G - coprime sequence (HDU-6025+前缀GCD+后缀GCD)

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

 题意：给你一个数列，去掉一个数，是剩下的数列的最大公约数最大；

解题思路：求一个数列的最大公约数我们可以先求前两个数的最大公约数，然后用所求的最大公约数与第三个数求最大公约数，就这样一直GCD到最后一个数，求得的最大公约数即为数列的最大公约数，因为数列的最大公约数与所求顺序无关，只要将所有前面所求的GCD与后面的数进行GCD，最后的GCD即为数列的最大公约数，所以当去掉第i个元素时，我们只需要求gcd(left[i-1],right[i+1])就行。代码中的ans初始化为第一个数和最后一个数的最大公约数的最大值，以方便后面的循环

AC代码：

#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN=1e5+10;
int gcd(int a,int b)
{
	return b?gcd(b,a%b):a;
}
int main()
{
	int a[MAXN],left[MAXN],right[MAXN];//left[]数组储存从左到右的gcd,right[]数组储存从右到左
                                        //的gcd，left[i]表示数列中0-i个数的最大公约数
                                        //right[i]表示i以后所有数的最大公约数
	int t,n,ans;
	cin>>t;
	while(t--)
	{
		cin>>n;
		for(int i=0;i<n;i++)
		cin>>a[i];
		left[0]=a[0];
		for(int i=1;i<n;i++)
			left[i]=gcd(left[i-1],a[i]);
		right[n-1]=a[n-1];
		for(int i=n-2;i>=0;i--)
			right[i]=gcd(right[i+1],a[i]);
		ans=max(left[n-2],right[1]);
		for(int i=1;i<n-1;i++)
			ans=max(ans,gcd(left[i-1],right[i+1]));
		cout<<ans<<endl;
	}
	return 0;
}