HDU - 6025 Coprime Sequence（给出一行数，去掉其中一个数，求这行数的最大的最大公约数（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：给你一行数，去掉一个数求最大的最大公约数；

思路：若求一行数的gcd（最大公约数），只要求出前两个的最大公约数，再用这两个的最大公约数,于后面一个的数gcd，求出来这个gcd，再与和它后面的一个数求gcd，依次类推；

但题意是除去一个数，求最大gcd，求出这个数的前缀gcd，和后缀gcd，然后再用这前缀gcd和后缀gcd求gcd，枚举每一个数，求得最大的gcd；

代码：

#include<stdio.h>     
#include<string.h>		 
#include<algorithm>	  // 若去掉一个数求最大gcd，求出这个数的前缀gcd，和后缀gcd，然后再用这前缀gcd和后缀gcd求gcd，枚举每一个数，求得最大的gcd 
using namespace std;
#define Max 100010
#define INF 0x3f3f3f3f

int a[Max],pre[Max],hou[Max];
int gcd(int a,int b)
{
	if(b==0)
		return a;
	else 
	{
		int tt = gcd(b,a%b);
		return tt;
	}
}

int main()
{
	int i,j,t,n;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=0;i<n;i++)
			scanf("%d",&a[i]);
		pre[0] = a[0];
		hou[n-1] = a[n-1];
		for(i=1;i<n;i++)
			pre[i] = gcd(pre[i-1],a[i]);  //前缀gcd 
		for(i=n-2;i>=0;i--)
			hou[i] = gcd(hou[i+1],a[i]);  //后缀gcd； 
		int ans = -INF;
		for(i=1;i<=n-2;i++)
			ans=max(ans,gcd(pre[i-1],hou[i+1])); //求出去掉这个数后的gcd; 
		ans=max(ans,hou[1]);        //  别忘了和去掉最前面的或最后面的比较； 
		ans = max(ans,pre[n-2]);
		printf("%d\n",ans);
	}
	return 0;
}