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
给你一堆数,是一个n*n矩阵包含的数,让你求一个arr 可以使当前arr中任何两个数求gcd 正好是给的矩阵的数,让你求出来这几个数。
很显然应该从最大的开始,因为最大的肯定是一个,然后不断枚举最大值求gcd 没求出来一个删除2个 因为 a b 和 b a是一样的
用map标记每个数字出现几次
map<int,int>m;
priority_queue <int,vector<int>,less<int> > q;
int ans[505];
int main ()
{
int n;
scanf("%d",&n);
n=n*n;
int k=0;
for(int i=0;i<n;i++)
{
int temp;
scanf("%d",&temp);
if(m[temp]==0)
q.push(temp);
m[temp]++;
}
while(!q.empty())
{
int u=q.top();
ans[k]=u;
if(m[u]>0)
{
m[u]--;
for(int i=0;i<k;i++)
{
int v=__gcd(u,ans[i]);
m[v]=m[v]-2;
}
k++;
}
if(m[u]==0)
q.pop();
}
for(int i=0;i<k;i++)
{
if(i)
printf(" ");
printf("%d",ans[i]);
}
printf("\n");
return 0;
}