Codeforces Round #323 (Div. 2) C. GCD Table 暴力

                                                C. GCD Table

time limit per test

2 seconds

The GCD table G of size n × n for an array of positive integers a of length n is defined by formula

Let us remind you that the greatest common divisor (GCD) of two positive integers x and y is the greatest integer that is divisor of both xand y, it is denoted as . For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:

Given all the numbers of the GCD table G, restore array a.

Input

The first line contains number n (1 ≤ n ≤ 500) — the length of array a. The second line contains n2 space-separated numbers — the elements of the GCD table of G for array a.

All the numbers in the table are positive integers, not exceeding 109. Note that the elements are given in an arbitrary order. It is guaranteed that the set of the input data corresponds to some array a.

Output

In the single line print n positive integers — the elements of array a. If there are multiple possible solutions, you are allowed to print any of them.

Sample test(s)

input

4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2

output

4 3 6 2

input

1
42

output

42 

input

2
1 1 1 1

output

1 1 

题意：给你一个矩阵n*n个数，随意的排列，由一个n的排列，相互去GCD得到的，问你这个n的排列
题解：必定为最大数，但取GCD可能会更大，贪心从大向小取

///1085422276
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<bitset>
#include<set>
#include<vector>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define TS printf("111111\n");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define mod 1000000007
#define inf 100000
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
//****************************************
map<int,int> m;

int gcd(int a,int b)
{
    return (b==0)? a : gcd(b,a%b);
}

vector<int> v;

int main()
{
    int n,k;
    cin >> n;
    for(int i=0;i<n*n;i++)
    {
        cin >> k;
        m[-k]++;
    }

    for(map<int,int>::iterator it=m.begin(); it!=m.end();it++)
    {
        int t=-it->first;
        while(it->second)
        {
            it->second--;
            for(int i=0;i<v.size();i++)
            {
                int h=gcd(v[i],t);
                m[-h]-=2;
            }

            v.push_back(t);
        }
    }


    for(int i=0;i<n;i++)
    {
        cout << v[i] << " ";
    }
    return 0;
}