【链接】 我是链接,点我呀:)
【题意】
【题解】
我们假设原数组是A 然后让A数组满足A[i](感性认知一下)每次都把A[i+1..n]这些数字产生的gcd都删掉
那么剩余的a[][]里面只有A[1..i]产生的gcd
那么因为A[i]是最大值,所以gcd(A[i],A[i])肯定也是a[][]里面的最大值,也即A[i]=剩余的a[][]里面的最大值
【代码】
import java.io.*;
import java.util.*;
public class Main {
static InputReader in;
static PrintWriter out;
public static void main(String[] args) throws IOException{
//InputStream ins = new FileInputStream("E:\rush.txt");
InputStream ins = System.in;
in = new InputReader(ins);
out = new PrintWriter(System.out);
//code start from here
new Task().solve(in, out);
out.close();
}
static int N = 500;
static class Task{
int n;
int ans[] = new int[N+10];
TreeMap<Integer, Integer> dic = new TreeMap<Integer,Integer>();
int _findmax() {
int x = dic.lastKey();
return x;
}
int gcd(int x,int y) {
if (y==0) return x;
else return gcd(y,x%y);
}
void _delete(int x) {
int y = dic.get(x);
y--;
if (y==0) dic.remove(x);else dic.put(x, y);
}
public void solve(InputReader in,PrintWriter out) {
n = in.nextInt();
for (int i = 1;i <= n*n;i++) {
int x = in.nextInt();
if (dic.containsKey(x)) {
int y = dic.get(x);
y++;
dic.put(x, y);
}else dic.put(x, 1);
}
for (int i = n;i >= 1;i--) {
ans[i] = _findmax();
_delete(gcd(ans[i],ans[i]));
for (int j = i+1;j<= n;j++) {
_delete(gcd(ans[i],ans[j]));
_delete(gcd(ans[i],ans[j]));
}
}
for (int i = 1;i <= n;i++)
out.print(ans[i]+" ");
out.println();
}
}
static class InputReader{
public BufferedReader br;
public StringTokenizer tokenizer;
public InputReader(InputStream ins) {
br = new BufferedReader(new InputStreamReader(ins));
tokenizer = null;
}
public String next(){
while (tokenizer==null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(br.readLine());
}catch(IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}