Count Squares

PROBLEM :

Given a sample space S consisting of all perfect squares starting from 1, 4, 9 and so on. You are given a number N, you have to print the number of integers less than N in the sample space S.

Input :
The first line contains an integer T, denoting the number of test cases.Then T test cases follow. The first line of each test case contains an integer N, denoting the number.

Output :
Print the answer of each test case in a new line.

Constraints :
1<=T<=100
1<=N<=10^18

Example
Input :
2
9
3

Output :
2
1

--------------------------------------------------------------------------------
SIMPLE c IMPLEMENTATION :
--------------------------------------------------------------------------------

#include <stdio.h>
int square(int n)
{
    int i=1,temp=0,count=0;
    while(temp<n)
    {
        temp=i*i;
        count=count+1;
        i++;
    }
    return count-1;
}
int main() {
int t,n;
scanf("%d",&t);
while(t--)
{
   scanf("%d",&n);
   printf("%d\n",square(n));
}
return 0;
}

--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION :
--------------------------------------------------------------------------------

#include<iostream>
#include<math.h>
using namespace std;
int main()
 {
long long t,no ;
cin>>t ;
while(t--)
{
   cin>>no ;
   cout<<floor(sqrt(no-1))<<endl ;
}
return 0;
}

---------------------------------------------------------------------------------

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