Do you know what is called Coprime Sequence''? That is a sequence consists of n 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.
Input
The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.
In each test case, there is an integer n(3≤n≤100000) in the first line, denoting the number of integers in the sequence.
Then the following line consists of n integers 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

利用那种前缀和 后缀和那种思想预处理 然后枚举一下删去的数

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=100050;
int a[N];
int pre[N];
int suf[N];
int gcd(int a,int b){
    return b==0 ? a : gcd(b,a%b);
}
int main(void){
    int t;
    int n;
    scanf("%d",&t);
    while(t--){
        memset(a,0,sizeof(a));
        memset(pre,0,sizeof(pre));
        memset(suf,0,sizeof(suf));
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
        //预处理前缀和后缀gcd
        pre[0]=a[0];
        for(int i=1;i<n;i++){
            pre[i]=gcd(pre[i-1],a[i]);
        }
        suf[n-1]=a[n-1];
        for(int i=n-2;i>=0;i--){
            suf[i]=gcd(suf[i+1],a[i]);
        }
        //枚举删除的数字
        int ans=1;
        for(int i=0;i<n;i++){
            ans=max(ans,gcd(pre[i-1],suf[i+1]));
        }
        printf("%d\n",ans);
    }
    return 0;
}