HDU-6025-Coprime Sequence（枚举+前后缀）

Coprime Sequence

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

解题思路：

简单思维题，枚举去掉每个n剩下的数的GCD值，那么可以开两个数组，一个往前gcd一个往后，这样删除一个数时就可以o1计算剩下的数的gcd值。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
const int N = 1e6+10;
ll front[N];
ll back[N];
ll a[N];

ll gcd(ll a1,ll b1)
{
    if (a1 < b1)
        swap(a1,b1);
    return b1 == 0 ? a1 : gcd(b1,a1%b1);
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        for(int i = 0 ; i < n ; i ++)
            cin>>a[i];
        front[0] = a[0];
        for(int i = 1 ; i < n ; i ++)
            front[i] = gcd(a[i],front[i-1]);
        back[n-1] = a[n-1];
        for(int i = n- 2 ; i >= 0 ; i --)
            back[i] = gcd(a[i],back[i+1]);
        ll ans = max(front[n-2],back[1]);
        for(int i = 1 ; i < n-1 ; i ++)
        {
            ans = max(ans,gcd(front[i-1],back[i+1]));
        }
        cout<<ans<<endl;
    }

}