Recently you have received two positive integer numbers xx and yy. You forgot them, but you remembered a shuffled list containing all divisors of xx (including 11 and xx) and all divisors of yy (including 11 and yy). If dd is a divisor of both numbers xx and yy at the same time, there are two occurrences of dd in the list.
For example, if x=4x=4 and y=6y=6 then the given list can be any permutation of the list [1,2,4,1,2,3,6][1,2,4,1,2,3,6]. Some of the possible lists are: [1,1,2,4,6,3,2][1,1,2,4,6,3,2], [4,6,1,1,2,3,2][4,6,1,1,2,3,2] or [1,6,3,2,4,1,2][1,6,3,2,4,1,2].
Your problem is to restore suitable positive integer numbers xx and yy that would yield the same list of divisors (possibly in different order).
It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers xx and yy.
Input
The first line contains one integer nn (2≤n≤1282≤n≤128) — the number of divisors of xx and yy.
The second line of the input contains nn integers d1,d2,…,dnd1,d2,…,dn (1≤di≤1041≤di≤104), where didi is either divisor of xx or divisor of yy. If a number is divisor of both numbers xx and yy then there are two copies of this number in the list.
Output
Print two positive integer numbers xx and yy — such numbers that merged list of their divisors is the permutation of the given list of integers. It is guaranteed that the answer exists.
Example
input
Copy
10 10 2 8 1 2 4 1 20 4 5
output
Copy
20 8
思路：先选出最大的，然后去遍历是他取模为0的并标记一下，其余的存在另一个数组里，在筛选出最大的
代码：
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int a[150];
int b[150];
int vis[10005];
int main()
{
int n;
cin>>n;
for(int t=0;t<n;t++)
{
scanf("%d",&a[t]);
}
sort(a,a+n);
int l=a[n-1];
int s=0;
for(int t=0;t<n;t++)
{
if(l%a[t]==0&&vis[a[t]]==0)
{
vis[a[t]]=1;
continue;
}
else
{
b[s++]=a[t];
}
}
sort(b,b+s);
cout<<l<<" "<<b[s-1]<<endl;
return 0;
}