Middle M2018 Spring C Introduction and Advanced Exercise Set Function Question 6-2 Use Function to Verify Goldbach's Conjecture (20 points)

This problem requires the realization of a simple function for judging prime numbers, and using this function to verify Goldbach's conjecture: any even number not less than 6 can be expressed as the sum of two odd prime numbers. A prime number is a positive integer that is only divisible by 1 and itself. Note: 1 is not a prime number, 2 is a prime number.

Function interface definition:

int prime( int p );
void Goldbach( int n );

The function returns 1 when primethe parameter passed in by the user pis a prime number, otherwise it returns 0; the function outputsGoldbach the prime number decomposition in the format " n= p + q" , where p ≤ q are all prime numbers. And because such a decomposition is not unique (for example, 24 can be decomposed into 5+19, and it can also be decomposed into 7+17), it is required to output the solution with the smallest p among all the solutions.n

Example of the referee test procedure:

#include <stdio.h>
#include <math.h>

int prime( int p );
void Goldbach( int n );

int main()
{
    int m, n, i, cnt;

    scanf("%d %d", &m, &n);
    if ( prime(m) != 0 ) printf("%d is a prime number\n", m);
    if ( m < 6 ) m = 6;
    if ( m%2 ) m++;
    cnt = 0;
    for( i=m; i<=n; i+=2 ) {
        Goldbach(i);
        cnt++;
        if ( cnt%5 ) printf(", ");
        else printf("\n");
    }

    return 0;
}

/* 你的代码将被嵌在这里 */

Input sample:

89 100

Sample output:

89 is a prime number
90=7+83, 92=3+89, 94=5+89, 96=7+89, 98=19+79
100=3+97,

 

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 1 int prime(int p)
 2 {
 3     int i;
 4     int flag;    /*素数1，非素数0*/
 5     if (p <= 1)
 6         flag = 0;
 7     else if (p == 2 || p == 3)
 8         flag = 1;
 9     else
10     {
11         for (i = 2; i <= sqrt(p); i++)
12             if (p%i == 0)
13             {
14                 flag = 0;
15                 break;
16             }
17         if (i == (int)sqrt(p) + 1)
18             flag = 1;
19     }
20     return flag;
21 }
22 
23 void Goldbach(int n)
24 {
25     int p, q;
26     if (n >= 6 && n % 2 == 0)
27         for (p = 2; p <= n / 2; p++)
28         {
29             q = n - p;
30             if (prime(p) && prime(q))
31             {
32                 printf("%d=%d+%d", n, p, q);
33                 break;
34             }
35         }
36 }