参考: https://blog.csdn.net/SSimpLe_Y/article/details/73804147
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
#include <map>
#include <set>
#include <string>
#include <vector>
#define maxn 0x3f3f3f3f
#define ll long long
using namespace std;
int main() {
int T,kase=0;
cin>>T;
while(T--){
ll n;
cin>>n;
ll ans=n-(int)sqrt(n)-(int)sqrt(n/2);
//sqrt(n)求n以内的完全平方数个数,1,4,9,16可以映射成1,2,3,4,这样就很明白了
printf("Case %d: %lld\n",++kase,ans);
}
return 0;
}