Problem:
Score: 300 points

Problem Statement

Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.

An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.

At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?

Constraints

All values in input are integers.

1 \leq A, B \leq 1000

0 \leq H \leq 11

0 \leq M \leq 59

Input

Input is given from Standard Input in the following format:

A B H M

Output

Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.

Sample Input 1

3 4 9 0

Sample Output 1

5.00000000000000000000

The two hands will be in the positions shown in the figure below, so the answer is 5 centimeters.

Sample Input 2

3 4 10 40

Sample Output 2

4.56425719433005567605

The two hands will be in the positions shown in the figure below. Note that each hand always rotates at constant angular velocity.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

/*
 *  Created on: 17 / 5 / 2020
 *  Author:     Eslam Waheed
 */
#include <bits/stdc++.h>
#include <cmath>
//#include <bits/BIGINT.h>
//#include <bits/Int128.h>
using namespace std;

//#define _CRT_SECURE_NO_WARNINGS
//#define __USE_MINGW_ANSI_STDIO
#define Eslam__Waheed ios_base::sync_with_stdio(false);cout.tie(NULL);cin.tie(NULL);

#define ll   long long
#define ull  unsigned ll
#define dd   double
#define ld   long dd
#define ss   string

#define pb      push_back
#define all(c)  c.begin(), c.end()
#define rall(c) c.rbegin(), c.rend()
#define stp(n)  cout << fixed << setprecision(n);
#define iq(v)   v.resize(unique(v.begin(),v.end())-v.begin())
#define sz(s)   (ll)(s.size())
#define endl    '\n'

#define loop(i,m,n)  for(ll i=m;i<n;i++)

ll gcd(ll a, ll b) { return !b ? abs(a) : gcd(b, a % b); }
ll lcm(ll a, ll b) { return abs(a / gcd(a, b)) * b; }

const ld PI = acos(-1);
//const ll mod = 1e15 + 7;
//__int128
/*
 you can ^_^ (•‿•)

( إن الله وملائكته يصلون على النبي يا أيها الذين آمنوا صلوا عليه وسلموا تسليما )
*/
int main()
{
    stp(20)
    Eslam__Waheed
    ld a, b, h, m; cin >> a >> b >> h >> m;
    if (m > h)
    {
        swap(m, h);
    }
    ll x = (12 - h) + m;
    ll y = h - m;
    dd z = min(x, y) * 30;
    //cout << cos(z) << endl;
    ld ans = sqrt((a * a) + (b * b) - (2 * a * b * cos((z/180)*PI)));
    cout << ans;
}