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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		double w = sc.nextInt();
		double h = sc.nextInt();
		double x = sc.nextInt();
		double y = sc.nextInt();
		if ((x == 0 && y == 0) || (x == w && y == h)) {
			double min = 0;
			System.out.print(String.format("%.9f", min));
			System.out.print(" " + 1);
		} else if (x == 0 || y == 0) {
			double min = 0;
			if (x == 0) {
				double min1 = y * w;
				double min2 = (h - y) * w;
				if (min1 > min2) {
					min = min2;
				} else {
					min = min1;
				}
			} else {
				double min1 = x * h;
				double min2 = (w - x) * h;
				if (min1 > min2) {
					min = min2;
				} else {
					min = min1;
				}
			}
			System.out.print(String.format("%.9f", min));
			System.out.print(" " + 0);
		} else {
			//縦
			double min0 = 0;
			double min1 = x * h;
			double min2 = (w - x) * h;
			if (min1 < min2) {
				min0 = min1;
			} else {
				min0 = min2;
			}
			//横
			double min5 = 0;
			double min3 = y * w;
			double min4 = (h - y) * w;
			if (min3 < min4) {
				min5 = min3;
			} else {
				min5 = min4;
			}
			if (min5 < min0) {
				System.out.print(String.format("%.9f", min0));
			} else {
				System.out.print(String.format("%.9f", min5));
			}
			if (min0 == min5) {
				System.out.print(" " + 1);
			} else {
				System.out.print(" " + 0);
			}

		}
	}
}