Coprime Sequence

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

Total Submission(s): 217    Accepted Submission(s): 126

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

//T 组数据，n 个数，开始时，n 个数 gcd 为 1 ，现可以任意抽走一个数，问剩下数最大 gcd 是多少?

// st 表很好做，O(n*logn) + n * O(n) 的时间

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define MX 100005

int n,m;
int dp[MX][20];
int mi[MX];
int data[MX];

int gcd(int a,int b)
{return b==0?a:gcd(b,a%b);}

void build()
{
    mi[0]=-1;
    for (int i=1;i<=n;i++)
    {
        mi[i]=((i&(i-1))==0) ? mi[i-1]+1:mi[i-1];   //注意位运算优先级很低
        dp[i][0]=data[i];
    }
    for (int k=1;k<=mi[n];k++)
    {
        for (int i=1; (i+(1<<k)-1)<=n ;i++)
        {
            dp[i][k]=gcd(dp[i][k-1],dp[i+(1<<(k-1))][k-1]);
        }
    }
}

int find_(int a,int b)
{
    int x = mi[b-a+1];
    return gcd(dp[a][x],dp[b-(1<<x)+1][x]);
}

int main()
{
    int T;
    cin>>T;
    while (T--)
    {
        scanf("%d",&n);
        for (int i=1;i<=n;i++)
            scanf("%d",&data[i]);
        build();
        int ans = 1;
        for (int i=1;i<=n;i++)
        {
            int res;
            if (i==1)
                res = find_(2,n);
            else if (i==n)
                res = find_(1,n-1);
            else
                res = gcd(find_(1,i-1),find_(i+1,n));
            ans = max(ans,res);
        }
        printf("%d
",ans);
    }
}

 强行上log n 啊，啊哈哈，其实，只要从左向右扫一遍，存储gcd前缀和，再从右向左扫一遍，存gcd后缀和，然后枚举要剔除的数即可