POJ-2739 Sum of Consecutive Prime Numbers

#include <iostream>#include <stdio.h>using namespace std;int main(){    int n;    int a[10000];    int k = 0;    for(int m = 2;m <= 10000;m++)    {        int t = 1;        for(int i = 2;i*i <= m;i ++)        {            if (m % i == 0)            {                t = 0;                break;            }        }        if(t)        {            a[k] = m;            k ++;        }    }    while(scanf("%d",&n)!=EOF&&n)    {        int x = 0;        for(int i=0;a[i]<=n;i++)        {            int sum=0;            for(int j=i;a[j]<=n;j++)            {                sum+=a[j];                if(sum==n)                {                    x++;                    break;                }                else if(sum>n)                    break;            }        }       printf("%d\n",x);    }    return 0;}

/*应该算是水题 枚举10000以内的素数存入数组再进行sum筛选。一开始把sum位置放错导致错误。*/