POJ2739 Sum of Consecutive Prime Numbers 确定某个数以内的所有素数

参考：https://www.cnblogs.com/baozou/articles/4481191.html

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=1e4+10;
bool prime[N];
void primemap()
{
    memset(prime,true,sizeof(prime));
    prime[0]=prime[1]=false;
    for (int i=2;i<N;i++)
    {
        if (prime[i])
        {
            for (int j=2*i;j<N;j+=i)
            {
                prime[j]=false;
            }
        }
    }
}
int solve(int x)
{
    int ans=0;
    for (int i=2;i<=x;i++)
    {
        if (prime[i])//遍历以素数开头的和，所以要先判断！
        {
            int sum=0;
            for (int j=i;j<=x;j++)
            {
                if (prime[j])
                {
                    sum+=j;
                }
                if (sum==x)
                {
                    ans++;
                    break;
                }
                if (sum>x)
                {
                    break;
                }
            }
        }
    }
    return ans;
}
int main()
{
    int in;
    primemap();
    while (cin>>in&&in)
    {
        cout<<solve(in)<<endl;
    }

    return 0;
}