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: Accepted)

import java.util.*;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;

import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
 
public class Main{
    static long mod=1000000007;
//    static int dx[]={-1,0,1,0};
//    static int dy[]={0,-1,0,1};

    public static void main(String[] args)   throws Exception, IOException{        
        Reader sc = new Reader(System.in);
        PrintWriter out=new PrintWriter(System.out);

//        int n = sc.nextInt();
        long w=sc.nextLong();
        long h=sc.nextLong();
        long x=sc.nextLong();
        long y=sc.nextLong();
        double ans=(double)w*h/2;
        int b=0;
        
        if(w%2==0 && h%2==0 && w/2==x && h/2==y)b=1; 
        
        out.println(ans+" "+b);
        out.flush();
    }

    static boolean validpos(int x,int y,int r, int c){
        return x<r && 0<=x && y<c && 0<=y;
    }

    static void db(Object... os){
        System.err.println(Arrays.deepToString(os));
    }
}
//class P {
//    int id, d;
//    P(int  id, int d) {
//        this.id=id;
//        this.d=d;
//    }
//}

//class P implements Comparable<P>{
//    int a,b;
//    double c;
//    P(int a, int b) {
//    	this.b=b;
//    	this.a=a;
//    	this.c=(double)b/a;
//    }
//
//    public int compareTo(P p){
//        return Double.compare(p.c,c);
//    }
//}

class Reader
{ 
    private BufferedReader x;
    private StringTokenizer st;
    
    public Reader(InputStream in)
    {
        x = new BufferedReader(new InputStreamReader(in));
        st = null;
    }
    public String nextString() throws IOException
    {
        while( st==null || !st.hasMoreTokens() )
            st = new StringTokenizer(x.readLine());
        return st.nextToken();
    }
    public int nextInt() throws IOException
    {
        return Integer.parseInt(nextString());
    }
    public int[] nextIntArray(int size) throws IOException{
        int r[] = new int[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextInt(); 
        }
        return r;
    }
    public long[] nextLongArray(int size) throws IOException{
        long r[] = new long[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextLong(); 
        }
        return r;
    }
    public long nextLong() throws IOException
    {
        return Long.parseLong(nextString());
    }
    public double nextDouble() throws IOException
    {
        return Double.parseDouble(nextString());
    }
}