ssu 499 Greatest Greatest Common Divisor

499. Greatest Greatest Common Divisor

Time limit per test: 0.5 second(s)
Memory limit: 262144 kilobytes

input: standard
output: standard

Andrew has just made a breakthrough in sociology: he realized how to predict whether two persons will be good friends or not. It turns out that each person has an inner friendship number (a positive integer). And the quality of friendship between two persons is equal to the greatest common divisor of their friendship number. That means there are prime people (with a prime friendship number) who just can't find a good friend, andWait, this is irrelevant to this problem. You are given a list of friendship numbers for several people. Find the highest possible quality of friendship among all pairs of given people.

Input

The first line of the input file contains an integer n () — the number of people to process. The next n lines contain one integer each, between 1 and (inclusive), the friendship numbers of the given people. All given friendship numbers are distinct.

Output

Output one integer — the highest possible quality of friendship. In other words, output the greatest greatest common divisor among all pairs of given friendship numbers.

Example(s)

sample input	sample output
4 9 15 25 16	5

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <map>
#include <cmath>
#include <queue>
#include <cstring>
#include <set>
#include <stack>
#include <string>
#define LL long long
#define maxn 100010
#define mod 1000000007
#define INF 2000000
#define MAX 16000010
#define eps 1e-6
using namespace std;

bool vi[maxn*10] ;
int Max ;
int max(int a ,int b){ return a > b ? a : b ;}

void insert( int n )
{
    int m ,i ,k ;
    m = sqrt(n+0.5) ;
    for( i = 1 ; i <= m ;i++)if(n%i == 0)
    {
        k = n/i ;
        if(vi[i])
        {
            Max = max(Max,i) ;
        }
        else vi[i] = true ;
        if(vi[k]&&k != i)
        {
            Max = max(Max,k) ;
        }
        else vi[k] = true ;
    }
}
int main()
{
    int i ,m , tt , j , n ;
    int L , R , mid ;
    // freopen("in.txt","r",stdin) ;
   // 枚举每个数的因子是可以过的  
    while( scanf("%d",&n) != EOF)
    {
        memset(vi,0,sizeof(vi)) ;
         Max = 0 ;
         for( i = 1 ; i <= n ;i++ ){
             scanf("%d",&m) ;
             insert(m) ;
         }
        printf("%d\n",Max) ;
    }
    return 0 ;
}