（数学+尺取法）2739

原题链接：http://poj.org/problem?id=2739

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题意：问一个数有几种方法用连续的素数和表示。

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分析：其实就是很简单，先打表，然后对prime数组跑一波尺取法，如果==n就ans++。

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代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <set>
#include <vector>
#include <queue>
#include <map>
#include <list>
#include <bitset>
#include <string>
#include <cctype>
#include <cstdlib>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;


const int inf = 1<<30;
const ll lnf = 1ll<<60;

//--------------------------

const int maxn=10010;
int prime[maxn];
bool vis[maxn];

int Euler_prime() {
    memset(vis, true, sizeof(vis));
    int tot = 0;
    for (int i = 2; i < maxn; i++) {
        if (vis[i]) prime[tot++] = i;
        for (int j = 0; j < tot&&prime[j] * i < maxn; j++) {
            vis[i*prime[j]] = false;
            if (i%prime[j] == 0) break;
        }
    }
    return tot;
}


void solve() {
    int n;
    int pn=Euler_prime();
    while(~scanf("%d",&n)&&n){
        int ans=0;
        int l=0,r=0;
        int res=0;
        while(r<pn){
            while(res<n){
                res+=prime[r];
                r++;
                if(r>=pn)break;
            }
            while(res>=n){
                if(res==n)ans++;
                res-=prime[l];
                l++;
            }
        }
        printf("%d
",ans);
    }
}



int main() {

#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif
    //iostream::sync_with_stdio(false);
    solve();
    return 0;
}