Codeforces Round #323 (Div. 1) A. GCD Table stl应用

A. 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.

Sample test(s)

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，已知，n个数，所有两两之间的最大公约数，要求，原n个数。

可以肯定，最大的数，一定是，原数中的一个，然后，除去与已知数的最大公约数就可以得到所有的数。

可以使用map set优化。这样复杂度为o(n * log(n));

#define N 505
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,a[N * N],s
,sn;
map<int,int> mymap;
map<int,int>::iterator it;
bool cmp(int a1,int b1){
return a1 > b1;
}
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
while(S(n)!=EOF)
{
int n2 = n * n;
mymap.clear();
FI(n2){
S(a[i]);
a[i] *= -1;
if(mymap.count(a[i]) == 0){
mymap[a[i]] = 1;
}
else
mymap[a[i]]++;
}
//printf("%d\n",mymap.size());
sn = 0;
while(!mymap.empty()){
it = mymap.begin();
int tt = it->first;
s[sn++] = tt;
FI(sn){
int g = -1 * gcd(-1 * tt,-s[i]);
if(i == sn-1)
mymap[g] -= 1;
else
mymap[g] -= 2;
if(mymap[g] <= 0){
mymap.erase(g);
}
}
}
FI(sn)
printf("%d ",-1 * s[i]);
printf("\n");
}
//fclose(stdin);
//fclose(stdout);
return 0;
}