1370A - Maximum GCD.cpp

/*
Let's consider all integers in the range from 1 to n (inclusive).

Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a,b), where 1=a<b=n.

The greatest common divisor, gcd(a,b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.

Input
The first line contains a single integer t (1=t=100)  — the number of test cases. The description of the test cases follows.

The only line of each test case contains a single integer n (2=n=106).

Output
For each test case, output the maximum value of gcd(a,b) among all 1=a<b=n.

Example
inputCopy
2
3
5
outputCopy
1
2
Note
In the first test case, gcd(1,2)=gcd(2,3)=gcd(1,3)=1.

In the second test case, 2 is the maximum possible value, corresponding to gcd(2,4).
*/



#include<bits/stdc++.h>
using namespace std;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
  	int t;
  	cin >> t;
  	
  	while (t--) {
  		int n;
  		cin >> n;
  		
  		cout << n / 2 << endl;
	  }
    
    
    return 0;
}