Problem 6-5 using the function to verify Goldbach's Conjecture

 1 #include <stdio.h>
 2 #include <math.h>
 3 
 4 int prime(int p);
 5 void Goldbach(int n);
 6 
 7 int main()
 8 {
 9     int m, n, i, cnt;
10 
11     scanf_s("%d %d", &m, &n);
12     if (prime(m) != 0) printf("%d is a prime number\n", m);
13     if (m < 6) m = 6;
 14      IF (m% 2 ) m ++ ;
 15      CNT = 0 ;
 16      for (I = m; I <= n-; I + = 2 ) {
 . 17          GOLDBACH (I);
 18 is          CNT ++ ;
 . 19          IF (CNT% . 5 ) the printf ( " , " );
 20 is          the else the printf ( " \ n- " );
 21 is      }
 22 is  
23 is      return  0 ;
 24  }
 25  
26 is  / * your code here will be embedded in* / 
27  int Prime ( int P)
 28  {
 29      IF (P == . 1 )
 30      {
 31 is          return  0 ;
 32      }
 33 is  
34 is      int is_prime = . 1 ;         // prime number flag, if a prime number, is. 1 
35      for ( int = I 2 ; I <P; I ++ )
 36      {
 37 [          IF (% I P == 0 )
 38 is          {
 39              is_prime = 0 ;
40             break;
41         }
42     }
43     return is_prime;
44 }
45 
46 void Goldbach(int n)
47 {
48     for (int i = 3; i <= n/2; i++)
49     {
50         if (prime(i) && prime(n - i))        //都为素数
51         {
52             printf("%d=%d+%d", n, i, n - i);
53             break;
54         }
55     }
56 }