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: Accepted)

import java.util.Scanner;
import java.util.ArrayList;

public class Main {
  public static void main(String[] args) {
    // 標準入力取得
    Scanner sc = new Scanner(System.in);    
    String iShortNeedle = sc.next();
    String iLongNeedle = sc.next();
    String h = sc.next();
    String m = sc.next();
    // 入力値加工
    double hour = Double.parseDouble(h);
    double min = Double.parseDouble(m);
    double wholeMin = (hour*60) + min;
    double shortNeedle = Double.parseDouble(iShortNeedle);
    double longNeedle = Double.parseDouble(iLongNeedle);

// check
//    System.out.println("shortNlength:" + shortNeedle + "cm");
//    System.out.println("longNlength:" + longNeedle + "cm");
//    System.out.println("hour:" + hour + "H");
//    System.out.println("min:" + min + "min");
//    System.out.println("wholeMin: "+ wholeMin + "min");

    double shortAngleByOneMin = (360.0/12.0)/60.0;
    double longAngleByOneMin = 360.0/60.0;
//    System.out.println("1分で短針が動く度数: "+ shortAngleByOneMin + "°");
//    System.out.println("1分で長針が動く度数: "+ longAngleByOneMin + "°");
//    System.out.println("XX分で短針が動く度数: "+ shortAngleByOneMin * wholeMin);
//    System.out.println("XX分で長針が動く度数: "+ longAngleByOneMin * min);

    // 各針の角度を計算する
    double rHour = 0; 
    double rMin = 0;
    if(wholeMin!=0){
      rHour = shortAngleByOneMin * wholeMin; 
    }
    if(min!=0){
      rMin = longAngleByOneMin * min;
    }
    double diffAngle = Math.abs(rHour - rMin);    

// check
//    System.out.println("rHour: " + rHour + "°");
//    System.out.println("rMin: " + rMin + "°");
//    System.out.println("diffAngle: " + diffAngle + "°");

    double aLength = 0;
    if(diffAngle!=0 || diffAngle!=180){
	  aLength = Math.sqrt( Math.pow(shortNeedle,2) + Math.pow(longNeedle,2) - 2 * shortNeedle * longNeedle * Math.cos(Math.toRadians(diffAngle)) );
    }else if(diffAngle==0){
      aLength = longNeedle - shortNeedle;    
    }else if(diffAngle==180){
      aLength = longNeedle + shortNeedle;
    }
    
//answer
    System.out.println(aLength);
  }
}