POJ 2739 Sum of Consecutive Prime Numbers

Description

Input

Output

Sample Input

2
3
17
41
20
666
12
53
0

Sample Output

1
1
2
3
0
0
1
2

真心水题！！！

AC代码如下：

///打表水过   16MS  164K

#include<iostream>
#include<cstdio>
#include<cstring>
#define M 10005
using namespace std;

int prime[M];
int tt=0;

void isp()
{
    int i,j;
    prime[tt++]=2;
    for(i=3;i<M;i+=2)
    {
        int flag=1;
        for(j=2;j*j<=i;j++)
        {
            if(i%j==0)
            {flag=0;break;}
        }
        if(flag)
            prime[tt++]=i;
    }
    return;
}

int main()
{
    isp();
    int n,i,j;
    int sum;
    while(~scanf("%d",&n)&&n)
    {
       int ans=0;
       for(i=0;i<tt;i++)
       {
           if(prime[i]>n)
            break;
            sum=0;
           for(j=i;j<tt;j++)
           {
               sum+=prime[j];
               if(sum==n)
                ans++;
               if(sum>n)
                break;
           }
       }
       printf("%d\n",ans);
    }
    return 0;
}

POJ 2739 Sum of Consecutive Prime Numbers

原文：http://blog.csdn.net/hanhai768/article/details/38020261