Codeforces Round #323 (Div. 2) C gcd

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本文介绍了一种算法，该算法可以根据给定的 GCD (最大公约数) 表格恢复原始数组。通过分析表格特性，采用排序及逐个元素验证的方法，确保了算法的有效性和正确性。

链接：戳这里

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：

给出n*n个数，去填一个n*n的表格，表格对应的gij=gcd(ai,aj)

要求你输出这n个数

思路：

分析一下，表格里面的数肯定是gij=gji的这是一定的，然后ai=gii这也是一定的

那么我们从大到小排个序，当前大的数肯定是需要的，那么怎么统计他产生的数呢

因为新加入的数肯定要和之前加的数算gcd，那么在减去2，for过去就可以了

代码：

#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
*/