Codeforces 582A GCD Table

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 12 2 1 2 3 2

output

4 3 6 2

input

1
42

output

42

input

2
1 11 1

output

1 1

解题思路：从大到小依次求解

#include <ctime>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
int ans[1010];
map<int, int> m;

int gcd(int x, int y) {
    return y == 0 ? x : gcd(y, x % y);
}

int main() {

    //freopen("aa.in", "r", stdin);

    int n, x;
    m.clear();
    scanf("%d", &n);
    for(int i = 1; i <= n*n; ++i) {
        scanf("%d", &x);
        m[x]++;
    }
    int cnt = 0;
    for(map<int, int>::reverse_iterator it = m.rbegin(); it != m.rend(); ++it) {
        if(it->second == 0) continue;
        while(it->second > 0) {
            ans[cnt] = it->first;
            it->second--;
            for(int j = 0; j < cnt; ++j) {
                int d = gcd(ans[j], it->first);
                m[d] -= 2;
            }
            cnt++;
        }
    }
    printf("%d", ans[0]);
    for(int i = 1; i < cnt; ++i) {
        printf(" %d", ans[i]);
    }
    printf("\n");
    return 0;
}