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.math.*;
import java.util.*;

public class Main {

	public static void main(String[] args){
		FastScanner sc = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		solve(sc,out);

		out.flush();
	}

	public static void solve(FastScanner sc, PrintWriter out) {
		double w = sc.nextDouble();
		double h = sc.nextDouble();
		double x = sc.nextDouble();
		double y = sc.nextDouble();
		
		double a = 0;
		double b = 0;
		
		a = Math.min((h - y) * w, y * w);
		b = Math.min((w - x) * h, x * h);
		
		if((h == y && w == x) || (x == w && y == 0) || (x == 0 && y == h) || (x == 00 && y == 0)) {
			out.println((h*w*1d/2d) + " " + 1);
		}else{
			out.println(new BigDecimal(Math.max(a, b)).toPlainString()+" "+ (a == b ? 1 : 0));
		}
	}

	public static long gcd(long a, long b) {
		if(a < b) return gcd(b, a);
		if(b == 0) return a;
		return gcd(b, a % b);
	}

	public static long lcm(long a, long b) {
		return a / gcd(a, b) * b;
	}
}

class Pair<K, V> extends AbstractMap.SimpleEntry<K, V> {

    public Pair(K key, V value) {
        super(key, value);
    }

    public Pair<V, K> swap() {
    	return new Pair<V, K>(getValue(), getKey());
    }
 }

class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;

    public FastScanner(InputStream in) {
        reader = new BufferedReader(new InputStreamReader(in));
        tokenizer = null;
    }

    public String next() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public String nextLine() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }

        return tokenizer.nextToken("\n");
    }

    public long nextLong() {
        return Long.parseLong(next());
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public double nextDouble() {
         return Double.parseDouble(next());
     }

    public int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = nextInt();
        return a;
    }

    public long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++)
            a[i] = nextLong();
        return a;
    }
}