Judge the ACing of the prime number and the acing of the prime number.

Judge the ACing of the prime number and the acing of the prime number.
1. Determine whether positive integer m is a prime number.

Int I, m; printf ("Enter a number:"); scanf ("% d", & m); for (I = 2; I <= m/2; I ++) if (m % I = 0) break; // if m can be divisible by an I, m is not a prime number, end the cycle if (I> m/2 & m! = 1) // If the loop ends normally, m cannot be divisible by any one I ("% d is a prime number! \ N ", m); else printf (" No! \ N "); return 0;

 

 

2. Use nested loops to obtain all prime numbers within 100

Int count, I, m, n; count = 0; // records the number of prime numbers, used to control the output format for (m = 2; m <= 100; m ++) {n = sqrt (m); for (I = 2; I <= n; I ++) if (m % I = 0) break; if (I> n) {// If m is a prime number printf ("% 6d", m); // output m count ++; // accumulate the number of output prime numbers if (count % 10 = 0) // if count is a multiple of 10, wrap printf ("\ n ");}} printf ("\ n"); return 0;

 

 

3. Use the function to calculate all prime numbers within 100

# Include <stdio. h> # include <math. h> int main (void) {int count, m; int prime (int m); // function declaration count = 0; // records the number of prime numbers, used to control the output format for (m = 2; m <= 100; m ++) {if (prime (m )! = 0) {// call prime (m) to determine whether m is a prime number printf ("% 6d", m); // output m count ++; // accumulate the number of output prime numbers if (count % 10 = 0) // if count is a multiple of 10, wrap printf ("\ n ");}} printf ("\ n");}/* defines the function for determining prime numbers. If m is a prime number, 1 is returned. Otherwise, 0 */int prime (int m) {int I is returned, n; if (m = 1) return 0; // 1 is not a prime number, returns 0 n = sqrt (m); for (I = 2; I <= n; I ++) {if (m % I = 0) {// return 0 if m is not a prime number; // return 0} return 1 ;}}