Problem:
Score : 300 points

Problem Statement

There is a rectangle in a coordinate plane. The coordinates of the four vertices are (0,0), (W,0), (W,H), and (0,H).
You are given a point (x,y) which is within the rectangle or on its border. We will draw a straight line passing through (x,y) to cut the rectangle into two parts. Find the maximum possible area of the part whose area is not larger than that of the other. Additionally, determine if there are multiple ways to cut the rectangle and achieve that maximum.

Constraints

1 \leq W,H \leq 10^9

0\leq x\leq W

0\leq y\leq H

All values in input are integers.

Input

Input is given from Standard Input in the following format:

W H x y

Output

Print the maximum possible area of the part whose area is not larger than that of the other, followed by 1 if there are multiple ways to cut the rectangle and achieve that maximum, and 0 otherwise.

The area printed will be judged correct when its absolute or relative error is at most 10^{-9}.

Sample Input 1

2 3 1 2

Sample Output 1

3.000000 0

The line x=1 gives the optimal cut, and no other line does.

Sample Input 2

2 2 1 1

Sample Output 2

2.000000 1

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        RemainderMinimization2019 solver = new RemainderMinimization2019();
        solver.solve(1, in, out);
        out.close();
    }

    static class RemainderMinimization2019 {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            HashSet<Long> set = new HashSet<>();
            long l = in.nextLong();
            long r = in.nextLong();

            if (r - l > 2019) {
                out.println(0);
                return;
            }
            long result = Long.MAX_VALUE;

            for (long i = l; i < r; i++) {
                for (long j = i + 1; j <= r; j++) {
                    result = Math.min(result, i * j % 2019);
                }
            }
            out.println(result);
        }

    }
}