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 Java (OpenJDK 1.8.0) (judge verdict: Accepted)

// created on : Tue Jun 23 2020
 
import java.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException {
        int a = in.nextInt();
        int b = in.nextInt();
        int h = in.nextInt();
        int m = in.nextInt();

        double angle = 30*h - 5.5*m;       // degrees

        // a^2 = b^2 + c^2 + 2bc cos(angle between b and c)
        double ans = a * a + b * b - 2 * a * b * (Math.cos(Math.toRadians(angle)));
        println(Math.sqrt(ans));

        in.close();
        out.close();
    }

    static int MIN = Integer.MIN_VALUE;
    static int MAX = Integer.MAX_VALUE;
    static Reader in = new Reader();
    static PrintWriter out = new PrintWriter(System.out);

    static int[] readInt(int N) throws IOException { 
    	int arr[] = new int[N];
    	for(int i = 0; i < N; i++){ 
    		arr[i] = in.nextInt(); 
    	} 
    	return arr;
    }
    
    static long[] readLong(int N) throws IOException { 
    	long arr[] = new long[N];
    	for(int i = 0; i < N; i++){ 
    		arr[i] = in.nextLong(); 
    	} 
    	return arr;
    }
    
    static void print(Object O) {  
    	out.print(O); 
    }
    
    static void println(Object O) { 
    	out.println(O); 
    }
    
    static void println(int arr[]) { 
    	for(int i = 0; i < arr.length; i++){ 
    		print(arr[i] + " "); 
    	} 
    	println(""); 
    }
    
    static void displayRuntime(long startTime){
        long stopTime = System.currentTimeMillis();
        long elapsedTime = stopTime - startTime;
        System.out.println("Runtime: " + elapsedTime / 1000D);
    }
}

class Reader {
    BufferedReader reader;
    StringTokenizer tokenizer;

    Reader() {
        reader = new BufferedReader(new InputStreamReader(System.in));
        tokenizer = new StringTokenizer("");
    }

    String next() throws IOException {
        while (!tokenizer.hasMoreTokens() ) { 
            tokenizer = new StringTokenizer(reader.readLine()); 
        }
        return tokenizer.nextToken();
    }

    int nextInt() throws IOException { 
        return Integer.parseInt(next()); 
    }
    
    double nextDouble() throws IOException { 
        return Double.parseDouble(next());
    }
    
    long nextLong() throws IOException { 
        return Long.parseLong(next()); 
    }
    
    String nextLine() throws IOException { 
        return reader.readLine(); 
    }
    
    void close() throws IOException { 
        reader.close(); 
    }
}