Time Limit: 2000/1000 MS(Java/Others) Memory Limit: 131072/131072 K(Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Problem Description
Do you know what is called``Coprime Sequence''? That is a sequence consists ofn 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 tryto 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 inputcontains an integerT(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 integersa1,a2,...,an(1≤ai≤109) , denoting the elements in the sequence.
Output
For each test case, print asingle 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
#include
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=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