cf 323 A. GCD Table （还原gcd_暴力）

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最新推荐文章于 2021-06-16 18:11:46 发布

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给定一个正整数数组a的GCD表格G，根据表格中的元素恢复原数组a。题目提供n（1≤n≤500）和G的n²个元素，所有元素不超过10^9，要求找出可能的数组a。

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

$\text{[math]}$

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 $\text{[math]}$ . For example, for array a = {4, 3, 6, 2} of length 4 the GCD table will look as follows:

$\text{[math]}$

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 



#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<vector>
using namespace std;
map<int,int>mm;
vector<int>ans;
int a[600*600];
int main()
{
	int n,i,j;cin>>n;
	for(i=1;i<=n*n;i++) {
		cin>>a[i];
		mm[a[i]]++;
	}
	sort(a+1,a+1+n*n);
	for(i=n*n;i>=1;i--) {
		if(mm[a[i]]==0) continue;
		mm[a[i]]--;
		
		for(j=0;j<ans.size();j++) {
			int t=__gcd(a[i],ans[j]);
			mm[t]-=2;
		}
		ans.push_back(a[i]);
	}
	for(i=0;i<n;i++) cout<<ans[i]<<" ";
	cout<<endl;
}










#include <iostream>
#include <fstream>
#include <set>
#include <map>
#include <string>
#include <vector>
#include <bitset>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <cassert>
#include <queue>


typedef long long ll;
typedef long double ld;

using namespace std;

vector<int> vv;

multiset<int> ss;

int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < n; ++j) {
			int a;
			scanf("%d", &a);
			ss.insert(a);
		}
	while (!ss.empty()) {
		int x = *ss.rbegin();
		printf("%d ", x);
		for (int i = 0; i < (int)vv.size(); ++i)
			ss.erase(ss.find(__gcd(x, vv[i]))), ss.erase(ss.find(__gcd(x, vv[i])));
		vv.push_back(x);
		ss.erase(ss.find(x));
	}
	return 0;
}




#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <vector>
const int N = 2005;
typedef long long LL;
using namespace std;
int n , res[N];
void work() {
    scanf("%d" , &n);
    map<int , int> Hash;
    for (int i = 0 ; i < n * n ; ++ i) {
        int x ;
        scanf("%d" , &x);
        ++ Hash[x];
    }    
    for (int i = 0 ; i < n ; ++ i) {
        res[i] = (-- Hash.end()) -> first;
        if (!-- Hash[res[i]])
            Hash.erase(res[i]);
        for (int j = 0 ; j < i ; ++ j) {
            int x = __gcd(res[i] , res[j]);
            if (!-- Hash[x])
                Hash.erase(x);
            if (!-- Hash[x])
                Hash.erase(x);            
        }
        printf("%d\n" , res[i]);
    }
}


int main() {
    work();
    return 0;
}