传送门
A. Maximum GCD
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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<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
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, 2is the maximum possible value, corresponding to gcd(2,4)

题意：输入m组数字n，判断m组数字中从1到n这些数字可以组成的整数对中最大的公约数。

思路：求最大公约数，直接输出n/2即可。

AC代码

#include<stdio.h>
#include<math.h>
int main()
{
    int m,n;
    scanf("%d",&m);
    while(m--)
    {
        scanf("%d",&n);
        printf("%d\n",n/2);
    }return 0;
}