poj 2739 Sum of Consecutive Prime Numbers

题意:一个数可由连续素数和构成.例如:41 =2+3+5+7+11+13, 41=11+13+17, and 41=41.有三种构成方式.对于20,20 = 7 + 13(7,13不连续)故不是,同样20 = 3+5+5+7,有重复也不是.

要求输出方式数.

附代码:

#include <iostream>#include <cstdio>using namespace std;bool p[10009];int prim[5000];int main(void){ int i,j,k; p[2] = 1; for(i=3; i<10000; i+=2) p[i] = 1; for(i=2; i<=10000/2; i++) if(p[i]) for(j=2; j<=10000/i; j++) p[i*j] = 0; for(k=0,i=2; i<10000; i++) if(p[i]) prim[k++] = i; int n; while(cin >> n && n) { int sum,count=0; for(i=0; prim[i]<=n; i++) { sum = 0; for(j=i; prim[j]<=n; j++) { if(sum+prim[j]<=n) { sum += prim[j]; if(sum==n) { count++; break; } } } } cout << count << endl; } return 0;}