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 <bits/stdc++.h>

#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define sz(a) (ll)(a).size()
#define last(a) (a).at((ll)(a).size()-1)
#define rep(i,n) for(ll i=0,i##_len=(n);i<i##_len;i++)
#define brep(i,a,b) for(ll i=(a),i##_len=(b);i<=i##_len;i++)
#define rrep(i,n) ll i=(n);ll now=-1;while(i--&&(now++||1))
#define xrep(i,n) for(ll i=1,i##_len=(n);i<=i##_len;i++)
#define Yes(n) cout<<((n)?YES:NO)<<'\n'
#define co(n) cout<<(n)<<'\n'
#define Sort(v) sort(all(v))
#define rSort(v) sort(all(v),greater<typeof(v[0])>())
#define Rev(v) reverse(all(v))
#define Unique(v) (v).erase(unique(all(v)),(v).end())
#define eb emplace_back
#define fi first
#define se second
#define index f89dsjj4
#define next ytjbmy8k
#define prev j4qzjkys

using namespace std;
using ll = long long;
using vll = vector<ll>;
using vbl = vector<bool>;
using vint = vector<int>;
using vvll = vector<vll>;
using vstr = vector<string>;

constexpr ll INF = 1e18;

// --- functions which take 1 argument --- //
static inline void ignore_ret(int x){x++;}
template<class T> inline int sgn(T x){return (x>0)-(x<0);}
template<class T> inline ll factorial(T n){if(n==0)return 1;ll r=1;for(T i=2;i<=n;i++)r*=i;return r;}
template<class T> inline map<T,T> factorize(T n){map<T,T>r;for(T i=2;i*i<=n;i++){while(n%i==0){r[i]++;n/=i;}}if(n!=1)r[n]=1;return r;}
template<class T> inline vector<T> divisor(T n){vector<T>r;for(T i=1;i*i<=n;i++){if(!(n%i)){r.eb(i);if(i*i!=n)r.eb(n/i);}}Sort(r);return r;}

// --- functions which take 2 arguments --- //
template<class T> inline bool chmax(T&a,const T&b){if(a<b){a=b;return 1;}return 0;}
template<class T> inline bool chmin(T&a,const T&b){if(a>b){a=b;return 1;}return 0;}
template<class T> inline T nPr(T n,T k){if(n<k||n<0||k<0)return 0;T r=1;for(T i=n-k+1;i<=n;i++)r*=i;return r;}
template<class T> inline T nCr(T n,T k){if(n<k||n<0||k<0)return 0;T r=1;k=min(k,n-k);for(T i=n-k+1;i<=n;i++)r*=i;return r/factorial(k);}

// --- functions which take vector(s) as argument(s) --- //
template<class T> inline void print(T v){cout<<"[ ";for(auto x:v)cout<<x<<' ';cout<<"]\n";}
template<class T> inline void print_2d(vector<T>v){for(T x:v)print<T>(x);}
template<class T> inline void print_d(T v,string d){rep(i,(ll)v.size()-1)cout<<v[i]<<d;cout<<v[(ll)v.size()-1]<<'\n';}
template<class T> inline void print_2d_d(vector<T>v,string d){for(T x:v)print_d<T>(x,d);}
template<class T> inline T min(vector<T>v){return *min_element(all(v));}
template<class T> inline T max(vector<T>v){return *max_element(all(v));}
template<class T> inline ll sum(T v){return reduce(all(v),0LL);}
template<class T> inline T gcd(vector<T>v){T r=v[0];for(T i=1;i<v.size();i++)r=gcd(r,v[i]);return r;}
template<class T> inline T lcm(vector<T>v){T r=v[0];for(T i=1;i<v.size();i++)r=lcm(r,v[i]);return r;}
template<class T> inline double abs(vector<T> v){return sqrt(reduce(all(v),0.0,[](T s,T v){return s+=v*v;}));}
template<class T> inline T vector_add(T u,T v){T r;rep(i,u.size())r.eb(u[i]+v[i]);return r;}
template<class T> inline T vector_subtract(T u,T v){T r;rep(i,u.size())r.eb(u[i]-v[i]);return r;}
template<class T> inline T dot_product(vector<T>u,vector<T>v){if(u.size()<v.size())return 0;T r=0;for(T i=0;i<u.size();i++)r+=u[i]*v[i];return r;}
template<class T> inline T cross_product(T u,T v){T r(3);r[0]=u[1]*v[2]-u[2]*v[1];r[1]=u[2]*v[0]-u[0]*v[2];r[2]=u[0]*v[1]-u[1]*v[0];return r;}

// --- functions which take set(s) as argument(s) --- //
template<class T> inline T min(set<T>v){return *min_element(all(v));}
template<class T> inline T max(set<T>v){return *max_element(all(v));}
template<class T> inline T gcd(set<T>v){T r=0;for(T x:v)r=(r==0)?x:gcd(r,x);return r;}
template<class T> inline T lcm(set<T>v){T r=0;for(T x:v)r=(r==0)?x:lcm(r,x);return r;}

// --- Template ends here --- //

void solve(double A, double B, ll H, ll M) {
  double angle_h = (H + M / 60.0) * 30, angle_m = M * 6;
  double ans_sq = pow(A, 2) + pow(B, 2) - 2*A*B*cos((angle_h - angle_m) * 3.14159265358979 / 180.0);
  cout << fixed << setprecision(15) << sqrt(ans_sq) << '\n';
}

int main() {
  double A;
  double B;
  ll H;
  ll M;
  cin >> A >> B >> H >> M;
  solve(A, B, H, M);
  return 0;
}