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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<iostream>
#include<stdio.h>
#include<vector>//s1.erase(n)=文字列s1の、n文字目以降削除
#include<algorithm>//辞書順=next_permutation
#include<set>
#include<string>//char 小文字のほうが32大きい
#include<sstream>
#include<string.h>
#include<complex>
#include<time.h>
#include<random>
#include<cmath>
#include<math.h>
#include<queue>
#include<time.h>
#include<iomanip>
#include<locale>
#include<utility>//swap=数値交換

#define rt "\n"
#define rep(i,n) for(int i=0;i<n;i++)
#define rop(i,n) for(int i=1;i<=n;i++)
#define drep(i,n) for(int i=n-1;0<=i;i--)
#define drop(i,n) for(int i=n;0<i;i--)
#define yes(ans) if(ans)cout<<"yes"<<rt;else cout<<"no"<<rt;
#define Yes(ans) if(ans)cout<<"Yes"<<rt;else cout<<"No"<<rt;
#define YES(ans) if(ans)cout<<"YES"<<rt;else cout<<"NO"<<rt;
#define sec(a,b,ans) if(ans)cout<<a<<rt;else cout<<b<<rt;
#define vcin(s,n) rep(i,n)cin>>s[i];//vcin(配列名),(繰り返し回数)
#define sort(s) sort(s.begin(),s.end())//標準=昇順
#define reve(s) reverse(s.begin(),s.end())
#define asas int ans=0
#define llcncn llint cnt=0
#define llasas llint ans=0
#define str string
#define please return
#define AC 0
#define ic_a int a;cin>>a;
#define ic_b int b;cin>>b;
#define ic_s string s;cin>>s;
#define Rapid_pleaseAC_fast cin.tie(0);ios::sync_with_stdio(false)
#define Pi 3.1415926535897932384626
#define nine 1000000000
using ll = long long;
using namespace std;
typedef vector<int> vint;
typedef vector<string> vstr;
typedef vector<char> vchar;
typedef vector<double> vdou;
typedef long long int llint;
typedef pair<int, int> pint;
typedef pair<llint, llint> pllint;
typedef vector<llint> vllint;
typedef vector<pint> vpint;
typedef vector<pair<llint, llint>> vpllint;
typedef vector<vector<int>> vvint;
typedef vector<vector<char>> vvchar;
typedef vector<vector<llint>> vvllint;
typedef vector<vector<string>> vvstr;
typedef vector<vector<bool>> vvbool;
typedef vector<vector<pint>> vvpint;
typedef vector<bool> vbool;
typedef long double ld;
const ld eps = 1.0e-12; // 許容される誤差
long long GCD(long long a, long long b) {
	if (b == 0) return a;
	else return GCD(b, a % b);
}

long long LCM(long long a, long long b) {
	return a * b / GCD(a, b);
}

unsigned GetDigit(unsigned num) {
	return std::to_string(num).length();
}

int tow(int n) {//2のn乗
	if (n == 0)return 1;
	int x = tow(n / 2);
	x *= x;
	if (n % 2 == 1)x *= 2;
	return x;//@domino
}

int keta(int n) {
	int sum = 0;
	while (n != 0) {
		sum += n % 10; n /= 10;
	}
	return sum;//Nanashi,tRue,wai
}

/*

 (char)toupper(a[n])=文字列のn文字目を大文字で出力

  pow(a,b)=aのb乗

  */
ld mysqrtl(ld x)
{
	ld a = sqrtl((double)x); // 近似値
	do {
		a = (a + x / a) / 2.0L;
	} while (fabsl(x - a * a) > eps);
	return a;
}
ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a % b);

}

ll lcm(ll a, ll b) {
	ll g = gcd(a, b);
	return a / g * b;
}

//平方根を求める。sqrtより精度が高い。
//epsを変えて誤差を調整する。
int main(void) {
	Rapid_pleaseAC_fast;
	llint ans = 0, n = 0,  count = 0, sum = 0,old=0;
    long double  a, b, h,m,dans,mkaku,hkaku,kaku1,kaku2,kaku,rad;
	bool check = false;
	string ss, sss;
	vector<llint> array;
	vector<bool> bo;
	cin >> a>>b>>h>>m;
	vector<llint> counter,rust;
	rad = (h / 12.0 + (m / 60.0) / 12.0 - m / 60.0) * 2 * 3.14159265358979323846;
	dans = sqrtl(a * a + b * b - 2 * a * b * cosl(rad));
	
	cout << std::fixed << std::setprecision(15) << dans << endl;
	
	please AC;
}