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.io.*;
import java.util.*;
class Main{
	static final long MOD = 1_000_000_007; // 10^9+7
    static final int MAX = 2_147_483_646; // intMax 
    static final int INF = 1_000_000_000; // 10^9  
    public static void main(String[] args) throws Exception {
		hayami saori = new hayami();
		int w = saori.saori_hayami();
		int h = saori.saori_hayami();
		int x = saori.saori_hayami();
		int y = saori.saori_hayami();
		int wmax = Math.max(x, w-x)*h;
		int hmax = Math.max(y, h-w)*x;
		if(hmax == wmax){
			System.out.println(wmax + " " + 1);
		} else {
			System.out.println(Math.max(wmax, hmax) + " " + 0);
		}

		
		saori.close();
    }
}

class hayami implements Closeable {
	private final InputStream in = System.in;
	private final byte[] hayami = new byte[1024];
	private int Hayami = 0;
	private int saori = 0;
	private boolean HayamiSaori() {
		if (Hayami < saori) {
			return true;
		}else{
			Hayami = 0;
			try {
				saori = in.read(hayami);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (saori <= 0) {
				return false;
			}
		}
		return true;
	}
	private int SaoriHayami() { 
		if (HayamiSaori()) {
            return hayami[Hayami++];
         }else{
             return -1;
         }
	}
	private static boolean hayami_saori(int hayami) { 
		return 33 <= hayami && hayami <= 126;
	}
	public boolean hayamisaori() { 
		while(HayamiSaori() && !hayami_saori(hayami[Hayami])) Hayami++; return HayamiSaori();
	}
	public String nextHayami() {
		if (!hayamisaori()) throw new NoSuchElementException();
		StringBuilder hayamin = new StringBuilder();
		int saori = SaoriHayami();
		while(hayami_saori(saori)) {
			hayamin.appendCodePoint(saori);
			saori = SaoriHayami();
		}
		return hayamin.toString();
	}
	public long saorihayami() {//nextLong
		if (!hayamisaori()) throw new NoSuchElementException();
		long hayami = 0;
		boolean misao = false;
		int saori = SaoriHayami();
		if (saori == '-') {
			misao = true;
			saori = SaoriHayami();
		}
		if (saori < '0' || '9' < saori) {
			throw new NumberFormatException();
		}
		while(true){
			if ('0' <= saori && saori <= '9') {
				hayami *= 10;
				hayami += saori - '0';
			}else if(saori == -1 || !hayami_saori(saori)){
				return misao ? -hayami : hayami;
			}else{
				throw new NumberFormatException();
			}
			saori = SaoriHayami();
		}
	}
	public int saori_hayami() {//nextInt
		long hayami = saorihayami();
		if (hayami < Integer.MIN_VALUE || hayami > Integer.MAX_VALUE) throw new NumberFormatException();
		return (int) hayami;
	}
	public double Hayamin() { //nextDouble
		return Double.parseDouble(nextHayami());
	}
	public void close() {
		try {
			in.close();
		} catch (IOException e) {
		}
    }
    
}