Leshan normal programming competition 2020-g: maximum common divisor [thinking]

Title Description

Give a number n, Any other integer a and b The greatest common divisor of is gcd(a, b), The solution comes from 1 To n The maximum value of the greatest common divisor of any two different integers in .
Review of the greatest common divisor ：
18 The number of divisors is 1,2,3,6,9,18;
24 The number of divisors is 1,2,3,4,6,8,12,24;
18 And 24 The greatest common divisor of is 6.

Input

Input includes multiple sets of data ;
The first line has only one integer t （1 ≤ t ≤ 100） Represents the number of groups of input data ;
Next input t That's ok , Each line contains only one positive integer n（2 ≤ n ≤ 106）.

Output

For each group of input , Output required gcd(a, b) The maximum of , among 1 ≤ a < b ≤ n.

The sample input

2
3
5

Sample output

1
2

Tips

For the first set of data ,gcd(1, 2) = gcd(2, 3) = gcd(1,3) = 1;
For the second set of data ,gcd(2, 4) = 2 Is the biggest possible result .

Ideas

1. When n In the case of an even number ：
[1, n] The greatest common divisor of two different numbers in must be n and n/2. So the answer is n/2.

2. When n In the case of an odd number ：
[1, n] The greatest common divisor of two different numbers in must be n-1 and (n - 1) / 2. So the answer is (n - 1) / 2.

AC Code

#include<cstdio>
using namespace std;
typedef long long ll;
const int MAXN = 1e5 + 5;
void solve() {
    
    int n;
    scanf("%d", &n);
    printf("%d\n", n / 2);
}
int main() {
    
    int t;
    scanf("%d", &t);
    for (int i = 0; i < t; ++i) {
    
        solve();
    }
    return 0;
}