2017中国大学生程序设计竞赛 - 女生专场 1003

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 545    Accepted Submission(s): 190

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

题意：给出一组数组，他们的最大公约数为1，现在删除一个数，使得最大公约数最大

解法：先知道，只有一种数字和两种数字的情况，后面记录一个数左边的最大公约数，这个数右边的最大公约数（左右都包括它自己），然后l[i-1]和r[i+1]再求一个最大公约数就是删除这个数字的情况

 

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
set<int>q;
int main()
{
    std::ios::sync_with_stdio(false);
    int t;
    int x[400000];
    int num;
    while(cin>>t)
    {
        while(t--)
        {
            q.clear();
            int a=0;
            int b=0;
            int n;
            cin>>n;
            for(int i=1; i<=n; i++)
            {
                cin>>x[i];
                if(x[i]%2)
                {
                    a++;
                }
                else
                {
                    b++;
                }
                q.insert(x[i]);
            }
            if(q.size()==1)
            {
                cout<<x[1]<<endl;
            }
            else if(q.size()==2)
            {
                if(x[1]==x[2])
                {
                    cout<<x[1]<<endl;
                }
                else
                {
                    cout<<x[2]<<endl;
                }
            }
            else
            {
                int l[400000],r[400000];
                l[1]=x[1];
                r[n]=x[n];
                for(int i=2;i<n;i++)
                {
                    l[i]=__gcd(l[i-1],x[i]);
                }
                for(int i=n-1;i>=1;i--)
                {
                    r[i]=__gcd(r[i+1],x[i]);
                }
                int ans=max(l[n-1],r[1]);
                for(int i=2;i<n;i++)
                {
                    ans=max(ans,__gcd(l[i-1],r[i+1]));
                }
                cout<<ans<<endl;
            }
        }
    }
    return 0;
}