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 Rust (1.42.0) (judge verdict: Accepted)

#[allow(unused_imports)]
use std::cmp::max;
#[allow(unused_imports)]
use std::cmp::min;
#[allow(unused_imports)]
use std::collections::HashMap;
#[allow(unused_imports)]
use std::f64::consts::PI;
#[allow(unused_imports)]
use std::fmt;
#[allow(unused_imports)]
use std::fmt::Debug;
#[allow(unused_imports)]
use std::io;
#[allow(unused_imports)]
use std::str::FromStr;

fn main() {
    let input = get_vec_input();
    let a: f64 = input[0];
    let b: f64 = input[1];
    let h: f64 = input[2];
    let m: f64 = input[3];

    let a_angle = ((h + m / 60.0) / 12.0) * 360.0;
    let b_angle = (m / 60.0) * 360.0;

    let diff_angle = (a_angle - b_angle).abs();
    let diff_angle = if diff_angle > 180.0 {
        360.0 - diff_angle
    } else {
        diff_angle
    };
    if diff_angle == 180.0 {
        println!("{}", a + b);
    } else {
        let answer = (a * a + b * b - 2.0 * a * b * get_radian_from(diff_angle).cos()).sqrt();
        println!("{}", answer);
    }
}

#[allow(dead_code)]
fn get_radian_from(degree: f64) -> f64 {
    (degree * PI) / 180.0
}

#[allow(dead_code)]
fn get_alphabet() -> Vec<char> {
    (0..26).map(|x| (x + b'a') as char).collect::<Vec<char>>()
}

#[allow(dead_code)]
fn get_binary_string(digit: usize) -> Vec<String> {
    (0..2i64.pow(digit as u32))
        .map(|x| format!("{:0>1$b}", x, digit))
        .collect::<Vec<String>>()
}

#[allow(dead_code)]
fn check_prime(value: i64) -> bool {
    if value == 2 {
        return true;
    }
    if value % 2 == 0 {
        return false;
    }
    for i in 3..(value as f64).sqrt().floor() as i64 {
        if value % i == 0 {
            return false;
        }
    }
    return true;
}

#[allow(dead_code)]
fn get_digit(value: i64) -> i64 {
    let mut digit = 0i64;
    let mut value = value;
    while value != 0 {
        value /= 10;
        digit += 1;
    }
    digit
}

#[allow(dead_code)]
fn sum_digit2(value: String) -> u32 {
    return value.chars().fold(0u32, |a, b| a + b.to_digit(10).unwrap());
}

#[allow(dead_code)]
fn sum_digit(value: i64) -> i64 {
    if value < 10 {
        return value;
    }
    return sum_digit(value / 10) + value % 10;
}

#[allow(dead_code)]
fn update_max<T>(left: &mut T, right: T)
where
    T: std::cmp::Ord + std::clone::Clone,
{
    if *left < right {
        *left = right.clone();
    }
}

#[allow(dead_code)]
fn update_min<T>(left: &mut T, right: T)
where
    T: std::cmp::Ord + std::clone::Clone,
{
    if *left > right {
        *left = right.clone();
    }
}

#[allow(dead_code)]
fn get_vec_input<T>() -> Vec<T>
where
    T: FromStr,
    T::Err: Debug,
{
    let mut input = String::new();
    io::stdin()
        .read_line(&mut input)
        .expect("Failed to read line");
    let vec: Vec<T> = input
        .split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect();
    vec
}

#[allow(dead_code)]
fn get_tuple_input() -> (u64, f32) {
    let mut s = String::new();
    io::stdin().read_line(&mut s).expect("");
    let mut iter = s.trim().split_whitespace();
    let c: u64 = iter.next().unwrap().parse().unwrap();
    let n: f32 = iter.next().unwrap().parse().unwrap();
    (c, n)
}

#[allow(dead_code)]
fn get_string_input() -> String {
    let mut s = String::new();
    io::stdin().read_line(&mut s).expect("");
    s.trim().to_string()
}

/// https://github.com/bluss/permutohedron/blob/master/src/lexical.rs
pub trait LexicalPermutation {
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the last ordered permutation.
    fn next_permutation(&mut self) -> bool;
}

impl<T> LexicalPermutation for [T]
where
    T: Ord,
{
    /// Original author in Rust: Thomas Backman <serenity@exscape.org>
    fn next_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 {
            return false;
        }

        // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i - 1] >= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the last-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Find the rightmost element larger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i - 1] {
            j -= 1;
        }

        // Step 3: Swap that element with the pivot
        self.swap(j, i - 1);

        // Step 4: Reverse the (previously) weakly decreasing part
        self[i..].reverse();

        true
    }
}