Coprime Sequence (prefix suffix GCD 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

Sample Output

1
2
2

Title:

Given a series of numbers whose greatest common divisor is 1, now you need to delete any number from these numbers to maximize the greatest common divisor of the remaining books

Ideas:

Traverse each number in turn, calculate the greatest common divisor x1 of a bunch of numbers on the left side of the number, and then calculate the greatest common divisor x2 of a bunch of numbers on the right side of the number, gcd(x1,x2) is the remaining number after deleting the current number The greatest common divisor of the numbers. The maximum gcd obtained by traversing each number is the required value.

The gcd on the left of each number and the gcd on the right can be preprocessed with O(n) time complexity, and traversing each number is O(n) in parallel with the previous number, so it can be solved linearly.

code show as below:

#include <bits/stdc++.h>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define f(i,a,b) for(int i=(a);i<=(b);++i)
#define ll long long
const int maxn = 100005;
const int mod = 1e9+7;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
#define rush() int T;scanf("%d",&T);while(T--)
int gcd(int a,int b)
{
return b?gcd(b,a%b):a;
}
int a[maxn];
int l[maxn];
int r[maxn];
int main()
{
int n;
rush()
{
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
l[0]=a[0];
r[n-1]=a[n-1];
for(int i=1;i<n;i++) //求前缀GCD
{
l[i]=gcd(l[i-1],a[i]);
}
for(int i=n-2;i>=0;i--) //求后缀GCD
{
r[i]=gcd(r[i+1],a[i]);
}
int ans = max (l[n-2],r[1]);
for(int i = 1 ;i < n-1 ;i++) //traverse to find the maximum value
{
ans=max(ans,gcd(l[i-1],r[i+1]));
}
printf("%d\n",ans);
}
return 0;
}

Code reference: https://blog.csdn.net/my_sunshine26/article/details/71320170