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

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C. 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.

Examplesinput
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 

一眼就看出出現奇數次的一定是答案,然後yy了一下,覺得出現兩次以上的偶數次的也是答案,五分鐘出了這題,結果就fst了(出題人用心險惡,好吧其實是我太弱)。
正解應該是顯然知道最大的數一定在答案中,每次引入答案中的數一定會和之前答案中有的數產生兩個新的gcd值,划去原來數組中的這兩個值,到最後就會留下答案了。
#include<iostream>#include<cstdio>#include<queue>#include<map>#include<algorithm>using namespace std;#define maxn 620int a[maxn * maxn];map<int, int> mp;int main() {    int n;    while(~scanf("%d", &n)) {        for(int i = 1; i <= n * n; i++){            scanf("%d", &a[i]);            mp[a[i]]++;        }        sort(a + 1, a + 1 + n * n);        int cnt = 0;        int ans[maxn];        for(int i = n * n; i >= 1; i--) {            if(!mp[a[i]]) continue;            mp[a[i]] --;            for(int j = 1; j <= cnt; j++) {                mp[__gcd(a[i], ans[j])] -= 2;            }            ans[++cnt] = a[i];        }        for(int i = 1; i <= cnt; i++) printf("%d ", ans[i]); puts("");    }}

 

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

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