Codeforces Round #323 (Div. 2) C gcd

連結：戳這裡

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.

Examples
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 

題意：

給出n*n個數，去填一個n*n的表格，表格對應的gij=gcd(ai,aj)

要求你輸出這n個數

思路：

分析一下，表格裡面的數肯定是gij=gji的 這是一定的  ，然後ai=gii這也是一定的

那麼我們從大到小排個序，當前大的數肯定是需要的，那麼怎麼統計他產生的數呢

因為新加入的數肯定要和之前加的數算gcd，那麼在減去2，for過去就可以了

程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n;
map<int ,int > mp;
int anw[1000100];
int gcd(int a,int b){
    return b==0?a:gcd(b,a%b);
}
int a[1000100];
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n*n;i++){
        scanf("%d",&a[i]);
        mp[a[i]]++;
    }
    sort(a+1,a+n*n+1);
    int num=0;
    for(int i=n*n;i>=1;i--){
        if(mp[a[i]]){
            mp[a[i]]--;
            for(int j=1;j<=num;j++){
                mp[gcd(a[i],anw[j])]-=2;
            }
            anw[++num]=a[i];
        }
    }
    for(int i=1;i<=n;i++) cout<<anw[i]<<" ";
    puts("");
    return 0;
}
/*
3
2 2 2 2 1 1 1 1 3
*/