coprime Sequence

Do you know what is called ``Coprime Sequence''? That is a sequence consists of nn positive integers, and the GCD (Greatest Common Divisor) of them is equal to 1.
``Coprime Sequence'' is easy to find because of its restriction. But we can try to maximize the GCD of these integers by removing exactly one integer. Now given a sequence, please maximize the GCD of its elements.

InputThe first line of the input contains an integer T(1≤T≤10)T(1≤T≤10), denoting the number of test cases.
In each test case, there is an integer n(3≤n≤100000)n(3≤n≤100000) in the first line, denoting the number of integers in the sequence.
Then the following line consists of nn integers a1,a2,...,an(1≤ai≤109)a1,a2,...,an(1≤ai≤109), denoting the elements in the sequence.OutputFor each test case, print a single line containing a single integer, denoting the maximum GCD.Sample Input

3
3
1 1 1
5
2 2 2 3 2
4
1 2 4 8

Sample Output

1
2
2
题解：有n个数，通过删除一个数后使他们的最大公约数最大。这（n-1)个数的最大公约数必定是最小的两个数的因子之一。

#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<map>
using namespace std;
map<int,int>::iterator it;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
       int n,x=0,t=0,i,j,a[100001],ans=1,l=2;
       scanf("%d",&n);map<int,int>mp;
       mp.clear();
       for(i=0;i<n;i++)
       {
           scanf("%d",&a[i]);
           if(a[i]==1)
               x++;
       }
       if(x>=2)
           printf("1
");
        else
        {
            sort(a,a+n);
            while(l--)
            {
            for(i=1;i<=sqrt(a[l]);i++)
            {
                if(a[l]%i==0)
                {
                  mp[i]++;
                  if(i*i!=a[l])
                  mp[a[l]/i]++;

                }
            }
            }
            for(i=2;i<n;i++)
              for(it=mp.begin();it!=mp.end();it++)
              {
                    if(a[i]%(it->first)==0)
                         it->second++;
              }
               for(it=mp.begin();it!=mp.end();it++)
                   if(it->second==n-1)
                       ans=max(ans,it->first);
                       printf("%d
",ans);
            }
    }
    return 0;
}