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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define ld long double
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
using P=pair<long long,long long>;
#define rep(i,n) for(long long i=0; i<(long long)n; i++)
#define FOR(i,a,b) for(long long i=a; i<b; i++)
#define all(x) (x).begin(), (x).end()
#define SZ(x) ((long long)(x).size())
#define COUT(x) cout << x << endl
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define vint vector<int>
#define vvint vector<vector<int>>
#define vstr vector<string>
#define vp vector<pair<int,int>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define SUM(x) accumulate(x.begin(),x.end(),0)
#define MAX(x) *max_element(x.begin(),x.end())
#define MIN(x) *min_element(x.begin(),x.end())
#define couty cout<<"Yes"<<endl
#define coutn cout<<"No"<<endl
#define coutY cout<<"YES"<<endl
#define coutN cout<<"NO"<<endl
#define yn(x) cout<<(x?"Yes":"No")<<endl
#define YN(x) cout<<(x?"YES":"NO")<<endl
long long gcd(long long a,long long b){return b?gcd(b,a%b):a;}
long long lcm(long long a, long long b){return a*b/gcd(a,b);}
const long long dx[4]={1,0,-1,0};
const long long dy[4]={0,1,0,-1};
const long long INF = 1e12;
const long long MOD = 1e9+7;

signed main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout<<fixed<<setprecision(15);

    ld w,h,x,y; cin >> w >> h >> x >> y;
    cout<<w*h/2<<' '<<(x*2==w && y*2==h)<<endl;
}