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+;
 bool prime[N];
 void primemap()
 {
     memset(prime,true,sizeof(prime));
     prime[]=prime[]=false;
     for (int i=;i<N;i++)
     {
         if (prime[i])
         {
             for (int j=*i;j<N;j+=i)
             {
                 prime[j]=false;
             }
         }
     }
 }
 int solve(int x)
 {
     int ans=;
     for (int i=;i<=x;i++)
     {
         if (prime[i])//遍历以素数开头的和，所以要先判断！
         {
             int sum=;
             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 ;
 }