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 11.0.6) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int a = getInt(sc);
		int b = getInt(sc);
		int h = getInt(sc);
		int m = getInt(sc);

		double angleA = (double) (360.0 / 720.0) * ((double) h * 60.0 + m);
		double angleB = (double) (360.0 / 60.0) * (double) m;
		double angle = Math.abs(angleA - angleB);
		if (angle == 0) {
			System.out.println(Math.abs(b - a));
		} else if (angle == 180) {
			System.out.println(a + b);
		} else {
			double cos = Math.abs(Math.cos(Math.toRadians(angle)));
			System.out.println(Math.sqrt(a * a + b * b - 2.0 * a * b * cos));
		}

	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をlongで返却
	 *
	 * @return
	 */
	public static long getLong(Scanner sc) {

		return Long.parseLong(sc.next());
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をintで返却
	 *
	 * @return
	 */
	public static int getInt(Scanner sc) {

		return Integer.parseInt(sc.next());
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をStringで返却
	 *
	 * @return
	 */
	public static String getString(Scanner sc) {

		return sc.next();
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をint配列で返却
	 *
	 * @return
	 */
	public static int[] getIntArray(Scanner sc) {
		String line = "";
		// 直前にnext()を実行していると改行コードだけ取ってしまうのでその対策
		while (line.equals("")) {
			line = sc.nextLine();
		}

		String[] strArray = line.split(" ");
		int[] intArray = new int[strArray.length];
		for (int i = 0; i < strArray.length; i++) {
			intArray[i] = Integer.parseInt(strArray[i]);
		}
		return intArray;
	}

}