Do you know what is called ``Coprime Sequence''? That is a sequence consists of nnpositive 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.
Input
The 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.
Output
For 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
PS：相信，题意大家都能读懂吧，就是去掉一个数是剩下的GCD最大。这里使用两个数组，一个前缀GCD数组和一个后缀GCD数组。细节看代码。
#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<stack>
#include<string>
const int maxn=1e6+10;
const int mod=1e9+7;
const int inf=1e8;
#define me(a,b) memset(a,b,sizeof(a))
typedef long long ll;
using namespace std;
int gcd(int a,int b)
{
return b==0?a:gcd(b,a%b);
}
int a[maxn],l[maxn],r[maxn];
int main()
{
int t;cin>>t;
while(t--)
{
int n;scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
l[1]=a[1],r[n]=a[n];
for(int i=2;i<=n;i++)
l[i]=gcd(l[i-1],a[i]);///保存前i项的GCD
for(int i=n-1;i>0;i--)
r[i]=gcd(r[i+1],a[i]);///保存后i项的GCD
int ans=max(l[n-1],r[2]);///因为数组是1到n，并且下面有i-1和i+1所以不能是max(l[n],r[1]);
for(int i=2;i<n;i++)
ans=max(ans,gcd(l[i-1],r[i+1]));///gcd(l[i-1],r[i+1])就表示除开第i项后数列的GCD、
cout<<ans<<endl;
}
return 0;
}