HDU6025 Coprime Sequence

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 72    Accepted Submission(s): 51

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

331 1 152 2 2 3 241 2 4 8

 

Sample Output

122

 

Source

2017中国大学生程序设计竞赛 - 女生专场

 

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题意就不多说了，作为菜鸡看到如此做法我一定要发表一下；

不多BB，看代码：

#include<iostream>
#include<cstring>
#include<stdio.h>
using namespace std;
int a[100005];//a[i]表示1到i所有数的GCD
int b[100005];//b[i]表示i到n所有数的GCD
int s[100005];
int gcd(int a,int b)
{
   if(b==0)
      return a;
   return gcd(b,a%b);
}
int main()
{
   ios::sync_with_stdio(false);
   int t;
   cin>>t;
   while(t--)
   {
      int n;
      cin>>n;
      for(int i=1;i<=n;i++)
      {
         cin>>s[i];
         if(i==1)
         {
            a[i]=s[i];
            continue;
         }
         if(i==n)
         b[i]=s[i];
         a[i]=gcd(a[i-1],s[i]);
      }
      for(int i=n-1;i>=1;i--)
      b[i]=gcd(b[i+1],s[i]);
      int ans=1;
      for(int i = 1; i <= n; ++i)
      {
         if(i == 1) ans=max(ans,b[2]);
         else if(i == n) ans=max(ans,a[n - 1]);
         else ans = max(ans, gcd(a[i - 1],b[i + 1]));
}
cout<<ans<<endl;
   }
}