CF582A GCD Table

题目链接：http://codeforces.com/problemset/problem/582/A

A. 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 bothx 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

42 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2

output

4 3 6 2

input

142

output

42 

input

21 1 1 1

output

1 1 

解题思路：在所有的数组成的集合中，最大的数一定为n个数中的一个，把这个数从集合中减去一个，并把这个数加入数组b中，然后找出第二大的数，将这个数与数组b中的数的最大公因数从集合中减去两个，再把这个数加入数组b中，接着重复上述步骤，找出第三，第四.....第n个数，数组b中的数即为答案。

#include <cstdio>#include <cstring>#include <map>#include <algorithm>using namespace std;#define N 250010int a[N],num[N],b[N];map<int,int>mp;int n,tot,x;int main(){    scanf("%d",&n);    for(int i=1;i<=n*n;++i)        scanf("%d",&b[i]);    mp.clear();    sort(b+1,b+n*n+1);    tot=0;    for(int i=1;i<=n*n;++i)    {        x=b[i];        if(mp.find(x)==mp.end())        {            mp[x]=++tot;            a[tot]=x;            num[tot]=1;        }        else ++num[mp[x]];    }    int z=1;    b[1]=a[tot];    printf("%d ",a[tot]);    --num[tot];    for(int i=tot;i>0;--i)    {        while(num[i]>0)        {            b[++z]=a[i];            --num[i];            printf("%d ",a[i]);            for(int j=1;j<z;++j)                num[mp[__gcd(b[z],b[j])]]-=2;        }    }    return 0;}