Coprime Sequence（前后缀GCD）

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 $T(1\leq T\leq10)$, denoting the number of test cases. 
In each test case, there is an integer $n(3\leq n\leq 100000)$ in the first line, denoting the number of integers in the sequence. 
Then the following line consists of $n$ integers $a_1,a_2,...,a_n(1\leq a_i\leq 10^9)$, 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

 

 

 

 

 

题目意思：给出n个数，去掉其中的一个数，得到一组数，使其最大公约数最大。第一组样例，去掉一个1，得到最大的最大公约数是1。第二组样例，去掉3，得到最大的最大公约数为2。第三组样例，去掉1，得到的最大的最大公约数是2。

 

解题思路：这里使用了之前我没有用到过的一种方法。依次遍历每个数，算出这个数左边的一堆数的最大公约数x1，再算出这个数右边的一堆数的最大公约数x2，gcd(x1,x2)即为删去当前数后剩下的数的最大公约数。遍历每个数得到的最大gcd即为所求。而每个数左边的gcd和右边的gcd可以用O(n)的时间复杂度预处理，遍历每个数是跟前面并列的O(n)复杂度，故可线性解决。

 

 #include<stdio.h>
 #include<algorithm>
 using namespace std;
 int gcd(int a,int b) ///基础 辗转
 {
     int r;
     while(b>)
     {
          r=a%b;
          a=b;
          b=r;
     }
     return a;
 }
 int a[],b[],c[];
 int main()
 {
     int t,n,i,ans;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         for(i=;i<n;i++)
         {
             scanf("%d",&a[i]);
         }
         sort(a,a+n);
         b[]=a[];
         c[n-]=a[n-];
         for(i=;i<n;i++)///求前缀GCD
         {
             b[i]=gcd(b[i-],a[i]);
         }
         for(i=n-;i>=;i--)///求后缀GCD
         {
             c[i]=gcd(c[i+],a[i]);
         }
         ans=max(c[],b[n-]);
         for(i=;i<n-;i++)
         {
             ans=max(ans,gcd(b[i-],c[i+]));
         }
         printf("%d\n",ans);

     }
     return ;
 }