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
1 //素数升级 2 //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。? 3 //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”。? 4 //如N = 5时输出:? 5 //5 = 1 + 1 + 3?????NO? 6 //5 = 1 + 2 + 2?????YES? 7 //5 = 1 + 3 + 1?????NO? 8 //5 = 2 + 1 + 2?????YES? 9 //5 = 2 + 2 + 1?????YES? 10 //5 = 3 + 1 + 1?????NO? 11 12 #include <stdio.h> 13 #include <stdlib.h> 14 #include <math.h> 15 16 int judgePrim(int n); 17 int main() 18 { 19 int sum = 0; 20 int m = 0; 21 int i; 22 printf("请输入一个数"); 23 scanf_s("%d", &m); 24 for (i = 1; i < m; i++) 25 { 26 for (int j = 1; j < m; j++) 27 { 28 int k = m - i - j; 29 if (i + j + k == m&&k>0) 30 { 31 sum = judgePrim(i) + judgePrim(j) + judgePrim(k); 32 printf("%d+%d+%d=%d\t", i, j, k, m); 33 if (sum >= 2) 34 { 35 printf("YES\n"); 36 } 37 else 38 { 39 printf("no\n"); 40 } 41 } 42 43 } 44 } 45 system("pause"); 46 return 1; 47 } 48 49 int judgePrim(int n) 50 { 51 if (n == 1) 52 { 53 return 0; 54 } 55 else if (n == 2 || n == 3) 56 { 57 return 1; 58 } 59 else 60 { 61 for (int i = 2; i <= (int)sqrt((double)n); i++) 62 { 63 if (n%i == 0) 64 return 0; 65 } 66 } 67 return 1; 68 }