“This is the 7th day of my participation in the Gwen Challenge in November. Check out the details: The Last Gwen Challenge in 2021.”
Recently, I want to review C language, so I will update an article about C language in nuggets every day! Freshmen who are just learning C, and those who want to review C /C++, don’t miss it! Solid foundation, slow down is fast!
10. Common divisor
Method 1: Violent solution
Greatest common divisor: may be the smallest of the two,
So let’s divide by the smallest of the two numbers and keep subtracting by 1
/ / method
int main(a)
{
int m = 0;
int n = 0;
int min = 0;
scanf("%d %d", &m, &n);
min = m < n ? m : n; // Find the smallest of the two numbers
while (min)
{
if (m % min == 0 && n % min == 0)
{
printf("The greatest common divisor of %d and %d is :%d\n",m, n, min);
break;
}
min--;
}
return 0;
}
Copy the codeMethod 2: Toss and turn division
int main(a)
{
int m = 0;
int n = 0;
scanf("%d %d", &m,&n);
int r = 0;
// Toss and turn division
while (r = m %n)
{
m = n; // Assign n to m
n = r; // Assign the remainder to n
}
printf("%d\n", n);
return 0;
}
Copy the code11. Find the least Common multiple
Method 1: Force solution
Smallest common multiple: Probably the largest of the two
I just have to go +1 down
int main(a)
{
int n = 0;
int m = 0;
scanf("%d %d", &n, &m);
int max = n > m ? n : m; // The largest of the two
while (1)
{
if (max % n == 0 && max % m == 0)
{
printf("%d %d least common multiple: %d\n", n, m, max);
break;
}
max++;
}
return 0;
}
Copy the codeMethod 2: least common multiple = product of two numbers/greatest common divisor
// Find the greatest common divisor
int MaxCommonDivisor(int m,int n)
{
int r = 0;
// Toss and turn division
while (r = m % n)
{
m = n; // Assign n to m
n = r; // Assign the remainder to n
}
return n;
}
int main(a)
{
int m = 0;
int n = 0;
scanf("%d %d", &m, &n);
int ret = MaxCommonDivisor(m, n);
printf("The least common multiple of %d %d is :%d, and the greatest common divisor is :%d \n", m, n, m * n / ret,ret);
return 0;
}
Copy the code12. 答 案 : Identify leap years between 1000 and 2000
Leap year:1.Can it be4Divisible, cannot be divided by100Aliquot &&2.Can it be400Aliquot | |Copy the codeint main(a)
{
int year = 0;
int count = 0; / / count
for (year = 1000; year <= 2000; year++)
{
if (year % 4= =0 && year % 100! =0 || year % 400= =0)
{
count++;
printf("%d ", year); }}printf("\n1000 to 2000, the number of leap years is :%d\n",count);
return 0;
}
//count:243
Copy the code// Can also be written as
int main(a)
{
int count = 0;
int year = 0;
for(year = 1000; year <=2000; year++) {if (year % 4= =0 && year % 100! =0)
{
count++;
printf("%d ", year);
}
else if (year % 400= =0)
{
count++;
printf("%d ",year); }}printf("\n1000 to 2000, the number of leap years is :%d\n", count);
return 0;
}
//count :243
Copy the codeError writing
// Error
int main(a)
{
int count = 0;
int year = 0;
for(year = 1000; year <=2000; year++) {if (year % 4= =0 )
{
if(year % 100! =0)
{
count++;
printf("%d ", year); }}else if (year % 400= =0)
{
count++;
printf("%d ",year); }}printf("\n1000 to 2000, the number of leap years is :%d\n", count);
return 0;
}
//count :240
// There are three years missing: 1200 1600 2000
If (y%4==0) else if (y%4==0) else if (y%4==0
Copy the code13. Programming problem: Print prime numbers within 100-200
Method 1: trial division
Prime number: not divisible by any number except 1 and itself
Determine if I is prime: Divide I by the number from 2 to i-1
// write method 1:
// Returns 1 for primes, 0 for non-primes
int is_prime(int n)
{
int i = 0;
for (i = 2; i < n; i++)
{
if (n %i == 0)
{
return 0; }}return 1;
}
/ / write 2
// Use a Boolean type -> reference to the header file stdbool
bool is_prime(int n)
{
int i = 0;
for (i = 2; i < n; i++)
{
if (n %i == 0)
{
return false; }}return true;
}
int main(a)
{
int i = 0;
int count = 0;
for (i = 100; i <= 200; i++)
{
if (is_prime(i) == 1)
{
count++;
printf("%d ", i); }}printf("The number of primes between \n100-200 is %d \n", count);
return 0;
}
// There are 21 primes between 100 and 200
Copy the codeMethod 2: Square it
If m is equal to a times b, at least one number of a and B is <=sqrt(m) such as:16 = 2*8 = 4*4As long as you have one number that divisible m you don't have to judge the other one so you just try to divide to the square root of msqrt() -> Square function references header file math.hCopy the code
int main(a)
{
int i = 0;
int count = 0;
for (i = 100; i <= 200; i++)
{
int flag = 1; // Assume a prime number
int j = 0;
// Try to divide to square I
for (j = 2; j <=sqrt(i); j++)
{
if (i % j == 0)
{
// If divisible, it is not prime
flag = 0;
break; }}if (flag == 1)
{
printf("%d ", i); count++; }}printf("The number of primes between \n100-200 is %d \n", count);
return 0;
}
Copy the codeMethod 3: optimization
Prime numbers only occur in odd numbers, so you can start with 101. Every time + = 2
int main(a)
{
int i = 0;
int count = 0;
for (i = 101; i <= 200; i+=2)
{
int flag = 1; // Assume a prime number
int j = 0;
// Try to divide to square I
for (j = 2; j <=sqrt(i); j++)
{
if (i % j == 0)
{
// If divisible, it is not prime
flag = 0;
break; }}if (flag == 1)
{
printf("%d ", i); count++; }}printf("The number of primes between \n100-200 is %d \n", count);
return 0;
}
Copy the code14. Program problem: find the minimum and maximum value of 10 numbers
// Error program
// Reason: set min and Max to 0, if we input negative values, the maximum value will be wrong
int main(a)
{
int arr[10] = { 0 };
int i = 0;
// Enter elements for the array
for (i = 0; i < 10; i++)
{
scanf("%d", &arr[i]);
}
int max = 0;
int min = 0;
for (int i = 0; i < 10; i++)
{
if (arr[i] > max)
{
max = arr[i];
}
if(arr[i] < min) { min = arr[i]; }}printf("min=%d max=%d", min, max);
return 0;
}
Copy the code/ / positive solutions
// Assume that an element in the array is the minimum. The maximum
int main(a)
{
int arr[10] = { 0 };
int i = 0;
// Enter elements for the array
for (i = 0; i < 10; i++)
{
scanf("%d", &arr[i]);
}
int max = arr[0];
int min = arr[0];
for (int i = 0; i < 10; i++)
{
if (arr[i] > max)
{
max = arr[i];
}
if(arr[i] < min) { min = arr[i]; }}printf("min=%d max=%d", min, max);
return 0;
}
Copy the code