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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<string>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define SORT(v,n) sort(v,v+n)
#define rep(i,n) FOR(i,0,n)
#define EPS (1e-7)
#define INF (1e9)

using namespace std;
typedef long long int ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef pair<int,int> PI;
typedef pair<ll,ll> PL;
typedef priority_queue<int> PQI;
typedef priority_queue<int,vector<int>,greater<int> > PQSI;
typedef priority_queue<ll> PQL;
typedef priority_queue<ll,vector<ll>,greater<ll> > PQSL;

const int MOD = 1000000007;
struct mint { int n; mint(int n_ = 0) : n(n_) {} };
mint operator-(mint a) { return -a.n + MOD * (a.n != 0); }
mint operator+(mint a, mint b) { int x = a.n + b.n; return x - (x >= MOD) * MOD; }
mint operator-(mint a, mint b) { int x = a.n - b.n; return x + (x < 0) * MOD; }
mint operator*(mint a, mint b) { return (long long)a.n * b.n % MOD; }
mint &operator+=(mint &a, mint b) { return a = a + b; }
mint &operator-=(mint &a, mint b) { return a = a - b; }
mint &operator*=(mint &a, mint b) { return a = a * b; }
istream &operator>>(istream &i, mint &a) { return i >> a.n; }
ostream &operator<<(ostream &o, mint a) { return o << a.n; }

vector<mint> F_{1, 1}, R_{1, 1}, I_{0, 1};
void check_fact(int n) {
  for (int i = I_.size(); i <= n; i++) {
    I_.push_back(I_[MOD % i] * (MOD - MOD / i));
    F_.push_back(F_[i - 1] * i);
    R_.push_back(R_[i - 1] * I_[i]);
  }
}

mint I(int n) { check_fact(abs(n)); return n >= 0 ? I_[n] : -I_[n]; }
mint F(int n) { check_fact(n); return n < 0 ? 0 : F_[n]; }
mint R(int n) { check_fact(n); return n < 0 ? 0 : R_[n]; }
mint C(int n, int r) { return F(n) * R(n - r) * R(r); }
mint P(int n, int r) { return F(n) * R(n - r); }
mint H(int n, int r) { return n == 0 ? (r == 0) : C(n + r - 1, r); }

template<typename T>
T gcd(T x, T y){
    if (y == 0) return x;
    else return gcd(y, x % y);
}

template<typename T>
T lcm(T x, T y){
    return (x / gcd(x, y)) * y;
}

template<typename T>
T ceil(T x, T y){
    return (x + (y - 1)) / y;
}

ll W1,H1,x,y;
int main(void){
    cin >> W1 >> H1 >> x >> y;
    cout << W1*H1/2.0 << " ";
    if(abs(x-W1/2.0) < EPS and abs(y-H1/2.0) < EPS) cout << 1 << endl;
    else cout << 0 << endl;

    return 0;
}