Codeforces 583c GCD Table

C. GCD Table

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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
x and 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

分析：因为最大的数和第二大的数一定在要求数组a中，如果不在a中，可能是比他还小的的数的最大公因数吗？所以找到这两个数的最大公因数，删除，再从数组中找最大的数，分别与a中的数求最大公因数，删除。。。直到a数组中有n个数。

#include <bits/stdc++.h>
using namespace std;

map<int, int> s;
map<int,int>::iterator it;
int main() {
ios::sync_with_stdio(false);
int N;
cin >> N;
for(int i = 0; i < N * N; i++) {
int a;
cin >> a;
++s[a];
}
vector<int> res;
for(int i = 0; i < N; i++) {
int Max = 0;
for(it=s.begin();it!=s.end();it++)
if(it->second > 0 && it->first > Max)
Max = it->first;
for(int j = 0; j <res.size(); j++)
s[__gcd(res[j], Max)] -= 2;
res.push_back(Max);
s[Max] -= 1;
}
for(int i = 0; i < res.size(); i++)
cout << res[i] << " ";
cout << endl;
return 0;
}