poj2739 Sum of Consecutive Prime Numbers 尺取法 素数打表

#include<cstdio>
#include<iostream>
using namespace std;
int prime[10005],ans[10005]={0},is_prime[10005];
int main(){
int p=0;
for(int i=0;i<=10000;i++)
is_prime[i]=1;
is_prime[1]=is_prime[1]=0;
for(int i=2;i<=10000;i++){
if(is_prime[i]){
prime[p++]=i;
for(int j
4000
=2*i;j<=10000;j+=i)
is_prime[j]=false;
}
}
int s=0,t=0,sum=0;
for(;;){
while(t<10000&&sum<=10000){
sum+=prime[t++];
ans[sum]++;
}
sum=0;s++;t=s;
if(prime[s]>10000)
break;
}
int n;
while((cin>>n)&&n!=0){
cout<<ans
<<endl;
}
}