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.
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.
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.
4
2 1 2 3 4 3 2 6 1 1 2 2 1 2 3 2
4 3 6 2
1
42
42
2
1 1 1 1
1 1
思路：
设数列X: a11, a12,...., ann;
由于gcd(a,b)<=min(a,b);
ans[N]存放已经选中的数,即array中一定存在的数；
首先从X中找到最大的一个值aij，然后对ans[N]中的每一个数，得到g = gcd(aij, ans[i]),
由于table矩阵是对称的，所以从X中删除2个值为 g 的数值！
最后将aij放入ans中！不断重复此过程，知道ans中数字个数为n；
#include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<map> #include<set> #include<algorithm> #define N 505 using namespace std; int n; map<int, int, greater<int> >mp;//key按照由大到小排序 int gcd(int a, int b){ return b==0 ? a : gcd(b, a%b); } int ans[N]; int main(){ cin>>n; int nn = n*n; for(int i=0; i<nn; ++i){ int x; cin>>x; mp[x]++; } int len = 0; for(map<int, int, greater<int> >::iterator it=mp.begin(); it!=mp.end();){ if(it->second == 0){//不为0，说明这个数还是array中的数字 ++it; continue; } --it->second; for(int i=0; i<len; ++i){ int gcdn = gcd(it->first, ans[i]); mp[gcdn]-=2; } ans[len++] = it->first; } for(int i=0; i<n; ++i){ if(i!=0) cout<<" "; cout<<ans[i]; } cout<<endl; return 0; }
Codeforces Round #323 (Div. 2) C.GCD Table
原文：http://www.cnblogs.com/hujunzheng/p/4854948.html