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;

}