（HDU 6025 女生专场）Coprime Sequence 水题

Coprime Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 557 Accepted Submission(s): 287

Problem Description
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

Source
2017中国大学生程序设计竞赛 - 女生专场

题意：
给你n个数，并且保证他们的GCD为1,问你删去哪一个数，可以使得剩下的数的GCD最大，问你最大的GCD为多少?

分析：
由于gcd()函数的时间复杂度为O(log n ),根据题目的数据的大小，我们知道要写出一个时间复杂度为O(n)的方法。

首先处理出所有前缀的gcd的值g1[i],和所有后缀的gcd的值g2[i].
然后枚举所要删去的数的位置i即可。 ans = max (ans,gcd(g1[i-1],g2[i+1]))

AC代码：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

const int maxn = 100010;
int g1[maxn];
int g2[maxn];
int a[maxn];

int _gcd(int a,int b)
{
    return a == 0 ? b : _gcd(b%a,a);
}

int main()
{
    int t,n,ans;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(i == 1) g1[i] = a[i];
            else g1[i] = _gcd(g1[i-1],a[i]);
        }
        for(int i=n;i>=1;i--)
        {
            if(i == n) g2[i] = a[i];
            else g2[i] = _gcd(g2[i+1],a[i]);
        }
        ans = 0;
        for(int i=1;i<=n;i++)
        {
            if(i == 1) ans = max(ans,g2[2]);
            else if(i == n) ans = max(ans,g1[n-1]);
            else ans = max(ans,_gcd(g1[i-1],g2[i+1]));
        }
        printf("%d\n",ans);
    }
    return 0;
}