素数应用

输入一个数 分解尾三个数想乘 如果三个数其中有两个为素数->yes 否则no

//素数升级
//功能：??已知N是正整数，?它可拆写成三个正整数N1、N2和N3之和的形式N=N1+N2+N3。?
//请编程序打印出满足上式的全部组合，并当N1、N2和N3中至少有两个是素数时打印“YES”否则打印“NO”。?
//如N = 5时输出：?
//5 = 1 + 1 + 3?????NO?
//5 = 1 + 2 + 2?????YES?
//5 = 1 + 3 + 1?????NO?
//5 = 2 + 1 + 2?????YES?
//5 = 2 + 2 + 1?????YES?
//5 = 3 + 1 + 1?????NO?

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

int judgePrim(int n);
int main()
{
    int sum = 0;
    int m = 0;
    int i;
    printf("请输入一个数");
    scanf_s("%d", &m);
    for (i = 1; i < m; i++)
    {
        for (int j = 1; j < m; j++)
        {
            int k = m - i - j;
            if (i + j + k == m&&k>0)
            {
                sum = judgePrim(i) + judgePrim(j) + judgePrim(k);
                printf("%d+%d+%d=%d	", i, j, k, m);
                if (sum >= 2)
                {
                    printf("YES
");
                }
                else
                {
                    printf("no
");
                }
            }
            
        }
    }
    system("pause");
    return 1;
}

int judgePrim(int n)
{
    if (n == 1)
    {
        return 0;
    }
    else if (n == 2 || n == 3)
    {
        return 1;
    }
    else
    {
        for (int i = 2; i <= (int)sqrt((double)n); i++)
        {
            if (n%i == 0)
                return 0;
        }
    }
    return 1;
}