POJ2739 Sum of Consecutive Prime Numbers

　题目大意：给出一个整数，如果有一段连续的素数之和等于该数，即满足要求，求出这种连续的素数的个数

　　水题：艾氏筛法打表+尺取法区间推进，0ms水过（注意循环的终止条件）

　　　　

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <fstream>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define in(n) scanf("%d",&(n))
#define in2(x1,x2) scanf("%d%d",&(x1),&(x2))
#define inll(n) scanf("%I64d",&(n))
#define inll2(x1,x2) scanf("%I64d%I64d",&(x1),&(x2))
#define inlld(n) scanf("%lld",&(n))
#define inlld2(x1,x2) scanf("%lld%lld",&(x1),&(x2))
#define inf(n) scanf("%f",&(n))
#define inf2(x1,x2) scanf("%f%f",&(x1),&(x2))
#define inlf(n) scanf("%lf",&(n))
#define inlf2(x1,x2) scanf("%lf%lf",&(x1),&(x2))
#define inc(str) scanf("%c",&(str))
#define ins(str) scanf("%s",(str))
#define out(x) printf("%d\n",(x))
#define out2(x1,x2) printf("%d %d\n",(x1),(x2))
#define outf(x) printf("%f\n",(x))
#define outlf(x) printf("%lf\n",(x))
#define outlf2(x1,x2) printf("%lf %lf\n",(x1),(x2));
#define outll(x) printf("%I64d\n",(x))
#define outlld(x) printf("%lld\n",(x))
#define outc(str) printf("%c\n",(str))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define mem(X,Y) memset(X,Y,sizeof(X));
typedef vector<int> vec;
typedef long long ll;
typedef pair<int,int> P;
const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
const bool AC=true;

bool is_prime[10000];
int prime[5000];
void table(){
    int k=0;
    fill(is_prime,is_prime+10000,true);
    rep(i,2,10000){
        if(is_prime[i]) prime[k++]=i;
        for(int j=2*i;j<=10000;j+=i)
            is_prime[j]=false;
    }
}
int main(){
    int n,s,t,sum,ans;
    table();
    while(in(n)==1){
        if(n==0) break;
        s=t=0;ans=0;
        sum=0;
        while(true){
            while(prime[t]<=n&&sum<n){
                sum+=prime[t++];
            }
            if(sum<n) break;
            else if(sum==n) ans++;
            sum-=prime[s++];
        }
        out(ans);
    }
    return 0;
}