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# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Text;

class Util{
	static void Main(){
		Sol mySol =new Sol();
		mySol.Solve();
	}
}

class Sol{
	public void Solve(){
		double[] ABHM = rda();
        double A = ABHM[0];
        double B = ABHM[1];
        double H = ABHM[2];
        double M = ABHM[3];

        double angleA = (H + M / 60.0) / 12.0;
        double angleB = M / 60.0;
        double angleDiff = 2.0 * Math.PI * (angleA - angleB);

        double ans;
        // if(angleDiff == 0) ans = 0;
        // else if(angleDiff == Math.PI) ans = A + B;
        // else 
        ans = Math.Sqrt(A * A + B * B - 2.0 * A * B * Math.Cos(angleDiff));

		Console.WriteLine(ans);
		Console.ReadLine();
	}

	static String rs(){return Console.ReadLine();}
	static int ri(){return int.Parse(Console.ReadLine());}
	static long rl(){return long.Parse(Console.ReadLine());}
	static double rd(){return double.Parse(Console.ReadLine());}
	static String[] rsa(char sep=' '){return Console.ReadLine().Split(sep);}
	static int[] ria(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>int.Parse(e));}
	static long[] rla(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>long.Parse(e));}
	static double[] rda(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>double.Parse(e));}
    static int[] generateNums(int num, int N){return Enumerable.Repeat(num, N).ToArray();}
    static int[] generateSuretsu(int N){return Enumerable.Range(0, N).ToArray();}
}

public static class Combination {
	//使い方:int[][] Is = Combination.Enumerate(nums, k, withRepetition:false).ToArray();
    public static IEnumerable<T[]> Enumerate<T>(IEnumerable<T> items, int k, bool withRepetition) {
        if (k == 1) {
            foreach (var item in items)
                yield return new T[] { item };
            yield break;
        }
        foreach (var item in items) {
            var leftside = new T[] { item };

            // item よりも前のものを除く （順列と組み合わせの違い)
            // 重複を許さないので、unusedから item そのものも取り除く
            var unused = withRepetition ? items : items.SkipWhile(e => !e.Equals(item)).Skip(1).ToList();

            foreach (var rightside in Enumerate(unused, k - 1, withRepetition)) {
                yield return leftside.Concat(rightside).ToArray();
            }
        }
    }
}