Nyoj154 smart Xiaoke [Mutual quality]

Intelligent cool time limit: 1000 MS | memory limit: 1000 kb difficulty: 3

Description

Xiao Ke is a girl who loves beauty. She has n New necklaces ranging from 1 to n. Each necklace has some differences in color. N necklaces are placed in sequence and placed in a circle. Xiao Ke will count to the K necklace from the last time he selected the necklace, and use this necklace as the necklace to be brought today. The direction is the same every time, now, I hope you can help Xiao Ke calculate the largest K to meet K <= n/2 at the same time, so that Xiao Ke will carry all the necklaces in the next n days.

For example, n = 7, K = 3

Days necklace No.

1 1

2 4

3 7

4 3

5 6

6 2

7 5

Input

The first line has an integer 0 <m <10000 indicates that there are M groups of test data, and the next m row has an integer in each line, indicating the number of Xiao Ke's necklaces 2 <= m <2 ^ 31

Output

Output M k values

Sample Input

227

Sample output

13

Source

[Iphxer] original

Uploaded

Iphxer

Maximum m of the output and N mutual quality meeting conditions

#include <stdio.h>int gcd(int a, int b) {return b ? gcd(b, a % b) : a;}int main() {int t, n, m;scanf("%d", &t);while(t--) {scanf("%d", &n);for(m = n / 2; m > 1; --m)if(gcd(m, n) == 1) break;printf("%d\n", m);}return 0;}

Nyoj154 smart Xiaoke [Mutual quality]