Codeforces Round #323 (Div. 2) 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.

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 //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 ++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 ++i){

 int gcdn = gcd(it- first, ans[i]);

 mp[gcdn]-=2;

 ans[len++] = it- first;

 for(int i=0; i ++i){

 if(i!=0) cout " ";

 cout ans[i];

 cout endl;

 return 0;

}

Codeforces Round #723 (Div. 2)B. I Hate 1111 Description You are given an integer x. Can you make x by summing up some number of 11,111,1111,11111,…? (You can use any number among them any number of times). For instance, 33=11+11+11 144=111+11+11+11