CodeForces 581C - GCD Table(思维)

C. GCD Table
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

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

题意:
一个数组里面有n个数,给出任意两个数的GCD,共有n²个数.
要求输出这n个数.

解题思路:
对于当前数列,最大的数必然是数组里面的数,把这个数提取出来,和已经提取出来的数求gcd,把数列里面等同与gcd的删除,循环这个过程.
要注意的是,删除的时候要删除两个gcd.

AC代码:

#include<bits/stdc++.h>using namespace std;const int maxn = 5e2+5;vector<int> vct;map<int,int> mp;int a[maxn*maxn];int gcd(int a,int b){return b?gcd(b,a%b):a;}int main(){    int n;    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);    for(int i = n*n;i >= 1;i--)    {        if(!mp[a[i]])   continue;        mp[a[i]]--;        for(int j = 0;j < vct.size();j++)            mp[gcd(a[i],vct[j])] -=2;        vct.push_back(a[i]);    }    for(int i = 0;i < vct.size();i++)   printf("%s%d",i?" ":"",vct[i]);    return 0;}