【POJ2739】Sum of Consecutive Prime Numbers

简单的素数打表，然后枚举。开始没注意n读到0结束，TLE了回。。下次再认真点。A过后讨论里面有个暴力打表过的，给跪了！

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <complex.h>
#include <cmath>
using namespace std;

const int N = 10005;

bool is[N]; int prm[N];
int getprm (int n) {
    int i, j, k = 0;
    int s, e = (int)(sqrt(0.0 + n) + 1);
    memset (is, 1, sizeof (is));
    prm[k ++] = 2; is[0] = is[1] = 0;
    for (i = 4; i < n; i += 2) is[i] = 0;
    for (i = 3; i < e; i += 2)
        if (is[i]){
            prm[k ++] = i;
            for (s = i * 2, j = i * i; j < n; j += s) {
                is[j] = 0;
        }
    }
    for (; i < n; i += 2) {
        if (is[i]) prm[k ++] = i;
    }
    return k;
}

int main () {
    getprm(10000);
    /*for (int i = 0 ; i < 100; ++ i) {
        cout << i << ":  ";
        cout << prm[i] << endl;
    }*/
    // 1229个素数 0-1w
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    while (~scanf ("%d", &n)) {
        if (n == 0) break;
        int sum, cnt = 0;
        for (int i = 0 ; prm[i] <= n; ++ i) {
            sum = 0;
            for (int j = i; prm[j] <= n; ++ j) {
                sum += prm[j];
                if (sum < n) {
                    continue;
                } else if (sum == n) {
                    cnt ++;
                    break;
                } else {
                    break;
                }
            }
        }
        printf ("%d
", cnt);
    }
    return 0;
}