Coprime Sequence

                                                       Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                                                                     Total Submission(s): 16    Accepted Submission(s): 13


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
 

Source
2017中国大学生程序设计竞赛 - 女生专场
 

题意：一共n个数，去掉其中一个，问剩余n-1个数gcd的最大值

解题思路：rmq或者求前缀和后缀


求前缀和后缀代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int a[100005],x[100005],y[100005];

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

int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
            
        x[0]=a[0];
        for(int i=1; i<n; i++)
            x[i]=gcd(x[i-1],a[i]);
            
        y[n-1]=a[n-1];
        for(int i=n-2; i>=0; i--)
            y[i]=gcd(y[i+1],a[i]);
        
        int ans=max(y[1],x[n-2]);
        for(int i=1; i<n-1; i++)
            ans=max(ans,gcd(x[i-1],y[i+1]));
        printf("%d\n",ans);
    }
    return 0;
}


RMQ：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int a[100006],n;
int dp[100006][20];

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

void init()
{
    for(int i=1;i<=n;i++) dp[i][0]=a[i];
    for(int i=1;(1<<i)<=n;i++)
        for(int j=1;j+(1<<i)-1<=n;j++)
            dp[j][i]=gcd(dp[j][i-1],dp[j+(1<<(i-1))][i-1]);

}

int query(int l,int r)
{
    int k=0;
    while(1<<(k+1)<=r-l+1) k++;
    return gcd(dp[l][k],dp[r-(1<<k)+1][k]);
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        init();
        int ma=-1;
        for(int i=2;i<n;i++)
        {
            int k=query(1,i-1),kk=query(i+1,n);
            ma=max(ma,gcd(k,kk));
        }
        ma=max(ma,query(2,n));
        ma=max(ma,query(1,n-1));
        printf("%d\n",ma);
    }
    return 0;
}