codeforce 1370A Maximum GCD(简单思维)

A. Maximum GCD
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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

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

Example
input
2
3
5
output
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).

链接：https://codeforces.ml/problemset/problem/1370/A
题意：求小于等于n 的a b两个数的最大公因数
思路：当n 为偶数的时 b = n , a = n/2这样a是最大公因数
当n为奇数时 如果令b = n，则a = n^1/2时a是最大公因数，但是a仍然小于 (n-1)/2 (可列方程推导一下，n>8^1/2+3)
ac代码如下：

#include
using namespace std;
int main()
{
     
	int t;
	cin>>t;
	while(t--)
	{
     
		int n;
		cin>>n;
		if(n%2==0)
		{
     
			cout<<n/2<<endl;
		} 
		else if(n%2==1)
		{
     
			cout<<(n-1)/2<<endl;
			
		}
	}
}