poj 2739 Sum of Consecutive Prime Numbers

悲剧,如此水题，竟搞了许久。唉，只因题目没看清。 

        准确的说：应该是准备工作没做好就开始敲代码了！！ 

        下次一定要想的很清楚了再code ing ! 

     

//144K	63MS
#include <stdio.h>
int prim[1300], n, p;
int sum[1300];
int ans;

void prime()
{
    int i, j, flag;
    n = 0;
    prim[n++] =2;
    for(i=3;i<10000; i+=2) {
        for(j=0,flag = 1; prim[j]*prim[j]<=i; j++)
        if(i %prim[j] == 0) {flag= 0; break;}
        if(flag) prim[n++] = i;
    }
    sum[0]=0;
    for(i=1; i<=n;i++) sum[i] =sum[i-1]+prim[i-1];
}

int main()
{
    int x, i, j;
    prime();
    while(scanf("%d",&x),x) {
        p = n-1; while( prim[p]> x)p--;
        p++;
        ans = 0;
        for(i=p; i>=1; i--)
           for(j=0;j<=i-1;j++)
               if(sum[i] - sum[j] == x) {
                    ans++;
                    //for(int k = i-1; k>=j;k--) printf("%d  ",prim[k]);
                    // printf("\n");
               }
        printf("%d\n", ans);
    }
    return 0;
}