HDU

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1098    Accepted Submission(s): 547

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

 

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jiangzijing2015

题意：给出一段序列，删去其中一个数使剩余数字的gcd最大；

#include <bits/stdc++.h>using namespace std;const int N = 100000 + 10;int t, n;int pre[N], suf[N], a[N];int gcd(int a, int b){    return b ? gcd(b, a%b) : a;}int main(){    scanf("%d", &t);    while(t--){        scanf("%d", &n);        for(int i = 1; i <= n; i++){            scanf("%d", &a[i]);        }        pre[0] = a[1];        for(int i = 1; i <= n; i++){            pre[i] = gcd(pre[i-1], a[i]);        }        suf[n+1] = a[n];        for(int i = n; i > 0; i--){            suf[i] = gcd(suf[i+1], a[i]);        }        int ans = 0;        pre[0] = suf[n-1];        suf[n+1] = pre[n-1];        for(int i = 1; i <= n; i++){            ans = max(ans, gcd(pre[i-1], suf[i+1]));        }        printf("%d\n", ans);    }}