Uva-1210 Sum of Consecutive Prime Numbers

題目鏈接：Sum of Consecutive Prime Numbers
題目大意：輸入一個數，求有多少種連續的素數和可以組成這個數。
解題思路：用篩法對素數打個表，然後用尺取法就行了。

代碼如下：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const int inf  = 0x3f3f3f3f;
const int maxn = 2e4 + 15;

int prime[maxn];
int flag[maxn];

void Sieve() {
    memset(flag, 0, sizeof flag);
    int t = 0;
    for(int i = 2; i <= 20000; i++) {
        if(!flag[i]) prime[t++] = i;

        for(int j = 0; j < t && i * prime[j] <= 20000; j++) {
            flag[i * prime[j]] = 1;
            if(i % prime[j] == 0) break;
        }
    }
    return;
}

int main(){
#ifdef NEKO
    freopen("Nya.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    int n; Sieve();
    while(cin >> n && n) {
        int ans = 0;
        int st = 0, ed = 0, sum = prime[0];
        while(1) {
            if(prime[ed] > n) break;
            if(sum == n) {
                ans++;
                sum -= prime[st++];
            }
            else if(sum < n) sum += prime[++ed];
            else sum -= prime[st++];    
        }
        cout << ans << endl;
    }
    return 0;
}