Blue Bridge Cup of Chocolate Java Implementation

Title: The title OJ

this problem is not difficult, as long as we make it clear that each square to be cut down, cut down the number is
(length / square side length) * (width / square side length) would solve half of
us have to look at Let's take a look at the data range 1e5. If we directly enumerate the length of the reduced square, O (n ^ 2) is definitely T.
So here we have to think that the positive side length of the reduction must be within 1 ~ the maximum side length of all chocolates.
Then we can dichotomize it, just find this value for the dichotomy.
Code:

import java.io.*;
import java.util.Arrays;
public class P7158分巧克力 {
	static int n,k;
	static Node a[];
	static boolean check(int x){
		int sum = 0;
		for(int i = 0;i < n;i++){
			sum += (a[i].x/x)*(a[i].y/x);
			if(sum >= k)
				return true;
		}
		return false;
	}
	public static void main(String[] args) throws IOException{
		StreamTokenizer re = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter pr = new PrintWriter(new OutputStreamWriter(System.out));
		re.nextToken(); n = (int)re.nval;
		re.nextToken(); k = (int)re.nval;
		a = new Node[n];
		int t = 0;
		for(int i = 0;i < n;i++){
			a[i] = new Node();
			re.nextToken(); a[i].x = (int)re.nval;
			re.nextToken(); a[i].y = (int)re.nval;
			if(a[i].x < a[i].y){ // 保证 x 为最大值
				t = a[i].x;
				a[i].x = a[i].y;
				a[i].y = t;
			}
		}
		Arrays.sort(a);
		int l = 1,r = a[0].x,mid = 0;
		while(l <= r){
			mid = (l+r)>>1;
			if(check(mid))
				l = mid+1;
			else
				r = mid-1;
		}
		System.out.println(l-1);
	}
}
class Node implements Comparable{
	int x,y;
	public int compareTo(Object b){
		Node a = (Node)b;
		return a.x - this.x;
	}
}