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)

import java.io.*;
import java.util.*;
import java.lang.*;

class Main {

	public static void main(String[] args) {
		final FastScanner sc = new FastScanner(System.in);
		double a = sc.nextInt();
		double b = sc.nextInt();
		double h = sc.nextInt();
		double m = sc.nextInt();
		double arad = -h * Math.PI / 6.0 - (m * Math.PI / 360) + Math.PI / 2;
		double brad = -m * Math.PI / 30.0 + Math.PI / 2;

		double ax = a * Math.cos(arad);
		double ay = a * Math.sin(arad);

		double bx = b * Math.cos(brad);
		double by = b * Math.sin(brad);

		double dx = ax - bx;
		double dy = ay - by;
		System.out.println(Math.sqrt(dx * dx + dy * dy));
	}

	static class FastScanner {
		private final InputStream in;
		private final byte[] buffer = new byte[1024];
		private int ptr = 0;
		private int buflen = 0;

		public FastScanner(InputStream source) {
			in = source;
		}

		private boolean hasNextByte() {
			if (ptr < buflen) {
				return true;
			} else {
				ptr = 0;
				try {
					buflen = in.read(buffer);
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (buflen <= 0) {
					return false;
				}
			}
			return true;
		}

		private int readByte() {
			if (hasNextByte()) return buffer[ptr++];
			else return -1;
		}

		private static boolean isPrintableChar(int c) {
			return 33 <= c && c <= 126;
		}

		public boolean hasNext() {
			while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
			return hasNextByte();
		}

		public String next() {
			if (!hasNext()) throw new NoSuchElementException();
			StringBuilder sb = new StringBuilder();
			int b = readByte();
			while (isPrintableChar(b)) {
				sb.appendCodePoint(b);
				b = readByte();
			}
			return sb.toString();
		}

		public long nextLong() {
			if (!hasNext()) throw new NoSuchElementException();
			long n = 0;
			boolean minus = false;
			int b = readByte();
			if (b == '-') {
				minus = true;
				b = readByte();
			}
			if (b < '0' || '9' < b) {
				throw new NumberFormatException();
			}
			while (true) {
				if ('0' <= b && b <= '9') {
					n *= 10;
					n += b - '0';
				} else if (b == -1 || !isPrintableChar(b)) {
					return minus ? -n : n;
				} else {
					throw new NumberFormatException();
				}
				b = readByte();
			}
		}

		public int nextInt() {
			long nl = nextLong();
			if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
				throw new NumberFormatException();
			return (int) nl;
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}