ccpc 女生专场 C - Coprime Sequence HDU - 6025

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.

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<utility>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#define maxn 110
#define INF 0x3f3f3f3f
#define LL long long
#define ULL unsigned long long
#define E 1e-8
#define mod 100000000
using namespace std;
#define raf(i,k,n) for(int i=k;i<=n;i++)
//Oo0Oooo00ooOoo00oO
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
int a[100005];
int dp1[100005],dp2[100005];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        for(int i = 0 ; i < n ; i++ )   cin>>a[i];
        dp1[0]  = a[0];
        for(int i = 1 ; i < n ; i++ )   dp1[i] = gcd(dp1[i-1],a[i]);
        dp2[n-1] = a[n-1];
        for(int i = n-2 ; i >= 0 ; i--)  dp2[i] = gcd(dp2[i+1],a[i]);
        int ma = max(dp2[1],dp1[n-2]);
        //cout<<"dp2 "<<dp2[n-2];
        for(int i=1;i<n-1;i++)

            ma = max(gcd(dp1[i-1] , dp2[i+1] ), ma);

        cout<<ma<<endl;
    }
    return 0;
}