HDU 6025 Coprime Sequence

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2336    Accepted Submission(s): 1075

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：输入n个数，随便删除一个数，求剩余n-1个数的最大公约数，问删除一个数后最大的最大公约数是几。

思路：依次从左边求开始出前i个数的GCD，在依次从右边开始求出前j个数的GCD。再一次模拟删除任意一个数，求剩余n-1个数的GCD，从其中去最大的GCD。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int GCD(int a,int b)
{
    return b==0?a:GCD(b,a%b);
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        int a[n+1];
        for(int i=1;i<=n;i++)
            cin>>a[i];
        int pre[n+1],ends[n+1],mid[n+1];
        pre[1]=a[1];
        for(int i=2;i<=n;i++)
            pre[i]=GCD(pre[i-1],a[i]);
        ends[n]=a[n];
        for(int i=n-1;i>=1;i--)
            ends[i]=GCD(ends[i+1],a[i]);
        int Max=max(pre[n-1],ends[2]);///比较是删除第一个数还是删除最后一个数后剩余n-1个数的GCD最大
        for(int i=2;i<=n-1;i++)
            Max=max(Max,GCD(pre[i-1],ends[i+1]));///删除2~n-1中的一个数的GCD
        cout<<Max<<endl;
    }
    return 0;
}