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=,t=,i,j,a[],ans=,l=;
        scanf("%d",&n);map<int,int>mp;
        mp.clear();
        for(i=;i<n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]==)
                x++;
        }
        if(x>=)
            printf("1\n");
         else
         {
             sort(a,a+n);
             while(l--)
             {
             for(i=;i<=sqrt(a[l]);i++)
             {
                 if(a[l]%i==)
                 {
                   mp[i]++;
                   if(i*i!=a[l])
                   mp[a[l]/i]++;

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