HDU-6025 Coprime Sequence

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 131072/131072 K (Java/Others)

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

Reference Code

#include
#include
const int maxn=1e5+5;
#define gcd std::__gcd
int head[maxn],tail[maxn],a[maxn];
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)scanf("%d",&a[i]);
        head[1]=a[0];
        for(int i=2;i<n;i++){
            head[i]=gcd(head[i-1],a[i-1]);
        }
        tail[n-2]=a[n-1];
        for(int i=n-3;i>=0;i--){
            tail[i]=gcd(tail[i+1],a[i+1]);
        }
        int ma=-1;
        for (int i=1;i<n-1;++i){
            ma=std::max(ma,gcd(head[i],tail[i]));
        }
        ma=std::max(ma,tail[0]);
        ma=std::max(ma,head[n-1]);
        printf("%d\n",ma);
    }
}

Tips

题意：
给定一个gcd为1的队列，求解删去一个数后的最大gcd。
题解：
head[i]存前(i-1)个数的gcd，tail[i]存后(n-i)个数的gcd。直接求解max(gcd(head[i],tail[i]))。