傳送門
題意:
思路:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int cnt;
int vis[40000010];
int pri[10000010];
ll sum[10000010];
void oula()
{
for(int i = 2; i <= 40000000; i++)
{
if(!vis[i])pri[++cnt] = i;
for(int j = 1; j <= cnt && pri[j]*i <= 40000000; j++)
{
vis[i*pri[j]] = 1;
if(i%pri[j] == 0)break;
}
}
}
int main()
{
oula();
for(int i = 1; i <= cnt; i++)sum[i] = sum[i-1]+pri[i];
int t;
cin>>t;
while(t--)
{
ll n;
cin>>n;
int ans = 0;
for(int i = 1,j = 0; i <= cnt && pri[i] <= n; i++)
{
while(sum[i]-sum[j] > n)j++;
if(sum[i]-sum[j] == n)j++,ans++;
}
cout<<ans<<endl;
}
}