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

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

Sample Output

1
2
2


去掉一個值，使得剩下的值gcd最大。
典型的預處理。將每個值的前綴和後綴求出來，掃一遍每個值，ans=max(ans,gcd(前綴，後綴))
 

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<string>
#include<stdlib.h>
#include<math.h>
#include<set>
#include<map>
#include<algorithm>
#define MAX 100005
#define INF 0x3f3f3f3f
using namespace std;

int a[MAX];
int pre[MAX],sub[MAX];
int gcd(int x,int y){
    if(y) return gcd(y,x%y);
    return x;
}
int main()  
{  
    int t,f,h,n,m,k,i,j;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        scanf("%d",&a[1]);
        pre[1]=a[1];
        for(i=2;i<=n;i++){
            scanf("%d",&a[i]);
            pre[i]=gcd(pre[i-1],a[i]);
        }
        sub[n]=a[n];
        for(i=n-1;i>=1;i--){
            sub[i]=gcd(sub[i+1],a[i]);
        }
        int maxx=(pre[n-1]>sub[2]?pre[n-1]:sub[2]);
        for(i=2;i<n;i++){
            if(gcd(pre[i-1],sub[i+1])>maxx) maxx=gcd(pre[i-1],sub[i+1]);
        }
        printf("%d\n",maxx);
    }
    return 0;  
}  

HDU - 6025 Coprime Sequence（gcd+前綴後綴）