# GCD is Funny

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description
Alex has invented a new game for fun. There are n integers at a board and he performs the following moves repeatedly:

1. He chooses three numbers ab and c written at the board and erases them.
2. He chooses two numbers from the triple ab and c and calculates their greatest common divisor, getting the number d (d maybe gcd(a,b)gcd(a,c) or gcd(b,c)).
3. He writes the number d to the board two times.

It can be seen that after performing the move n2 times, there will be only two numbers with the same value left on the board. Alex wants to know which numbers can left on the board possibly. Can you help him?

Input
There are multiple test cases. The first line of input contains an integer T (1T100), indicating the number of test cases. For each test case:

The first line contains an integer n (3n500) — the number of integers written on the board. The next line contains n integers: a1,a2,...,an (1ai1000)— the numbers on the board.

Output
For each test case, output the numbers which can left on the board in increasing order.

Sample Input
3
4
1 2 3 4
4
2 2 2 2
5
5 6 2 3 4

Sample Output
1 2
2
1 2 3

Alex发明了一个有趣的游戏. 一开始他在黑板上写了n个正整数, 然后他开始重复进行如下的操作:

1. 他选择黑板上三个数字a, b和c, 把他们从黑板上擦掉.
2. 他从这三个数a, b和c中选择了两个数, 并计算出他们的最大公约数, 记这个数为d (d 可以是gcd(a,b), gcd(a,c)或者gcd(b,c)).

3. 他在黑板上写下两次数字d.

考虑到, 最后留下来的数一定是某个子集的gcd. 我们只要在一开始丢掉了一个数, 考虑留下来两个数是x,x, 那么又选了另一个数y的话, 我们只要丢掉其中一个x就能获得了两个gcd(x,y), 也就是说接下来每次操作我们都有了一个额外的数用来丢弃, 且不会改变子集gcd的种类数.

所以我们可以先求出给出数字的两两GCD，然后用已有的GCD数列再一次与原数求一次GCD，依次输出.

AC代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<iostream>
using namespace std;
int num[1100],a[1100];
int gcd(int a,int b)
{
if(b==0)return a;
return gcd(b,a%b);
}
int main()
{
int T,n;
cin>>T;
while(T--)
{
cin>>n;
memset(num,0,sizeof(num));
for(int i=0; i<n; i++)
cin>>a[i];
for(int i=0; i<n; i++)
for(int j=i+1; j<n; j++)
num[gcd(a[i],a[j])]=1;      //记录初始两两的GCD
int flag=1,cnt=n;
while(flag&&--cnt>2)
{
flag=0;
for(int i=1; i<=1000; i++)      //对已有的GCD再次与数字进行GCD运算
if(num[i])
for(int j=0; j<n; j++)
{
int t=gcd(i,a[j]);
if(!num[t])num[t]=1,flag=1;
}
}
int t=0;
for(int i=1; i<=1000; i++)          //输出
if(num[i])
{
if(t)printf(" %d",i);
else
{
printf("%d",i);
t=1;
}
}
printf("\n");
}
}