【Codeforces 582A】GCD Table

标签：数组   put   puts   write   HERE   数字   string   ati   ring   

【链接】 我是链接,点我呀:)
【题意】

给你一个数组A[]经过a[i][j] = gcd(A[i],A[j])的规则生成的二维数组
让你求出原数组A

【题解】

我们假设原数组是A
然后让A数组满足A[i]<Ai+1
然后我们要先想到一个不等式
a[i][j]=gcd(A[i],A[j])<=A[min(i,j)]
而miin(i,j)<=n
则a[i][j]<=A[n]
所以a[i][j]里面的最大值就是A[n]
之后,我们把gcd(A[n],A[n])删掉
这样剩余的n*n-1个数字里面
min(i,j)<=n-1
则a[i][j]<=A[n-1]
所以剩余的a[i][j]里面的最大值就是A[n-1];
然后我们应该把gcd(A[n-1],A[n-1]),gcd(A[n],a[n-1]),gcd(A[n-1],A[n])这些数字都删掉
这样的话剩余的数字里面就会满足min(i,j)<=n-2了
然后我们就能够继续取a[][]最大值来获取A[n-2]了.

(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());
        }
    }
}

【Codeforces 582A】GCD Table

标签：数组   put   puts   write   HERE   数字   string   ati   ring   

原文地址：https://www.cnblogs.com/AWCXV/p/10482345.html