HDU 6025 Coprime Sequence（技巧暴力）

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2181    Accepted Submission(s): 1019

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

331 1 152 2 2 3 241 2 4 8

 

Sample Output

122

 

Source

2017中国大学生程序设计竞赛 - 女生专场

题意：给你n个数，求去掉任意一个数后剩下数的最大GCD。

三个for循环：

1、for 1~n 前缀GCD qian[i]

2、for n~1 后缀GCD hou[i]

3、for 1~n max(GCD(qian[i-1],hou[i+1]));

我当时傻傻的用2000+b的RMQ也搞过去了，写的超级麻烦，醉。

AC代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define ll long long
using namespace std;
ll gcd(ll x,ll y)
{
    if(y==0) return x;
    else return gcd(y,x%y);
}
ll a[100005],b[100005];
ll c[100005];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ll n;
        ll t;
        scanf("%lld",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&c[i]);
        }
        t=c[1];
        for(int i=1;i<=n;i++)
        {
            t=gcd(t,c[i]);
            a[i]=t;
        }
        t=c[n];
        for(int i=n;i>=1;i--)
        {
            t=gcd(t,c[i]);
            b[i]=t;
        }
        ll maxn=max(b[2],a[n-1]);
        for(int i=2;i<=n-1;i++)
        {
            maxn=max(maxn,gcd(a[i-1],b[i+1]));
        }
        cout<<maxn<<endl;
    }
}