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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
 
#ifndef bhartiya
    #pragma GCC optimize("Ofast")
    #pragma GCC optimize("unroll-loops")
    // #pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#endif
 
 
typedef long long ll;
typedef pair<ll, ll> pll;
typedef gp_hash_table<long long, long long> umap;
typedef tree<int, null_type, less<int>, rb_tree_tag,
		tree_order_statistics_node_update> oset;
// not that imp
typedef pair<pll, ll> plll;
typedef vector<ll> vl;
typedef vector<pll> vll;
 
#define inf 200000000000000ll
#define mod 1000000007ll
#define eps 1e-7
#define PI 3.1415926535897932385
// #define PI acos(-1)
 
#define pb push_back
#define bitc  __builtin_popcountll
#define mp make_pair
#define ff first
#define ss second
#define all(ar) ar.begin(), ar.end()
 
#define fr(i,a,b) for (ll i = (a), _b = (b); i <= _b; i++)
#define rep(i,n) for (ll i = 0, _n = (n); i < _n; i++)
#define repr(i,n) for (ll i = n - 1; i >= 0; i--)
#define frr(i,a,b) for (ll i = (a), _b = (b); i >= _b; i--)
#define foreach(it,ar) for (auto it = ar.begin(); it != ar.end(); it++)
#define fill(ar,val) memset(ar, val, sizeof(ar))
 
#ifdef bhartiya
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
	cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
	const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
 
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
inline bool equals(double a, double b) {return fabs(a - b) < 1e-9;}
ll gcd(ll a, ll b){ return b==0 ? a : gcd(b, a%b); }
 
const ll maxn = 500005;
ll n;
 
int main() 
{ 
	ios_base::sync_with_stdio(false);
    cin.tie(0), cout.tie(0), cout.precision(10); //cout<<fixed;
    cin.exceptions(cin.failbit);
    #ifdef bhartiya
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
    int t = 1;
    // cin>>t;
    while(t--){
        ll w,h,x,y;
        cin>>w>>h>>x>>y;
        double ans = w*h;
        ans/=2.0;
        cout<<ans<<" ";
        if(x*2 == w && y*2 == h){
            cout<<1;
        }
        else{
            cout<<0;
        }
    }
    end_routine();
}