follows: suppose we have three kinds of moon cakes with inventories of 1.8, 150,000 and 100,000 tons respectively, and total selling prices of 7.5, 7.2 and 4.5 billion yuan respectively. If the market’s maximum demand is only 200,000 tons, then our maximum profit strategy should be to sell all 150,000 tons of type 2 mooncakes and 50,000 tons of type 3 mooncakes, and get 72 + 45/2 = 9.45 billion yuan.

Input format: Each input contains one test case. Each test case starts with a positive integer N not more than 1000 to represent the number of mooncakes and a positive integer D not more than 500 (in tons) to represent the maximum market demand. The following line gives N positive numbers to represent the inventory of each mooncake (in tons); The last line gives N positive numbers for the total price of each mooncake (in hundreds of millions of yuan). Numbers are separated by Spaces.

Output format: For each group of test cases, output the maximum revenue in one line, in hundreds of millions of yuan and accurate to two decimal places.

Example Value: 3 20 18 15 10 75 72 45 Example value: 94.50

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct love{
	double ax;		// Get the unit price of one ton of each mooncake
	double k;		// Get the inventory of each mooncake
	double m;		// The total price of each mooncake
}stu[1000];
int comp(const void* a,const void* b)
{
	struct love *aa = (struct love *)a;
	struct love *bb = (struct love *)b;
	return ((bb->ax)-(aa->ax));		// Members are sorted by unit price
}
int main(a)
{
	int i,j;
	double money=0;
	double n;
	double d;		// Maximum demand
	scanf("%lf",&n);		// There are several kinds of moon cakes
	scanf("%lf",&d);		// Maximum demand
	// Enter the inventory of mooncakes
	for(i=0; i<n; i++) {scanf("%lf",&stu[i].k);
	}
	// Enter the total price of mooncakes
	for(i=0; i<n; i++)scanf("%lf",&stu[i].m);	
	for(i=0; i<n; i++) stu[i].ax = stu[i].m / stu[i].k;// Get the unit price of each mooncake
	qsort(stu,n,sizeof(stu[0]),comp);
	/*printf("%lf %d %d\n",stu[0].ax,stu[0].k,stu[0].m); / / 0 subscript for maximum printf (" % lf % d % d \ n ", stu [1]. The ax, stu [1]. K, stu. [1] m); / / 0 subscript for maximum printf (" % lf % d % d \ n ", stu [2]. The ax, stu [2]. K, stu. [2] m); //0 subscript is Max */
	for(i=0; i<n; i++) {if(d<=stu[i].k)// If the maximum demand is not greater than the current inventory
		{
			money+=stu[i].ax*d;// Maximum revenue is the current unit price multiplied by maximum demand
			break;// Make sure to exit in time
		}
		else
		{
			money+=stu[i].m;// Otherwise, the maximum revenue is added to the current total price
			d-=stu[i].k;// Quantity demanded minus current inventory}}printf("%.2lf\n",money);
}
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66. Single digit statistics (15 points)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main(a)
{
	int i,j;
	int x[10] = {0};
	int xing;
	char s[1001];
	gets(s);
	int k = strlen(s);	// Get the length
	for(i=0; i<k; i++) { xing=s[i]- 48;	// Convert a number in a string to an integer number
		x[xing]++;		// use it as an array subscript, corresponding to +1
	}
	for(i=0; i<10; i++)if(x[i]! =0)
		printf("%d:%d\n",i,x[i]);
	return 0;
}
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67、 D进制的A+B (20 分)

Input two non-negative 10 base integers A and B (≤2 30 −1), and output the D (1

Input format: Input gives three integers A, B, and D in A row.

Output format: Output the D base number of A+B.

Example Value: 123 456 8 Example value: 1103

#include<stdio.h>
int main(a)
{
	int i,j;
	long  a,b,d;
	long  sum=0;
	int jin[31],k=0;
	scanf("%ld %ld %ld",&a,&b,&d);
	sum = a+b;
	if(sum==0)    // If the sum is equal to 0, output 0 directly, otherwise enter the loop into the pit
	{
	  printf("0");  
	  return 0;
	}
	while(sum! =0)			// Decimal to octal is divided by 8, the remainder of each octal digit is the remainder of each octal digit. When sum=0, the loop is broken
	{
		jin[k++]=sum%d;
		sum/=d;
	}
	for(i=k- 1; i>=0; i--)// When an array outputs octal numbers, pay attention to the reverse output
		printf("%d",jin[i]);
	return 0;
}
		
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68, The minimum number of groups (20 points)

#include<stdio.h>
int main(a)
{
	int i,j;
	int n;
	int flag = 0;
	int a[10] = {0};
	for(i=0; i<10; i++) {scanf("%d",&a[i]);      // If there is a number at the beginning of 1, print it first. Then print it again
		if(i>=1&& a[i]! =0 && flag>=0)
		{
			printf("%d",i);
			a[i]-=1;
			flag=- 1;
			//	printf("\n-1\n");}}//printf("%d\t%d\t%d\n",a[0],a[1],a[2]);
	for(i=0; i<10; i++) {if(a[i]! =0)
		{
			while(a[i]--)
			printf("%d",i); }}return 0;
}
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69, Check id card (15 points)

A valid ID card number consists of a 17-digit area, date and sequence number plus a one-digit verification code. The calculation rules of the verification code are as follows:

First, the first 17 digits are weighted and summed. The weight assignment is: {7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2}; Then modulo the calculated sum with respect to 11 to get the value Z; Finally, the Z value and the check code M are corresponding according to the following relation:

Z: 0 1 2 3 4 5 6 7 8 9 10 M: 10 X 9 8 7 6 5 4 3 2 Now given some ID numbers, please verify the validity of the verification code and output the number in question.

Input format: The first line indicates that N (≤100) is the number of id numbers to be entered. Then N lines, each line gives an 18-digit ID number.

Output format: Output one problematic ID number in each line according to the input sequence. It does not check whether the first 17 bits are reasonable, but only whether the first 17 bits are all digits and the last 1 bit is calculated correctly. If All numbers are normal, All passed is displayed.

Example 1:4 320124198808240056 12010X198901011234 110108196711301866 37070419881216001X Example 1: 12010X198901011234 110108196711301866 37070419881216001X Example 2:2 320124198808240056 110108196711301862 Example 2: All passed

#include <stdio.h>  
#include <stdlib.h>  
   
int charToInt(char c)
{  
     return (int)(c-'0');  
 }  
 int main(int argc, char *argv[]) 
 {  
     int n,i,j,k=0,s,  
     a[]={7.9.10.5.8.4.2.1.6.3.7.9.10.5.8.4.2};  
     char cur,b[11] = {'1'.'0'.'X'.'9'.'8'.'7'.'6'.'5'.'4'.'3'.'2'},  
     c[20];  
     scanf("%d",&n);  
     for(i = 0; i < n; i++) { s=0;  
         scanf("%s",&c);  
         for(j= 0; j<17; j++) {if(c[j]>='0'&&c[j]<='9')
			 {  
                 s+=charToInt(c[j])*a[j];  
			 }
			 else
			 {  
                 printf("%s",c);  
                 if(i<n- 1)
				 {  
                     printf("\n");  
                 }  
                 s=- 1;  
                 break; }}if(b[s%11]==c[j]&&s! =- 1)
		 {  
             k++;  
         }
		 else if(b[s%11]! =c[j]&&s! =- 1)
		 {  
             printf("%s",c);  
             if(i<n- 1)
			 {  
                 printf("\n"); }}}if(k==n)
	 {  
         printf("All passed");  
     }  
     return 0;  
}  
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70. Code with Obama (15 points)

#include<stdio.h>
int main(a)
{
	int i,j,k;
	int n;
	char c;
	int flag=1;
	scanf("%d %c",&n,&c);
	for(i=0; i<n; i++) {printf("%c",c);
	}
	printf("\n");
	for(i=0; i<n4 -; i++) {if(flag==- 1)
		{
			//printf("\n");
			flag=1;
		}
		else
		{
			for(j=0; j<n; j++) {if(j==0||j==n- 1)
				printf("%c",c);
				else
				printf("");
			}
			printf("\n");
			flag=- 1; }}for(i=0; i<n; i++) {printf("%c",c); }}Copy the code

71. Stroke (15 points)

Scratch boxing is an interesting part of ancient Chinese wine culture. The method that two people draw fist on wine table is: each mouth shout out a number, at the same time with hand draw out a number. If the sum of the numbers is exactly equal to the sum of the numbers shouted out by the two people, the winner, the loser fined a drink. If two players win or two players lose, the next round continues until there is a single winner.

Here is the record of a and B, please count how many glasses of wine they drank at last.

Input format: Input a positive integer N (≤100) in the first line, followed by N lines, each line gives a record of a round of strokes, the format is:

A shout a row b shout B row where shout refers to the number shouted, row refers to the number drawn, both are positive integers not more than 100 (both hands row together).

Output format: Output the number of cups a and B drink successively in a line, separated by a space.

Example Input: 5 8 10 9 12 5 10 5 10 3 8 5 12 12 18 1 13 4 16 12 15 Example output: 12

1046 scratch (15 points)

#include<stdio.h>
int main(a)
{
	int i,j;
	int sum=0;
	int a,b,c,d;
	int jia=0,yi=0;
	int n;
	scanf("%d",&n);
	for(i=0; i<n; i++) {scanf("%d%d%d%d",&a,&b,&c,&d);
		sum=a+c;
		if(sum==b&&sum==d)
			continue;
		if(b==sum)
		yi++;
		else if(d==sum)
		jia++;
		sum=0;
	}
	printf("%d %d\n",jia,yi);
	
	
	
}
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72, The sum of the number of combinations (15 marks)

Given N non-zero ones digits, any two of them can be combined to form a two-digit number. Find the sum of all possible 2-digit numbers. For example, given 2, 5, 8, the combination can be 25, 28, 52, 58, 82, 85, their sum is 330.

Input format: Input is given N (1 < N < 10) in one line, followed by N different non-zero digits. Numbers are separated by Spaces.

Output format: Outputs the sum of all possible 2-digit numbers.

Example Value: 3 2 8 5 Example value: 330

Sum of the Number of combinations (15 points)

#include<stdio.h>
int main(a)
{
	int i,j;
	int n;
	int sum=0;
	int a[11];
	scanf("%d",&n);
	for(i=0; i<n; i++) {scanf("%d",&a[i]);
	}
	for(i=0; i<n; i++) {for(j=0; j<n; j++)if(i! =j) sum+=a[i]*10+a[j];
	}
	printf("%d\n",sum);
	
	
	
}
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73, True or false (15 points)

You should write a simple program to help the teacher judge the questions and count the scores of the students to judge the questions.

Input format: Input in the first line gives two positive integers N and M, not more than 100, respectively the number of students and the number of test questions. The second line gives M positive integers up to 5, the full score for each question. The third line gives the correct answer to each question, with 0 for “no” and 1 for “yes”. And then N rows, each row gives a student’s answer. Numbers are separated by Spaces.

Output format: Output each student’s score in the order of input, each score in a line.

Example: 3 6 2 13 3 4 5 0 0 1 0 1 0 11 0 0 11 0 1 0 11 0 0 11 0 0 11 0 0 11 example: 13 11 12

1061 True Or False questions (15 points)

#include<stdio.h>
int main(a)
{
	int i,j;
	int n,m;
	int a[100];
	int c[100];
	int b[100];
	scanf("%d%d",&n,&m);
	for(i=0; i<m; i++) {scanf("%d",&a[i]);		// Enter the score for each question
	}
	for(i=0; i<m; i++)scanf("%d",&c[i]);
	int x,sum=0;
	for(j=0; j<n; j++) {for(i=0; i<m; i++) {scanf("%d",&x);
			if(c[i]==x) { sum+=a[i]; }}printf("%d\n",sum);
		sum=0; }}Copy the code

74. Image Filtering (15 points)

1066 Image Filtering (15 points) Image filtering is to color the non-important pixels in the image into the background color, so that the important parts are highlighted. Given a black and white image, you are asked to replace all pixel colors within a specified range of gray values with a specified color.

Input format: input gives the resolution of an image in the first line, that is, two positive integers M and N (0

Output format: output filtered images as required. That is, output M lines, each line N pixel gray value, each gray value occupies 3 bits (for example, black to be displayed as 000), separated by a space. There must be no extra space at the beginning or end of a line.

Example: 3 5 100 150 0 3 189 254 101 119 150 233 151 99 100 88 123 149 0 255 Example: 003 189 254 000 000 000 233 151 099 000 088 000 000 000 255

1066 Image Filtering (15 points)

#include<stdio.h>
int main(a)
{
	int i,j;
	int m,n,a,b,c;
	int temp;
	int te[600] [600];
	scanf("%d%d%d%d%d",&m,&n,&a,&b,&c);
	if(a>b)
	{
		temp=a;
		a=b;
		b=temp;
	}
	//printf("%d %d\n",a,b);
	for(i=0; i<m; i++) {for(j=0; j<n; j++) {scanf("%d",&te[i][j]);
			if(te[i][j]>=a&&te[i][j]<=b) { te[i][j]=c; }}}for(i=0; i<m; i++) {for(j=0; j<n- 1; j++) {printf("%03d ",te[i][j]);
		}
		printf("%03d\n",te[i][j]);
	}
	//printf("sda");
	
	
	
}
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75. Wifi Password (15 points)

Here is A photo circulating on Weibo: “Dear students, in view of the need to use Wifi sometimes and the fear of delaying your parents’ study, now set the Wifi password to the following math problem answer: A-1; B – 2; C – 3; D – 4; Please write your own answers. change every other day. Thanks for your cooperation!! ~ – Teachers are trying their best to promote students learning. This requires you to write a program to translate the answers to a series of questions into wifi passwords according to the corresponding relationship given on the paper. Here we simply assume that each multiple choice question has four choices, with one and only one correct answer.

Input format: The first line gives a positive integer N (≤ 100), followed by N lines. Each line gives four choices of a question according to the format of number – answer. T indicates correct choice, F indicates wrong choice. Options are separated by Spaces.

Output format: Output wifi password in one line.

Example Input: 8 A-T B-F C-F D-F C-T B-F A-F D-F A-F D-F C-F B-T B-T A-F C-F D-F B-F D-T A-F C-F A-T C-F B-F D-F D-T B-F C-F A-F C-T A-f B-F D-F Example output: 13224143

#include<stdio.h> 
#include<string.h> 
int main(a) { 
		int N; 
		scanf("%d", &N); / / 8
		getchar();
		for (int i = 0; i < N; i++)// One problem at a time
		{ 
			char str[16]; 
			gets(str); // A whole line a-t b-f c-f d-f
			for (int j = 2; j < strlen(str); j += 4) 
				if (str[j] == 'T') 
					printf("%d", str[j - 2] - 'A' + 1); 
	} 
		return 0; 
}
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76. Checking passwords (15 points)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main(a)
{
	int i,j;
	int n,k;
	char s[100];
	int flag=1;
	int shu=0,zi=0,luan=0,duan=0;
	scanf("%d",&n);
	getchar();
	for(i=0; i<n; i++) { gets(s);int len = strlen(s);
		if(len<6)
		{
			printf("Your password is tai duan le.\n");
			continue;
		} 
		for(j=0; j<len; j++) {if((((s[j]>='a'&&s[j]<='z')||(s[j]>='A'&&s[j]<='Z'))||(s[j]>='0'&&s[j]<='9')) = =0&&s[j]! ='. ')
			{
				flag=1;
				break;
			}
			if(flag==2 -&&(s[j]>='a'&&s[j]<='z')||(s[j]>='A'&&s[j]<='Z'))
			{
				flag=4 -;
			}
			else if(flag==- 3&&s[j]>='0'&&s[j]<='9')
			{
				flag=4 -;
			}
			else if(flag! =4 -)
			{
				if(s[j]>='0'&&s[j]<='9')
				{
					flag=2 -;
				}
				else if((s[j]>='a'&&s[j]<='z')||(s[j]>='A'&&s[j]<='Z'))
				{
					flag=- 3; }}}//printf("flag = %d\n",flag);
		if(flag==1)
			printf("Your password is tai luan le.\n");
		else if(flag==4 -)
			printf("Your password is wan mei.\n");
		else if(flag==2 -)
			printf("Your password needs zi mu.\n");
		else if(flag==- 3)
			printf("Your password needs shu zi.\n");
		/*else if(flag==-1) printf("Your password is tai luan le.\n"); * /
		flag=1; }}Copy the code

77. I won’t tell you (15 points)

When you are doing your homework, the child next to you asks you: “What’s five times seven?” You should smile politely and tell him, “Fifty-three.” So they’re asking you, for any given pair of positive integers, to print their product backwards.

Input format: The first line gives two positive integers, A and B, not more than 1000, separated by Spaces.

Output format: Output the product of A and B backwards in one line.

Example Value: 5 7 Example value: 53

1086 won’t tell you (15 points)

#include <stdio.h>
int main(a)
{
	int n1=0,n2=0,mul,count=0,i=0,flag=0; 
	int a[10];
	scanf("%d%d",&n1,&n2);
	mul=n1*n2;
	while(mul>0){
		a[count++]=mul%10;
		mul=mul/10;
	}
	for(i=0; i<count; i++){if(a[i]! =0){
			flag=1;
		}
		if(flag==1) {printf("%d",a[i]); }}return 0;
}
 
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78, How many different values are there (20 points)

When the natural numbers n take 1, 2, 3… How many different values of ⌊ N /2⌋+⌊ N /3⌋+ ⌋ N /5 (Note: ⌊x⌋ is an integer function, representing the largest natural number not exceeding x, i.e., the integer part of x.)

Input format: Input gives a positive integer N (2≤N≤10 4).

Output format: Output the number of different values taken by the formula in the problem in a single line.

Example Value: 2017 Example value: 1480

How many different values are there for 1087 (20 points)

#include<stdio.h>
int main(a)
{
	int i,j;
	int n;
	int num;
	int a[10010],k=0;
	scanf("%d",&n);
	for(i=1; i<=n; i++) { num = i/2+i/3+i/5;
		//printf("num = %d\n",num);
		if(i==1)
			a[k++]=num;
		for(j=0; j<k; j++) {if(num==a[j])
				break;
		}
		if(j==k)
			a[k++]=num;
	}
	printf("%d\n",k);
}
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79, Dangerous goods packing (25 points)

When shipping goods in containers, we must take special care not to pack incompatible goods in one container. For example, oxidizer must not be packed with flammable liquid, otherwise it is easy to cause explosion.

Given a list of incompatible items, you are asked to check the list of items in each container to see if they fit in the same container.

Input format: The first line of input gives two positive integers: N (≤10 4) is the logarithm of a pair of incompatible items; M (≤100) is the singular of the container list.

The data is then presented in two chunks. The first block has N rows, and each row gives a pair of incompatible items. The second block has M lines, and each line gives a list of one box of goods in the following format:

K G [1] [2] G… G[K] K (≤1000) is the number of items, and G[I] is the number of items. For simplicity, each item is represented by a five-digit number. Separate the two numbers with a space.

Output format: Judge whether the cargo list of each container can be safely transported. Print Yes on a line if there are No incompatible items, No otherwise.

Example Input: 6 3 20001 20002 20003 20004 20005 20006 20003 20001 20005 20004 20004 20006 4 00001 20004 00002 20003 5 98823 20002 20003 20006 10010 3 12345 67890 23333 Example: No Yes Yes

1090 Packing of dangerous goods (25 points)

#include<stdio.h>
struct wu{
 	int wu1;
	int wu2;
}W[20010];
	int main(a){
	 int N,M;
	 scanf("%d%d",&N,&M);
	 int c[100010] = {0};
	 int e;
	 for(int i=0; i<N; i++){int a,b;
	  scanf("%d%d",&a,&b);
	  c[a]++;
	  c[b]++;
	  W[i].wu1=a;
	  W[i].wu2=b;
	 }
	 for(int j=0; j<M; j++){int K;
	  scanf("%d",&K);
	  int flag=0;
	  int b[100010] = {0}; 
	  for(int i=0; i<K; i++){int d;
	   scanf("%d",&d);
	   if(flag==1) {continue;
	   }
	   else if(b[d]==1){
	    flag=1;
	    continue;
	   }
	   else if(c[d]! =0){
	    e=c[d];
	    for(int i=0; i<N; i++){if(d==W[i].wu1){
	      b[W[i].wu2]=1;
	      c[d]--;
	     }
	     else if(d==W[i].wu2){
	      b[W[i].wu1]=1;
	      c[d]--;
	     }
	     if(c[d]==0) {break; } } c[d]=e; }}if(flag==1) {printf("No\n");
	  }
	  else{
	   printf("Yes\n"); }}return 0;
	}
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80, N- Defensive number (15 points)

If the last digits of a number K multiplied by N equals K, the number is called “n-defensive”. For example, 3×92 2 =25392, and the last two digits of 25392 are exactly 92, so 92 is a 3-defensive number.

Write a program to determine whether a given number is n-defensive about some N.

Input format: In the first line, the positive integer M (≤20) is given. In the following line, the positive integer M (≤ 1000) to be detected is given.

Output format: Output the smallest N and NK 2 values in a line for each number to be detected if it is n-defensive, separated by a space; Otherwise, output No. Notice that they make sure that N is less than 10.

Example: 392 5 233 Example: 3 25392 1 25 No

N- Defensive number (15 points)

#include<stdio.h>
#include<math.h>
int funtion(int k)
{
	int mix=1;
	while(k)
	{
		mix*=10;
		k/=10;
	}
	return mix;
}
int main(a)
{
	int t;
	int n;
	int i,j;
	int x;
	int avg;
	int ge,shi,bai;
	int flag;
	int sum=0;
	scanf("%d",&n);
	for(i=0; i<n; i++) {scanf("%d",&x);
		t = funtion(x);
		sum = pow(x,2);
		flag=0;
		for(j=1; j<10; j++) { avg = sum*j;/*ge = avg%10; shi = avg/10%10; bai = avg/100%10; * /
			if((avg-x)%t==0)
			{
				printf("%d %d\n",j,avg);
				flag=1;
				break;
			}
			/*else if(x==ge+shi*10) { printf("%d %d\n",j,avg); flag=-1; break; } else if(x==ge+shi*10+bai*100) { printf("%d %d\n",j,avg); flag=-1; break; } * /
		}
		if(flag==0)
			printf("No\n");
		//flag=1;	}}Copy the code

81, Programming in C is fun! (5 points)

Programming in C is fun! .

Input format: No input for this topic.

Output format: Output the phrase “Programming in C is fun!” in a single line .

#include<stdio.h>
int main(void)
{
	printf("Programming in C is fun! \n");
}
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82. Taxi Fare (15 points)

#include<stdio.h>
int main(a)
{
	double km;			/ / km
	int m=0;			/ / minute
	double money=0;		/ / money
	int yu;
	scanf("%lf %d",&km,&m);
	if(m<5)				// <5 minutes
	{
		if(km<=3)		/ / 0 ~ 3 km
			money=10;
		else if(km<=10)	/ / 3 ~ 10 kilometers
			money=10+(km- 3) *2;
		else			// More than 10 km
			money=10+7*2+(km- 10) *3;
	}
	else
	{
		if(km<=3)		/ / 0 ~ 3 km
			money=10+(m/5) *2;
		else if(km<=10)	/ / 3 ~ 10 kilometers
			money=10+(km- 3) *2+(m/5) *2;
		else			// More than 10 km
			money=10+7*2+(km- 10) *3+(m/5) *2;
	}
	printf("%1.lf\n",money);
	return 0;
}
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83, Finding the Partial Sum of power Series Expansion (20 marks)

We know that e x can be expanded as a power series 1+x+x 2/2! +x 3 /3! +. + x k/k! +… Given a real number x, we are asked to use this power series partial sum to approximate e x, and the sum continues until the absolute value of the last term is less than 0.00001.

Input format: input gives a real number x∈[0,5] in a single line.

Output format: Output the partial sum of the power series in one line, with four decimal points reserved.

Example Value: 1.2 Example value: 3.3201

#include<stdio.h>
#include<math.h>
double jiecheng(double n);
int main(a)
{
	double x;		// Enter a real number
	double n=0;		// I'll give you a formula to get the idea, bro
	double c=0;		// to indicate the power
	double sum=0;	/ / o combined
	double result;	// The absolute value of the last term
	scanf("%lf",&x);
	/* * * * * * * * * * * * * * * * * * * * * * * * * + (x ^ 3/3! + (x ^ n/n! * /
	/* There are 5 brackets in each bracket, but all 5 are equal! Do not believe you in the book to calculate ah ^_^ ha ha! * /
	/* Push to the formula: [x^0/0!] + "x 1/1 ^!" + (x ^ 2/2! + "x ^ 3 + 3!" + (x ^ n/n! * /
	do
	{
		result=pow(x,c)/(jiecheng(n));
		sum+=result;
		n++;
		c++;
	}while((jiecheng(n))>=0.00001);
	result=pow(x,c)/(jiecheng(n));
	sum+=result;
	printf("%.4f\n",sum);
	return 0;
}
double jiecheng(double n)
{
	int mix=1;
	for(int i=1; i<=n; i++) mix=mix*i;return mix;
}
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84, What is a computer? (5 points)

#include<stdio.h>
int main(a)
{
	printf("What is a computer?");
	return 0;
}
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85. Assign x squared to Y (5 marks)

Assuming the value of x is 3, compute the square of x and assign the value to Y, and output the values of x and Y as “y = x ∗ x” and “x ∗ x = y”, respectively.

Input format: This topic has no input

Output format: Output x=3 in the following format:

y = x * x

x * x = y

#include<stdio.h>
int main(a)
{
	int x=3;
	int y;
	printf("%d = %d * %d\n%d * %d = %d\n",x*x,x,x,x,x,x*x);
	return 0;
}
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86. Calculate the running time of the train (15 minutes)

#include <stdio.h>
#include <stdlib.h>
int main(a) 
{
    int h;		/ / hour
	int m;		/ / minute
	int num1;	// Departure time
	int num2;	// Time of arrival
    scanf("%d%d",&num1,&num2);
    h=num2/100-num1/100;            // Take the first two bits (hour)
    m=num2%100-num1%100;            //4位整型取余后2位(分钟)
    if(m<0)                  //num1, num2 are both on the same day, so h must be greater than 0
    {
        m=60+m;                 // When m is less than 0, borrow an hour from h to fill in m
        h=h- 1;
    }
    printf("%02d:%02d\n",h,m);	
	return 0;
}
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87, Calculate the salary (15 points)

The wages of employees of a company are calculated as follows: If they work no more than 40 hours in a week, they will be paid according to normal working hours; Working hours in excess of 40 hours will be paid at 1.5 times normal working hours. Employees are divided into new employees and old employees according to the time they have been in the company. Those who have been in the company for at least 5 years are old employees, and those who have been in the company for less than 5 years are new employees. The regular wage for new employees is 30 yuan/hour, and that for old employees is 50 yuan/hour. Please calculate the salary of employees according to this method of payment.

Input format: Enter two positive integers in one line, which are the number of years an employee has been employed and the number of weeks he/she has worked, separated by Spaces.

Output format: Output the employee’s weekly salary in one line, accurate to 2 decimal places.

Input example 1:5 40 Output example 1:2000.00 Input example 2:3 50 Output example 2:1650.00

#include<stdio.h>
int main(a)
{
	int year;					// Year of employment
	int h;						/ / hour
	double money;
	scanf("%d %d",&year,&h);	// Input function
	if(year<5)					// New employees
	{	
		if(h<=40)				// Normal working condition
			money=h*30;
		else					// Overtime
			money=40*30+(h- 40) *1.5*30;
	}
	else						// The situation of the old employees
	{
		if(h<=40)				// Normal working condition
			money=h*50;
		else					// Overtime
			money=40*50+(h- 40) *1.5*50;
	}
	printf("%.2lf\n",money);	
	return 0;
}
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88, Day K candle picture (15 points)

#include<stdio.h>
int main(a)
{
	double o;		//Open
	double h;		//High
	double l;		//Low
	double c;		//Close
	scanf("%lf%lf%lf%lf",&o,&h,&l,&c);
	if(c<o)
	{
		printf("BW-Solid");
		if(l<o && l<c && h>o && h>c)
			printf(" with Lower Shadow and Upper Shadow");
		else if(l<o && l<c)
			printf(" with Lower Shadow");
		else if(h>o && h>c)
			printf(" with Upper Shadow");
	}
	else if(c>o)
	{
		printf("R-Hollow");
		if(l<o && l<c && h>o && h>c)
			printf(" with Lower Shadow and Upper Shadow");
		else if(l<o && l<c)
			printf(" with Lower Shadow");
		else if(h>o && h>c)
			printf(" with Upper Shadow");
	}
	else if(c==o)
	{
		printf("R-Cross");
		if(l<o && l<c && h>o && h>c)
			printf(" with Lower Shadow and Upper Shadow");
		else if(l<o && l<c)
			printf(" with Lower Shadow");
		else if(h>o && h>c)
			printf(" with Upper Shadow");
	}
	return 0;
}
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89. Are you too fat (10 points)

It is said that a person’s standard weight should be the number of kilograms obtained by subtracting 100 from his height in centimetres and multiplying by 0.9. Real weight and standard weight error within 10% are perfect shape (that is, the real weight – standard weight | | < standard weight (10%). We know that a kilo is twice as much as a kilo. Given a group of people’s height and actual weight, please tell them whether they are too fat or too thin.

Input format: The first line of input gives a positive integer N (≤ 20). Then N lines, each of which gives two integers, are a person’s height H (120 < H < 200; Unit: cm) and true weight W (50 < W ≤ 300; Unit: city jin), which is separated by a space.

Output format: output a conclusion for each person: if it is perfect figure, output You are Wan Mei! ; If You are too fat, output You are tai Pang le! ; Otherwise output You are tai shou le! .

Example: 3 169 136 150 81 178 155 Example: You are wan MEI! You are tai shou le! You are tai pang le!

#include<stdio.h>
#include<math.h>
int main(a)
{
	int n;		// Enter a positive integer
	int h;		/ / height
	int w;		// True weight
	int x;		// Standard weight
	/* Standard weight = (height -100) *0.9*0.2 */
	scanf("%d",&n);
	if(n<1 || n>20)		return 0;   //n<=20
	for(int i=1; i<=n; i++) {scanf("%d %d",&h,&w);
		if(h<120 || h>=200 || w<50 || w>300)	return 0;     // 120 < H < 200 and 50 < W ≤ 300
		x = (h- 100.) *0.9*2;
		if(fabs(w-x) < x*0.1)
			printf("You are wan mei! \n");
		else if((fabs(w-x) >= x*0.1) && w<x)
			printf("You are tai shou le! \n");
		else if((fabs(w-x) >= x*0.1) && w>x)
			printf("You are tai pang le! \n");
	}
	return 0;
}
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90. Find the smallest string (15 points)

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
int main(void)
{
	char a[10000] [80];
	int n;
	int i,j;
	int min;
	char temp[80];
	scanf("%d",&n);
	getchar();
	for(i=0; i<n; i++) gets(a[i]);for(i=0; i<n- 1; i++) { min=i;for(j=i+1; j<n; j++)if(strcmp(a[min],a[j])>0)
				min=j;
			if(min! =i) {strcpy(temp,a[min]);
				strcpy(a[min],a[i]);
				strcpy(a[i],temp); }}printf("Min is: %s",a[0]);
	return 0;
}
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91. Bubble sorting of Strings (20 points)

We’ve seen bubble sort for ordering N integers from smallest to largest. For any given K (<N), output the intermediate result sequence after scanning the KTH time.

Input format: N and K are given in line 1 (1≤K

Output format: Output the intermediate result sequence after the bubble sort scanning for the KTH time, each line contains a string.

Example: 6 2 Best cat east a free day Example: Best a cat day east free

7-1 Bubble sort for strings (20 points)

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(a)
{
	int n,k;			//n→ have several strings, k→ scan several times
	int i,j;			// For loops and array subscripts
	char a[100] [11];	// Used to store strings
	char temp[11];
	scanf("%d%d",&n,&k);	// Input n and k
	getchar();
	for(i=0; i<n; i++)// Provide input string function
		scanf("%s",a[i]);

	for(i=0; i<k; i++) {for(j=0; j<n- 1-i; j++) {if(strcmp(a[j],a[j+1) >0)
			{
				strcpy(temp,a[j]);
				strcpy(a[j],a[j+1]);
				strcpy(a[j+1],temp); }}}for(i=0; i<n; i++)printf("%s\n",a[i]);
	return 0;
}
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92, The organization of the week information (10 points)

Enter a positive integer repeat (0<repeat<10) and repeat the following operation:

Define an array of Pointers to organize the following week’s information, enter a string, look in the table, if present, output the string in the table ordinal, otherwise output -1.

Sunday Monday Tuesday Wednesday Thursday Friday Saturday

Input and output Examples: Note in parentheses. No input and output is required

Example Value (repeat=3) : 3 Tuesday Wednesday Year Example value: 3 4-1

7-2 Organization week information (10 points)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(a)
{
	char *p[] ={"Sunday"."Monday"."Tuesday"."Wednesday"."Thursday"."Friday"."Saturday"};
	int repeat;		// Enter several numbers
	int i,j;		// For loops and array subscripts
	//int flag = 0; // Check if the sentence is output
	char tmp[50];	// The input string
	scanf("%d",&repeat);
	getchar();
	for(i=0; i<repeat; i++) { gets(tmp);for(j=0; j<7; j++) {if(strcmp(tmp,p[j])==0)
			{	printf("%d\n",j+1);	break;	}
			else if(j==6)
				printf("-1\n"); }}return 0;
}
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93. Even and Odd (10 points)

Given N positive integers, how many odd and even integers are there?

Input format: The first line of input gives a positive integer N (≤1000); Line 2 gives N non-negative integers, separated by Spaces.

Output format: Output the number of odd numbers and even numbers successively in a line. Separate the names with one space.

Example Value: 9 88 74 101 26 15 0 34 22 77 Example value: 3 6

7-3 Odd and Even Split (10 points)

#include <stdio.h>
#include <math.h>
int main(a)
{
	int n;			// There are several numbers
	int a[1001];	// Space to store data
	int i;		// For loops and array subscripts
	int ji=0,ou=0;	//ji→ count odd numbers, ou→ count even numbers
	scanf("%d",&n);
	for(i=0; i<n; i++)scanf("%d",&a[i]);
	for(i=0; i<n; i++) {if(a[i]%2= =0)
			ou++;
		else
			ji++;
	}
	printf("%d %d\n",ji,ou);
	return 0;
}
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94. Pirates’ Spoils (25 points)

P pirates stole D diamonds and went to the high seas to divide the spoils. They agreed on the following strategy:

First, P pirates draw lots to determine the number 1-P. Then, pirate No. 1 proposes a distribution plan (the plan should give the specific amount of each pirate’s share). If an absolute majority (i.e., more than half) including Pirate No. 1 agrees, the distribution plan will be implemented, otherwise, Pirate No. 1 will be thrown into the sea to feed sharks; Then, pirates no. 2, no. 3, and so on, put forward similar proposals in turn, until a plan can be approved by an absolute majority, or there is only one pirate left, who monopolizes all the diamonds. Write a program to give the number of diamonds in pirate 1’s diamond allocation plan.

Three incidental assumptions:

“Smart” and “greedy” assume that each pirate will always act in his own best interest; “Human” assumes that pirates won’t deliberately kill their friends if they can get as many diamonds as possible; “Unbiased” assumes that there is no personal animosity between pirates and that diamonds are distributed to other pirates in the order of small numbers first. Input format: Input gives two positive integers D and P (3≤P≤D≤100) in one line.

Output format: Output the number of diamonds in pirate 1’s diamond allocation scheme.

Example Value: 10 7 Example value: 6

7-9 Pirates’ Spoils (25 points)

#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(a)
{
	int d,p;					//d→ number of diamonds, p→ number of diamonds
	scanf("%d %d",&d,&p);		// Provide input diamond, number of functions
    if(p == 3) {printf("%d",d- 1);
    }
    else{
        printf("%d",d-(p/2+1));	When there are more than three people, the first pirate gets gold: D - (P / 2 + 1)
    }
 
 
    return 0;
}
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95. Speeding judgment (10 points)

Simulated traffic police radar. Enter the speed of the car. If the speed exceeds 60 MPH, “Speeding” will be displayed, otherwise “OK” will be displayed.

Input format: Input gives a non-negative integer not more than 500 in a line, that is, the speed measured by radar.

Output format: Outputs the speedometer display results in a line in the format of Speed: V-s, where V is the Speed and S is the Speeding or OK.

Example Value 1: Speed: 40 Example value 1: Speed: 40 -OK Example value 2:75 Example value 2: Speed: 75-restricted

7-4 Speeding judgment (10 points)

#include<stdio.h>
int main(a)
{
	int sudu;
	scanf("%d",&sudu);
	if(sudu<=60) {printf("Speed: %d - OK\n",sudu);
	}else{
		printf("Speed: %d - Speeding\n",sudu);
	}
	return 0;
}
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96, JMU-DS – Order table interval element deletion (15 marks)

If a linear table L is stored in a sequential storage structure, all of its elements are integers. Design an algorithm to delete all elements whose element values are between [x,y]. The time complexity of the algorithm is O(n) and the space complexity is O(1).

Input format: three rows of data, the first row is the number of elements in the order table, the second row is the elements in the order table, and the third row is X and y.

Output format: delete all elements between [x,y].

Example Value: 10 5 1 9 10 67 12 8 33 6 2 3 10

Example Output: 1 67 12 33 2

Jmu-ds – Order table interval Element deletion (15 marks)

import java.util.Scanner;
public class Main {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int g=0,t=0;
		int x,y;
		int L[]={1.2.3.4.5.6.7.8.9.21213.123.1123.1.132123.123
				,123.132.23.123.12.32.123.123.12.132.123.12.32
				,123.12.132.123.1.23.132.123.312.321
				,12312.32.123.123.1.1231.231.23.123.1321.32
				,12.132.123.132.123.123.132.12.123.123.123.123.1.231.231.231.231.231.21.321.231.231.3
				,1.1.3.1321.1.1.132.123.1.1321.1.32.32.123.123.123.132.23.1321.321.1.23.1
				,123.13.1231.32.231.1.321.123.1.132.123.123.132.123.132.12.132.123.132.132.321.132
		  ,123.23.132.132.132.132.13.213.2.132.1.231.32.132.132.132.132
		  ,23.132.132.132.132.13.213.2.132.1.231.32.132.132.132.13
		  ,23.132.132.132.132.13.213.2.132.1.231.32.132.132.132.13};
		int i,j;
		for(i=0; i<n; i++) L[i]=scanner.nextInt(); x = scanner.nextInt(); y = scanner.nextInt();for(i=0; i<n; i++) {if(L[i]>=x && L[i]<=y)
			{	L[i]=0;}
		}
		for(i=0; i<n; i++)if(L[i]! =0)
				g++;
		for(i=0; i<n; i++) {if(L[i]! =0) {if(t! =g-1)
				System.out.print(L[i] + "");
				elseSystem.out.print(L[i]); t++; }}}}Copy the code

97, Find the specified character (15 marks)

This requires writing a program to find a specified character from a given string.

Input format: The first line of input is a character to be looked up. The second line is a non-empty string (no more than 80 characters) that ends with a carriage return.

Output format: If found, output the maximum subscript of the character in the string in the format “index = subscript” (the subscript starts from 0). Otherwise, “Not Found” is printed.

Input Example 1: M Programming Example 1: index = 7 Input Example 2: A 1234 Output example 2: Not Found

#include<stdio.h>
int main(void)
{
	char c,b[10010];	// The character to be queried, the input character string
	int index=- 1;		/ / tag
	scanf("%c",&c);
	getchar();		// Eats a space, otherwise it will be considered the content of the stringgets(b); Affected by the gets input, the strings all start with \0As the endfor(int i=0; b[i]! ='\ 0'; i++)// So it starts at 0 and ends at = \0
		if(b[i]==c)	// When a character to be queried occurs, it is marked once until the end of the loop
			index=i;
	if(index==- 1)
		printf("Not Found\n");
	else
		printf("index = %d\n",index);
}
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98. Four Operations on complex numbers (15 points)

#include<stdio.h>
void shizi(double x,double y);		// Print the formula
void result(double x,double y);		// Print the result
struct Love{		
	double shi,xu;	// Represents the real part, imaginary part
}love1,love2;
int main(a)
{	
	scanf("%lf%lf%lf%lf",&love1.shi,&love1.xu,&love2.shi,&love2.xu);
	// The following calculation is based on the real and imaginary parts of the complex number:
	// Find the real and imaginary parts of the addition
	double jia_shi = love1.shi + love2.shi;
	double jia_xu  = love1.xu  + love2.xu;

	// Find the real and imaginary parts of the subtraction method
	double jian_shi = love1.shi - love2.shi;
	double jian_xu  = love1.xu  - love2.xu;

	// Find the real and imaginary parts of the multiplication
	double cheng_shi = love1.shi*love2.shi-love1.xu*love2.xu;
	double cheng_xu  = love1.xu*love2.shi+love1.shi*love2.xu;
	
	// Find the real and imaginary parts of division
	double chu_shi = (love1.shi*love2.shi+love1.xu*love2.xu)/(love2.shi*love2.shi+love2.xu*love2.xu);
	double chu_xu  = (love1.xu*love2.shi-love1.shi*love2.xu)/(love2.shi*love2.shi+love2.xu*love2.xu);
	
	// Since the output of the problem can be divided into several cases, we can judge the case output in the function
	
	/ / add
	shizi(love1.shi,love1.xu);	/ / formulas
	printf("+");
	shizi(love2.shi,love2.xu);	/ / formulas
	printf("=");
	result(jia_shi,jia_xu);		/ / the result

	/ / subtraction
	shizi(love1.shi,love1.xu);	/ / formulas
	printf("-");
	shizi(love2.shi,love2.xu);	/ / formulas
	printf("=");
	result(jian_shi,jian_xu);		/ / the result
	
	/ / the multiplication
	shizi(love1.shi,love1.xu);	/ / formulas
	printf("*");
	shizi(love2.shi,love2.xu);	/ / formulas
	printf("=");
	result(cheng_shi,cheng_xu);		/ / the result

	/ / division
	shizi(love1.shi,love1.xu);	/ / formulas
	printf("/");
	shizi(love2.shi,love2.xu);	/ / formulas
	printf("=");
	result(chu_shi,chu_xu);		/ / the result
	return 0;
}
void shizi(double x,double y)		// Print the formula
{
	if(y<0)							// Do not add '+' when the imaginary part is negative
		printf("(%.1lf%.1lfi)",x,y);
	else
		printf("(%.1lf+%.1lfi)",x,y);	// The reverse is true
}
void result(double rshi,double rxu)
{	
	// If the imaginary part does not exist, just print the real part
	if(rxu<=0.05&&rxu>=0.05)	// If there is neither segment, output 0.0
		printf("%.1lf\n",rshi);
	else if(rshi<=0.05&&rshi>=0.05)	// If the real part does not exist, output a single imaginary part
		printf("%.1lfi\n",rxu);
	else if(rxu<0)
		printf("%.1lf%.1lfi\n",rshi,rxu);	// Do not add the '+' sign when the imaginary part is negative
	else
		printf("%.1lf+%.1lfi\n",rshi,rxu);	// The reverse is true
}
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99. Output triangle character array (15 points)

Write A program to output an array of n lines of triangular characters starting with A capital letter A.

Input format: Input is given a positive integer n (1≤n<7) on a single line.

Output format: Output n lines of triangular character array starting with capital letter A. The format is shown in the output sample, where each letter is followed by a space.

Example: 4 Example output: A B C D E F G H I J

Output triangle character array (15 points)

#include<stdio.h>
int main(a)
{
	int n,i,j;
	char x=64;  // Start at A, the ASCLL of A is 65, +1 into the j loop
	scanf("%d",&n);
	for(i=0; i<n; i++)// Control the number of rows
	{
		for(j=0; j<n-i; j++)// Control print
		{ 
			x += 1; // Each time +1, A→65, B→66
			printf("%c ",x);
		}
		printf("\n");
	}
	return 0;
}
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100. Find the longest string (15 points)

In this case, we need to write a program to output the longest string of N characters.

Input format: The first line of input gives a positive integer N; Each of the following N lines gives a non-empty string of less than 80 length, with no newlines, Spaces, or tabs.

Output format: Output the longest string in one line in the following format:

If The longest is The same length, The first input string is output.

Example: 5 Li Wang Zhang jin xiang The longest is: Zhang

Find the longest string (15 points)

#include<stdio.h>
#include<string.h>
int main(a)
{
	int n,i;
	char a[100],b[100];	//a→ take input, b→ take the longest string
	scanf("%d",&n);
	getchar();	// Eat the space
	for(i=0; i<n; i++){ gets(a);if(i==0)
			strcpy(b,a);	// The first one needs to be copied to prevent cases where the first one is the longest
		if(strlen(b)<strlen(a))	// If the new one is longer than the old one, copy it to cover the old one
			strcpy(b,a);
	}
	printf("The longest is: ");
	puts(b);
	return 0;
}
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101, The classification of integers (20 points)

Given N positive integers, you are required to obtain the following three results:

A1 = The largest integer divisible by 3 A2 = The number of integers with an integer K that can be expressed as 3K+1 A3 = The average (to 1 decimal place) of all integers with an integer K that can be expressed as 3K+2 The input first gives a positive integer N on the first line, followed by N positive integers on the next line. All the digits cannot exceed 100. The digits in the same row are separated by Spaces.

Output format: Outputs the values A1, A2, and A3 in sequence in one line, separated by one space. If a number does not exist, NONE is printed.

Example 1:8 5 8 7 6 9 1 3 10 Example 1:9 3 6.5 Example 2:8 15 18 7 6 9 1 3 10 Example 2:18 3 NONE

Classification of integers (20 points)

#include<stdio.h>
int main(a)
{
	int n,a2=0,a3=0,i,x,max=0,sum=0;
	scanf("%d",&n);
	for(i=0; i<n; i++){scanf("%d",&x);
		if(x%3= =0&&x>max)	// get A1
			max=x;	
		if((x- 1) %3= =0)		// The number of valid ==A2
			a2++;
		if((x2 -) %3= =0)		// The number of valid integers and / ==A3{ a3++; sum+=x; }}if(max==0)	printf("NONE ");	// Output HONE if a1, A2, or a3 is not present
	else		printf("%d ",max);
	if(a2==0)	printf("HONE ");
	else		printf("%d ",a2);
	if(sum==0)	printf("HONE");
	else		printf("%.1f",sum*1.0/a3);
	return 0;
}
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102. Length of Service (20 points)

Given the seniority of N employees in a company, the number of employees in each seniority segment is output in ascending order.

Input format: input a positive integer N (≤10 5), that is, the total number of employees; N integers are then given, i.e. the length of service of each employee, ranging from [0, 50].

Output format: Displays the number of employees of each seniority in ascending order, in the format of Seniority: Number of employees. Each term is one row. If the number of people is 0, the item is not printed.

Input example: 8 10 2057 7 52 Output example: 0:1 2:3 5:2 7:1 10:17 7-8 Statistical service years (20 points)

#include<stdio.h>
int main(a)
{
	int i,n,a,b[52] = {0};
	scanf("%d",&n);
	for(i=0; i<n; i++) {scanf("%d",&a);
			b[a]++;
	}
	for(i=0; i<52; i++)if(b[i]! =0)
			printf("%d:%d\n",i,b[i]);

	return 0;
}
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103, 7-10 Array looping left (20 points)

#include<stdio.h>
int main(void){
	
	int a[101];
	int n,m,i;
	while(scanf("%d%d",&n,&m)! =EOF){for( i=0; i < n; i++)scanf("%d",&a[i]);
		
		for( i =m; i<n+m; i++){if( i >=n)
			  printf("%d",a[ i%n]);
			else printf("%d",a[i]);
			
			if( i == n+m - 1)
			  printf("\n");
			else printf(""); }}return 0;
}
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Find the reciprocal KTH term of a chained linear list (20 marks)

Given a series of positive integers, design an algorithm as efficient as possible to find the number in the KTH position to the last.

Input format: Input begins with a positive integer K, followed by a number of positive integers, and ends with a negative integer (the negative number is not counted in the sequence and is not processed).

Output format: Outputs the data in the KTH position. If the location does not exist, output an error message NULL.

Input example: 4 1 2 3 4 5 6 7 8 9 0-1 Output example: 7

Find the reciprocal KTH term of a chained linear list (20 marks)

#include<stdio.h>
#include<string.h>
#define maxn 0x7fffffff
int a[maxn];
int main(a)
{
    int k;
    scanf("%d",&k);
    int count=0,x;
    while(1)
    {
        scanf("%d",&x);
        if(x<0)break;
        else{ a[count]=x; count++; }}if(count-k<0)
     printf("NULL\n");
    else
     printf("%d\n",a[count-k]); 
    return 0;
}
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105, Square Circle to the right (20 points)

In this case, we need to write a program to move each element in a given n× N square matrix to the right by m positions. That is, the 0, 1,…, and N −1 columns are transformed into the N − M, N − M +1,…, N −1, 0, 1,…, and N − M −1 columns.

Input format: The first line of input gives two positive integers m and n (1≤n≤6). Then there are n rows, each row of n integers, representing a square matrix of order n.

Output format: Output the moved square matrix according to the input format: that is, output N lines with n integers in each line and a space after each integer.

Example Value: 2 3 1 2 3 4 5 6 7 8 9 Example value: 2 3 1 5 6 4 8 9 7

7-12 Square Circle to the right (20 points)

#include <stdio.h>
// Time: 20:08:02, April 23rd, 2018
// If you can control the index of the column, you can do it
// The loop prints the two-dimensional array after the loop moves to the right. Because the loop moves to the right, the value m needs to be controlled between 0 and n-1.
int main(a)
{
	int a[10] [10];
	int i, j, n, m;
	scanf("%d%d",&m,&n);
	for (i = 0; i<n; i++)
	{
		for (j = 0; j<n; j++)
		{
			scanf("%d", &a[i][j]);
		}
	}
	m %= n; // the value of m is less than n, and the value of 0 to n-1 is easy to control the following subscript.
	for (i = 0; i<n; i++)
	{
		for (j = 0; j<n; j++)
		{
			printf("%d ", a[i][(n - m + j) % n]); A [I][(n-m+j) % n] a[I][(n-m+j) % n]
		}
		printf("\n");
	}
	return 0;
}
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106, The minimum number of groups (20 points)

Given a number of numbers from 0 to 9. You can arrange the numbers in any order, but you must use them all. The goal is to make the resulting number as small as possible (note that 0 is not the first digit). For example, given two zeros, two ones, three fives, and one eight, the smallest number we can get is 10015558.

Given a number, write a program to output the smallest number that can be composed.

Input format: Input gives 10 non-negative integers in a row, in order to indicate that we have the numbers 0, 1,… The number of 9’s. Integers are separated by a space. The total number of 10 digits cannot exceed 50, and there must be at least one non-0 digit.

Output format: Outputs the smallest number that can be formed in a line.

Input example: 2 2 00 0 3 001 0 Output example: 10015558

Minimum number of groups 7-3 (20 points)

#include<stdio.h>
int main(int argc, char const *argv[])
{
	int i,j;
	int x[9];
	for(i=0; i<10; i++)scanf("%d",&x[i]);
	for(i=1; i<10; i++)if(x[i])
		{	printf("%d",i);  x[i]--;  break;	}
	for(i=0; i<10; i++)while(x[i])
		{	printf("%d",i);	x[i]--;		}
	return 0;
}
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107. Monkeys eat peaches (15 points)

A monkey picked a number of peaches on the first day, immediately ate half, not satisfied, and eat one more; The next morning he ate half of the remaining peaches and one more. After that, I ate half of what I had left the day before and one more every morning. To the N day morning to eat again, see only a peach left. Q: How many peaches were picked on the first day?

Input format: Input is given a positive integer N (1

Output format: Output the number of peaches picked on the first day in one line.

Example Input: 3 Example output: 10

7-1 Monkeys eating peaches (15 points)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main(a)
{
	int i,j;
	int n;
	int sum=2;
	scanf("%d",&n);
	for(i=1; i<n; i++) {if(i==n- 1)
			sum*=2;
		else
			sum=sum*2+1;
	}
	printf("%d\n",sum);
	return 0;
}
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108, String uppercase conversion (15 points)

In this case, you are required to write a program to convert all lowercase letters of a string ending with “#” to uppercase letters and all uppercase letters to lowercase letters. Other characters are output unchanged.

Input format: Enter a string (no more than 30 characters) ending with “#”.

Output format: Output the result string after case conversion in one line.

Example: Hello World! 123# output example: hELLO wORLD! 123

7-2 String uppercase conversion (15 points)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main(a)
{
	int i,j;
	char a[10010];
	gets(a);
	int k = strlen(a);
	a[k- 1] = '\ 0';
	for(i=0; a[i]! ='\ 0'; i++) {if(((a[i]>='a'&&a[i]<='z') || (a[i]>='A'&&a[i]<='Z')) = =0)
			continue;
		else if(a[i]>='a'&&a[i]<='z')
			a[i]-=32;
		else
			a[i]+=32;
	}
	puts(a);
	return 0;
}
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109. Calculation of piecewise Functions [3] (10 marks)

The following piecewise function f(x) is computed:

Input format: Input gives the real number X on one line.

Output format: F (x) = result in a line, where x and result have a decimal place.

Example 1: F (10.0) = 0.1 Example 2:234 Example 2: F (234.0) = 234.0

Calculation of piecewise Functions [3] (10 marks)

#include<string.h>
#include<stdlib.h>
int main(a)
{
	float x;
	float result=0;
	scanf("%f",&x);
	if(x! =10)
		result = x;
	else
		result = 1/x;
	printf("f(%.1f) = %.1f\n",x,result);
	return 0;
}
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110, Find integer (10 marks)

The problem asks to find the given X from the N integers in the input. If found, output the position of X (counting from 0); If Not, “Not Found” is displayed.

Input format: Input gives two positive integers N (≤20) and X in the first line, and N integers in the second line. The number must be no more than a long integer, separated by Spaces.

Output format: Print the position of X in a line, or “Not Found”.

Example 1:5 7 3 5 7 1 9 Example 1:2 Example 2:5 7 3 5 8 1 9 Example 2: Not Found

7-6 Finding integers (10 points)

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
int main(a)
{
	int n,x,i;
	int c[10010];
	scanf("%d%d",&n,&x);
	for(i=0; i<n; i++) {scanf("%d",&c[i]);
		if(c[i]==x)
			break;
	}
	if(i==n)
	printf("Not Found\n");
	else
	printf("%d\n",i);
	return 0;
}
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111. Tourism Planning (25 points)

With a road map, you’ll know the length of the highway between cities and how much tolls will be charged. Now you need to write a program to help inquiring travelers find the shortest route between their origin and destination. If several paths are shortest, the cheapest path needs to be output.

Input format: Input description: The first line of the input data contains four positive integers N, M, S, and D, where N (2≤N≤500) is the number of cities. Assume that the number of cities ranges from 0 to (N−1). M is the number of highways; S is the city number of the departure point; D is the city number of the destination. In the following M lines, each line gives the information of one expressway, namely, city 1, city 2, expressway length and toll, separated by Spaces. The numbers are all integers and not more than 500. The input guarantees the existence of a solution.

Output format: The length of the output path and the total amount of charges in one line, separated by Spaces, no extra space at the end of the output.

Example: 4 5 0 30 1 1 20 1 3 2 30 0 3 4 10 0 2 2 20 2 3 1 20 Example: 3 40

7-7 Tourism Planning (25 points)

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
int main(a)
{
	int cheng1[5010];	1 / / city
	int cheng2[5010];	2 / / city
	int chang[5010];		// The length of the expressway
	int money[5010];		// Total fees
	int i,j;
	int N;	// City number
	int M;	// The number of highways
	int S;	// City number of departure
	int D;	// Destination city number
	int MIN=0;	// Shortest path
	int result=0;	/ / their
	int shangMIN=0;
	int shangResult=0;
	int index;
	scanf("%d%d%d%d",&N,&M,&S,&D);
	for(i=0; i<M; i++)// Enter information for each highway
		scanf("%d%d%d%d",&cheng1[i],&cheng2[i],&chang[i],&money[i]);
	for(i=0; i<M;) {if((cheng1[i]==0)&&(cheng2[i]==N- 1)) { result=money[i]; MIN=chang[i]; index=1; }else
		{
			MIN=chang[i]+chang[i+1];
			result=money[i]+money[i+1];
			index=2;
		}
		if(i==0)
		{
			shangMIN = MIN;
			shangResult=result;
		}
		else
		{
			if(MIN==shangMIN && result<shangResult) { shangMIN=MIN; shangResult=result; }}if(index==1)
			i++;
		else if(index==2)
			i=i+2;
	}
	printf("%d %d\n",shangMIN,shangResult);
	return 0;
}
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Number of Black Holes (20 points)

Black hole numbers, also known as trap numbers and “Kaprekar problems”, are a class of numbers with peculiar transformation properties.

Any three-digit number with different digits will always yield 495 after a finite number of rearrangement operations. The resulting 495 is the number of three-position black holes. The so-called “rearrangement difference” operation is the rearrangement of the largest number of the number minus the rearrangement of the smallest number. (6,174 is the number of four-digit black holes.)

For example, for the three-digit number 207:

The difference of the first rearrangement is 720-27 = 693; The second rearrangement is 963-369 = 594; Third rearrangement: 954-459 = 495; It’s going to stay at 495. If all three digits of a three-digit number are the same, it is 0 after one conversion.

Any input of a three – digit, programming gives the process of rearrangement difference.

Input format: Input is given a three-digit number in a line.

Output format: Output the process of rearranging the difference in the following format:

Ordinal number: Maximum number rearranged – Minimum number rearranged = Difference Ordinal number starts at 1 until 495 appears to the right of the equals sign.

Example: 123 Example: 1: 321-123 = 198 2: 981-189 = 792 3: 972-279 = 693 4: 963-369 = 594 5: 954-459 = 495

Number of Black Holes (20 points)

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
int number[6] = {0};
int MAX,MIN;
int sum=0;
int swap2(int x,int y)
{
	if(x<y)
		return x;
	else
		return y;
}
int swap(int x,int y)
{
	if(x>y)
		return x;
	else
		return y;
}
int main(a)
{
	int n;
	int ci=1;
	int i=0,j;
	int index;
	int ge,shi,bai;
	int result;
	scanf("%d",&n);
	index = n;
	while(1)
	{
		ge=index%10;
		shi=index/10%10;
		bai=index/100;
			number[0] = ge + shi*100 + bai*10;
			number[1] = shi + ge*10 + bai*100;
			number[2] = shi + ge*100 + bai*10;
			number[3] = bai + shi*10 + ge*100;
			number[4] = bai + shi*100 + ge*10;
			number[5] = index;
			MAX=number[0];
			MIN=number[0];
		for(i=1; i<6; i++) MAX = swap(number[i],MAX);for(i=1; i<6; i++) MIN = swap2(number[i],MIN); sum = MAX-MIN;printf("%d: %d - %d = %d\n",ci,MAX,MIN,sum);

		ci++;
		index = sum;
		if(sum==495)
			break;
	}
	return 0;
}
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113. The Hare and the Tortoise (20 points)

Tortoise and rabbit race, running field is a rectangular runway, runway can rest anywhere. A tortoise can run 3 meters per minute, and a hare 9 meters per minute. The rabbit too slow tortoise run, feel sure to beat the tortoise, so every 10 minutes run back to look at the tortoise, if found himself more than the tortoise, on the roadside rest, rest 30 minutes each time, otherwise continue to run for 10 minutes; But the tortoise worked very hard. He ran all the time and didn’t rest. Suppose the tortoise and the rabbit at the same starting point at the same time start to run, I would like to ask T minutes after the tortoise and rabbit who ran faster?

Input format: Input gives the race time T (minutes) in one line.

Output format: Output the result of the race in one line: turtle wins and output @ @, rabbit wins and output _, tie and output –; Followed by a space to print the distance the winner has run.

Example Input: 242 Example output: @_ @726

7-9 The Tortoise and the Hare (20 points)

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
int main(a)
{
	int fen;
	int i,j;
	int gui=0,tu=0;
	int index=0;	// Mark the rabbit rest time
	int swap=1;		// Decide whether the rabbit will continue to run
	scanf("%d",&fen);
	for(i=1; i<=fen; i++) {if(index==30)
		{	index=0;	swap=1;	}
		if(swap! =- 1)
		{
			gui+=3;
			tu+=9;
		}
		else
		{
			gui+=3;
			index++;
		}
		if(i%10= =0)
		{
			if(tu>gui)
				swap=- 1; }}if(gui>tu)
	printf("@_@ %d\n",gui);
	else if(gui==tu)
	printf("-_- %d\n",gui);
	else
	printf("^_^ %d\n",tu);
	//printf("gui = %d,tu = %d\n",gui,tu);
	return 0;
}
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114, Spiral square (20 points)

The so-called “spiral square matrix” means that for any given N, the numbers from 1 to N×N are filled into the N×N square matrix in a clockwise spiral direction starting from the first box in the upper left corner. The problem requires the construction of such a spiral matrix.

Input format: Input is given a positive integer N (<10) on one line.

Output format: output N×N spiral matrix. Each row contains N digits, each of which contains 3 digits.

Example: 5 Example: 12 3 4 5 16 17 18 19 6 15 24 25 20 7 14 23 22 21 8 13 12 11 10 9

7-4 Spiral square (20 points)

#include<stdio.h>
int main(a)
{
	int n;
	int h=0,l=0;
	int result=1;
	int t;
	int a[100] [100] = {0};
	scanf("%d",&n);
	a[h][l]=1;
	while(n*n! =result) {for(t=result;;)		/ / cross
	{
		if((l+1<n && ! a[h][l+1])! =true)
			break;
		a[h][++l] = ++result;
	}
	for(t=result;;)		/ / vertical
	{
		if((h+1<n && ! a[h+1][l])! =true)
			break;
		a[++h][l] = ++result;
	}	
	for(t=result;;)		/ / cross
	{
		if((l- 1> =0 && !a[h][l- 1])! =true)
			break;
		a[h][--l] = ++result;
	}
	for(t=result;;)		/ / vertical
	{
		if((h- 1> =0 && !a[h- 1][l])! =true)
			break; a[--h][l] = ++result; }}for(h=0; h<n; h++) {for(l=0; l<n; l++)printf("%3d",a[h][l]);
		printf("\n");
	}
	return 0;
}
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115. Delete substrings from a string (20 points)

Enter two strings S1 and S2, and delete all substrings S2 that appear in string S1, that is, the result string cannot contain S2.

Input format: Two non-empty character strings with a length of no more than 80 characters and ending with carriage return are given in two lines, corresponding to S1 and S2.

Output format: Output the result string after deleting all substring S2 that occurs in string S1 on one line.

Example: Tomcat is a male CCatat CAT Example: Tom is a male

Delete substrings from a string (20 points)

#include<stdio.h>
#include<string.h>
char a[85],b[85],l,m;
void dete1(char *p,char *q){
   p=strstr(p,q);
   while(*(p+l)! ='\ 0'){
   	*p=*(p+l);
   	*p++;
   }
   *p=*(p+l);
   return ;
}
int main(a){
	int i,j,k;
    gets(a);
    gets(b);
	l=strlen(b);
	while(strstr(a,b)! =NULL){
		 m=strlen(a);
         dete1(a,b); 
	   }
     printf("%s\n",a);
	return 0;
} 
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116, A-B (20 points)

They ask you to calculate A−B. The trouble is that both A and B are strings — that is, A−B is formed by removing all the characters contained in B from string A.

Input format: Input gives strings A and B in sequence on 2 lines. Both strings must be no longer than 10 or 4, and each string must consist of visible ASCII characters and whitespace characters, ending with a newline character.

Output format: Print the result string of A−B in one line.

Example: I love GPLT! It’s a fun game! Aeiou Example output: I LV GPLT! It ‘s fn gm!

#include<stdio.h>
#include<string.h>
int main(a)
{
	int i=0,j=0;			// For loops and array subscripts
	char str1[10010];		// Store two strings A and B
	char str2[10010];		// Used to store the string letters to be deleted
	int length;				// To store the length of a string
	gets(str1);
	gets(str2);
	length = strlen(str2);	// Just remember the length of the str2 string
	while(str1[i]! ='\ 0')	// start with str1
	{
		for(j=0; j<length; j++)// Put this letter in str2 and see if there is a corresponding letter, if there is a corresponding letter
		{					// Then the for loop will jump out prematurely, and j cannot be equal to length,
			if(str1[i]==str2[j])// Remember the key words in advance! So we can be sure from this that if the for loop executes
				break;			// At the end of the day, there is no corresponding letter in str2,
		}						// We need to display it.
		if(j==length)
			printf("%c",str1[i]);
		i++;
	}
	printf("\n");
	return 0;
}
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117, Guess the number (20 points)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct E{
char name[11];
int num;
}a[10001];
int main(a){
    int n,sum=0;
    scanf("%d",&n);
    for(int i=0; i<n; i++){scanf("%s%d",a[i].name,&a[i].num);
        sum+=a[i].num;
    }
    int avg=(sum/n)/2,minm=11111,tag;
    for(int i=0; i<n; i++){if(abs(avg-a[i].num)<minm){
            minm=abs(avg-a[i].num); tag=i; }}printf("%d %s\n",avg,a[tag].name);
    return 0;
}
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118, 40059 Four Operations (15 points)

Write a priority-free pocket calculator that can add, subtract, multiply and divide.

Input format: for example, keyboard 10.83+0.10/10=.

Output format: output result 1.093000 after program calculation (operation formula ends with ‘=’).

Example: 34.5-2.4 x 3.1/11= Example: 9.722727

#include<math.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(a)
{
	int i,j;
	double num1,num2;
	char c;
	scanf("%lf",&num1);
	for(i=0;; i++) {scanf("%c",&c);
		if(c=='=')
		break;
		scanf("%lf",&num2);
		if(c==The '-')
			num1-=num2;
		else if(c=='+')
			num1+=num2;
		else if(c==The '*')
			num1*=num2;
		else if(c=='/')
		{
// if(num2==0)
// num1=0;
// elsenum1/=num2; }}printf("%lf\n",num1);
		return 0;
}
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Who is the tallest in the dormitory (20 points)

#include<math.h>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct Student{
	int number;
	char name[18];
	int height;
	int weight;
}stu[10010] = {0};
int main(a)
{
	int n;
	int i;
	char temp1[20];
	int index[100],k=0;
	int temp2;
	int number_new;
	char name_new[18];
	int height_new;
	int weight_new;

	scanf("%d",&n);
	for(i=0; i<n; i++) {scanf("%d%s%d%d",&number_new,&name_new,&height_new,&weight_new);
		index[k++]=number_new;
		if(stu[number_new].height == 0)		// Use the serial number to find whether the height has been saved
		{
			stu[number_new].number=number_new;
			strcpy(stu[number_new].name,name_new);
			stu[number_new].height=height_new;
			stu[number_new].weight=weight_new;
		}
		else
		{
			if(height_new>stu[number_new].height)
			{
				stu[number_new].number=number_new;
				strcpy(stu[number_new].name,name_new); stu[number_new].height=height_new; stu[number_new].weight=weight_new; }}}//printf("\n");
	for(i=0; i<k; i++) {if(stu[index[i]].number! =- 1)
		{
		printf("%06d %s %d %d\n",stu[index[i]].number,stu[index[i]].name,stu[index[i]].height,stu[index[i]].weight);
		stu[index[i]].number=- 1; }}//printf("%06d %s %d %d\n",stu[0].number,stu[0].name,stu[0].height,stu[0].weight);
	return 0;
}
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120. Stepped electricity price (15 points)

In order to encourage residents to save electricity, the power company of a province implemented the “stepped electricity price”. The electricity price of one household with one meter is divided into two “steps” : the price of electricity within 50 KWH (including 50 KWH) is 0.53 yuan/KWH; If the electricity consumption exceeds 50 KWH, the price will be increased by 0.05 yuan/KWH. Please write a program to calculate the electricity charge.

Input format: Enter the monthly electricity consumption of a user in one line (unit: KWH).

Output format: Output the electricity fee payable by the user (yuan) in one line, with two decimal digits reserved, for example, “cost = value of electricity fee payable”. If the power consumption is less than 0, “Invalid Value!” is displayed. .

Example Value 1: 10 Example value 1: cost = 5.30 Example value 2: 100 Example value 2: cost = 55.50

#include<stdio.h>
int main(a)
{
	double n;
	scanf("%lf",&n);
	if(n<0)
		printf("Invalid Value! \n");
	else if(n<=50)
		printf("cost = %.2lf\n",n*0.53);
	else if(n>50)
		printf("cost = %.2lf\n", (50*0.53+(n- 50) *0.05));
	return 0;
}
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Creation is not easy, useful please point a thumbs-up, thank you!

Original is not easy, useful please point a praise, thank you!

121. Speedway penalties (15 points)

#include<stdio.h>
#include<math.h>
int main(a)
{
	float chesu,xiansu;
	scanf("%f%f",&chesu,&xiansu);
	// If (chesu
	if(chesu-xiansu<xiansu*0.1)
		printf("OK\n");
	else if(chesu<xiansu*1.5)
		printf("Exceed %1.f%%. Ticket 200\n",((chesu-xiansu)/xiansu)*100);
	else
		printf("Exceed %1.f%%. License Revoked\n",((chesu-xiansu)/xiansu)*100);
	return 0;
}
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122, catch the mouse ah ~ lost or earned? (20 points)

A mouse disaster, now reward catch mice, every catch a reward of 10 yuan, so began to fight with the mouse wits: every day in the corner can choose the following three operations: place a mouse trap with a piece of cheese (T), or place a piece of cheese ©, or nothing (X). The traps are reusable, regardless of cost, and each piece of cheese costs 3 yuan. What about the smart mouse? They might send a rat to the corner every day to see what’s in the corner:

If there was nothing (X), the mice were Unhappy and had a period of unhappiness lasting up to a day (the next day). Rats are not sent out during periods of displeasure. When the period of unhappiness was over, rats were sent. With a trap, the mouse is enticed to eat the cheese and killed, and the mice have two days of grief (the second and third days). Do not send mice during the period of sorrow. When the sad period was over, rats were sent out. In this case, catching a mouse would earn you $10, but it would also cost you a piece of cheese. Note that if a trap is placed on a given day and the mouse does not appear, no cheese is consumed. If there is cheese ©, the mouse will be Happy to eat the cheese (Happy! , there will be as long as two days (the second and third days) of excitement. During the period of excitement, even if there is added unhappiness or sadness, the rats are bound to be sent. In this case, no mouse was caught and a piece of cheese was used. Note that if the cheese is placed on a given day and the mouse does not appear, the cheese can be used again without cost. Now you are given a sequence of operations for several consecutive days, and it is known that rats will be sent on the first day. Please judge the state of rats every day and calculate profits.

Input format: Input gives contiguous C or T or X strings of up to 70 characters on a single line, ending with $. Each character in the string represents that day’s operation (i.e. X: nothing; T: Rat traps; C: Cheese). The problem guarantees at least one day of operation input.

Output format: Requires a continuous string output on the first line, corresponding to the input, giving the status of the mouse:

! D means that the mice were killed and U means that the mice were not found

• The second line should output an integer for profit. (If there is a loss, it is negative)

Example 1: TXXXXC$Example 1: D – U-! 4 Input Example 2: CTTCCX$Output Example 2:! DD – U 11

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main(a)
{
	int money=0;
	int i,j;
	int flag=0,index=0,kong=1;
	int pai=1;
	char s[72];
	gets(s);
	//int len = strlen(s);
	//s[len-1]='\0';
	for(i=0; s[i]! ='$'; i++) {if(flag<=0 && index==0)
		{
			if(s[i]=='X')
			{	printf("U");	index=- 1; pai-=1;	}
			else if(s[i]=='T')
			{	
				if(pai>=1)
				{
					printf("D");	index=2 -;
					money+=7;
					pai-=2; }}else if(s[i]=='C')
			{	
				if(pai>=1)
				{
					printf("!");	flag=2;
					money-=3; }}}else if(flag>0 && kong>=0)
		{
			if(s[i]=='X')
			{	printf("U");		}
			else if(s[i]=='T')
			{	printf("D"); kong--; money+=7;	}
			else if(s[i]=='C')
			{	printf("!"); money-=3; } flag--; }if(kong==- 1)
		{	kong=0;	index=2 -;	}
		while(index<0 && s[i+1]! ='$')
		{
			if(index! =0)
			{
				printf("-");
				++index;
			}
				i=i+1;
				if(pai<1)
				pai++;
		}
		if(pai<1)
		pai++;
	}
	printf("\n%d\n",money);
	return 0;
}
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123. Find a ball with a scale (10 points)

#include<stdio.h>
int max;
void swap(int a,int b,int c)
{
	max=a;
	if(max! =b&&max==c) { max=b; }else if(max!=c&&max==b)
	{	max=c;	}
}
int main(a)
{
	int a,b,c;
	scanf("%d%d%d",&a,&b,&c);
	swap(a,b,c);
	if(max==a)
		printf("A\n");
	else if(max==b)
		printf("B\n");
	else 
		printf("C\n");
	return 0;
}
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124, Find the set of integers that meet the given conditions (15 marks)

Given A positive integer A not exceeding 6, consider four consecutive numbers starting with A. Print all 3-digit numbers that are composed of them without duplicates.

Input format: Input gives an A on A line.

Output format: Output the 3-digit numbers that meet the conditions. Each line contains six integers from the smallest to the largest. Integers are separated by Spaces, but there can be no extra Spaces at the end of the line.

Example Input: 2 Example output: 234 235 243 245 253 254 324 325 342 345 352 354 423 425 432 435 452 453 523 524 532 534 542 543

#include<stdio.h>
int main(a)
{
	int i,j,k;
	int huan_hang=0;	// Control whitespace and line breaks
	int t,n;
	scanf("%d",&t);
	n=t;
	for(i=n; i<=n+3; i++) {for(j=n; j<=n+3; j++) {if(i! =j) {for(k=n; k<=n+3; k++) {if(j! =k&&i! =k) { huan_hang++;// Control whitespace and line breaks
						if(huan_hang<=5)
							printf("%d%d%d ",i,j,k);
						else
							printf("%d%d%d",i,j,k);
						if(huan_hang%6= =0)
						{
							printf("\n");
							huan_hang=0;
						}
					}
				}
			}
		}
	}
	return 0;
}
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125. Count 24 points (25 points) with playing cards

#include<stdio.h>
#define N 4

// The following function is used to assign four different symbols
static int result(int arr[],char op[]);

// The function that computes the result according to different symbols of different priorities
static double Result(double x, double y, int op);

// The following five functions assign different priorities
//((A op B) op C) op D
static double result1(double num1,double num2,double num3,double num4,int op1,int op2,int op3);
//(A op (B op C)) op D
static double result2(double num1,double num2,double num3,double num4,int op1,int op2,int op3);
//A op (B op (C op D))
static double result3(double num1,double num2,double num3,double num4,int op1,int op2,int op3);
//A op ((B op C) op D)
static double result4(double num1,double num2,double num3,double num4,int op1,int op2,int op3);
//(A op B) op (C op D)
static double result5(double num1,double num2,double num3,double num4,int op1,int op2,int op3);

int main(a)
{			
		int num[N],i,j,k,t;
		char op[5] = {'right'.'+'.The '-'.The '*'.'/'};	// Store all four operators in an array
		for(i=0; i<N; i++)// Enter four numbers at will
			scanf("%d",&num[i]);
		
		// Combine the four digits into different digits
		for(i=0; i<N ; i++) {for(j=0; j<N; j++) {if(i==j) continue;
				for(k=0; k<N; k++) {if(i==k || j==k) continue;
					for(t=0; t<N; t++) {if(i==t || j==t || k==t) continue;
						int zuhe_num[N] = {num[i],num[j],num[k],num[t]};	/* The four numbers appear in different order, because of the different operators, so the calculation results are naturally different */
						int swap =	result(zuhe_num,op);	// If there is a set of data that is valid for 24, then it ends
						if(swap)// If there is a solution, end the program directly
						{
							return 0;
						}
						
					}
				}
			}	
		}
		/ / there is no solution
		printf("1");	
}
static int result(int zuhe_num[],char op[])	// The following function is used to assign four different symbols
{
		double sum =0;
		// enumerate all combinations of operations
		for(int i=1; i<=N; i++) {for(int j=1; j<=N; j++) {for(int k=1; k<=N; k++) { sum = result1(zuhe_num[0],zuhe_num[1],zuhe_num[2],zuhe_num[3],i,j,k);
					if(sum==24)		// If the conditions are met, the output is direct, and return 1 to end the program, and so on
					{
						//((A op B) op C) op D
						printf("((%d%c%d)%c%d)%c%d\n",zuhe_num[0],op[i],zuhe_num[1],op[j],zuhe_num[2],op[k],zuhe_num[3]);
						return 1;
					}
					sum = result2(zuhe_num[0],zuhe_num[1],zuhe_num[2],zuhe_num[3],i,j,k);
					if(sum==24)
					{
						//(A op (B op C)) op D
						printf("(%d%c(%d%c%d))%c%d\n",zuhe_num[0],op[i],zuhe_num[1],op[j],zuhe_num[2],op[k],zuhe_num[3]);
						return 1;
					}
					sum = result3(zuhe_num[0],zuhe_num[1],zuhe_num[2],zuhe_num[3],i,j,k);
					if(sum==24)
					{
						//A op (B op (C op D))
						printf("%d%c(%d%c(%d%c%d))\n",zuhe_num[0],op[i],zuhe_num[1],op[j],zuhe_num[2],op[k],zuhe_num[3]);
						return 1;
					}
					sum = result4(zuhe_num[0],zuhe_num[1],zuhe_num[2],zuhe_num[3],i,j,k);
					if(sum==24)
					{
						//A op ((B op C) op D)
						printf("%d%c((%d%c%d)%c%d)\n",zuhe_num[0],op[i],zuhe_num[1],op[j],zuhe_num[2],op[k],zuhe_num[3]);
						return 1;
					}
					sum = result5(zuhe_num[0],zuhe_num[1],zuhe_num[2],zuhe_num[3],i,j,k);
					if(sum==24)
					{
						//(A op B) op (C op D)
						printf("(%d%c%d)%c(%d%c%d)\n",zuhe_num[0],op[i],zuhe_num[1],op[j],zuhe_num[2],op[k],zuhe_num[3]);
						return 1; }}}}// If there is no solution
		return 0;
}
static double Result(double num1, double num2, int op)		// The function that computes the result according to different symbols of different priorities
{
	if(op==1)		// Make a judgment based on the passed symbol, and perform the corresponding calculation
		return num1+num2;
	else if(op==2)
		return num1-num2;
	else if(op==3)
		return num1*num2;
	else if(op==4)
		return num1/num2;
	else
		return 0;
}
// The following five functions assign different priorities
static double result1(double num1,double num2,double num3,double num4,int op1,int op2,int op3)	
{														//((A op B) op C) op D
		double r1,r2,r3;
		r1 = Result(num1,num2,op1);
		r2 = Result(r1,num3,op2);
		r3 = Result(r2,num4,op3);
		return r3;
}
static double result2(double num1,double num2,double num3,double num4,int op1,int op2,int op3)
{														//(A op (B op C)) op D
		double r1,r2,r3;
		r1 = Result(num2,num3,op2);
		r2 = Result(num1,r1,op1);
		r3 = Result(r2,num4,op3);
		return r3;
}
static double result3(double num1,double num2,double num3,double num4,int op1,int op2,int op3)
{														//A op (B op (C op D))
		double r1,r2,r3;
		r1 = Result(num3,num4,op3);
		r2 = Result(num2,r1,op2);
		r3 = Result(num1,r2,op1);
		return r3;	
}
static double result4(double num1,double num2,double num3,double num4,int op1,int op2,int op3)
{														//A op ((B op C) op D)
		double r1,r2,r3;
		r1 = Result(num2,num3,op2);
		r2 = Result(r1,num4,op3);
		r3 = Result(num1,r2,op1);
		return r3;
}

static double result5(double num1,double num2,double num3,double num4,int op1,int op2,int op3)
{														//(A op B) op (C op D)
		double r1,r2,r3;
		r1 = Result(num1,num2,op1);
		r2 = Result(num3,num4,op3);
		r3 = Result(r1,r2,op2);
		return r3;
}
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126, Approximate PI (15 marks)

In this case, a program is required to approximate PI according to the following formula until the last term is less than the given precision EPS.

Input format: The input gives the precision EPS in one line. The input can be read using the following statements:

scanf(“%le”, &eps); Output format: In a single line, print an approximation of π (with 5 decimal places reserved) in the following format:

PI = Approximate Value Input example: 1E-5 Output example: PI = 3.14158

#include<stdio.h>
double fenzi(int n);
double fenmu(int n);
int main(a)
{
	int n=0;
	double sum=0;
	double eps;
	scanf("%le",&eps);
	do
	{
		sum+=fenzi(n)/fenmu(n);
			  		n++;
	}while((fenzi(n)/fenmu(n))>=eps);
	if(sum>=eps)
	sum+=fenzi(n)/fenmu(n);
	printf("PI = %.5lf", sum * 2);
	return 0;
}
double fenzi(int n)
{
	if(n==0)
	return 1;
	
	return n*fenzi(n- 1);
}
double fenmu(int n)
{
	if(n==0)
	return 1;
	
	return (n*2+1) * fenmu(n- 1);
}
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127. A Simple Calculator for Two Numbers (10 points)

#include<stdio.h>
int main(int argc,char* argv[])
{
	int num1,num2;
	char c;
	scanf("%d %c %d",&num1,&c,&num2);
	switch(c)
	{
		case '+':printf("%d\n",num1+num2);break;
		case The '-':printf("%d\n",num1-num2);break;
		case The '*':printf("%d\n",num1*num2);break;
		case '/':printf("%d\n",num1/num2);break;
		case The '%':printf("%d\n",num1%num2);break;
		default :printf("ERROR\n");
	}
	return 0;
}
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