HDU - 6025 Coprime Sequence（gcd+前缀后缀）

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

去掉一个值，使得剩下的值gcd最大。
典型的预处理。将每个值的前缀和后缀求出来，扫一遍每个值，ans=max(ans,gcd(前缀，后缀))

#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[]);
        pre[]=a[];
        for(i=;i<=n;i++){
            scanf("%d",&a[i]);
            pre[i]=gcd(pre[i-],a[i]);
        }
        sub[n]=a[n];
        for(i=n-;i>=;i--){
            sub[i]=gcd(sub[i+],a[i]);
        }
        int maxx=(pre[n-]>sub[]?pre[n-]:sub[]);
        for(i=;i<n;i++){
            if(gcd(pre[i-],sub[i+])>maxx) maxx=gcd(pre[i-],sub[i+]);
        }
        printf("%d\n",maxx);
    }
    return ;
}