HDU 6025 Coprime Sequence

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 44    Accepted Submission(s): 34

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

题意：给出一列数，去掉其中一个数，使得所有数的最大公因数最大。

枚举每一个数字，尝试去掉，然后求所有数的最大公因数，去掉就是求公因数时标记跳过。提取出前两个因子，和第三个求因子，得出的最大共因子，再和第四个数比较。。。中间适当做优化，有几种情况可以跳出循环。

#include
#include
#include
using namespace std;
inline int GCD(int a,int b)
{
    int t;
    if(b>a)
    {
        t=a;
        a=b;
        b=t;
    }
    while(b)
    {
        t=b;
        b=a%b;
        a=t;
    }
    return a;
}

int main()
{
    int a[100001],b[100001];
    int t,n,i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0; i