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: Wrong Answer)

#include <bits/stdc++.h>
#include <stdio.h>

using namespace std;
#define inputInt(a) int a; cin >> a; 
#define inputInt2(a,b) int a; int b; cin >> a >> b; 
#define inputInt3(a,b,c) int a; int b; int c; cin >> a >> b >> c;
#define inputLong(a) long a;cin >> a; 
#define inputIntArray(a,N) int a[N];for(int i=0;i<N;i++){cin >> a[i];}
#define inputLongArray(a,N) long a[N];for(int i=0;i<N;i++){cin >> a[i];}
#define inputIntArray2(a,b,N) int a[N]; int b[N]; for(int i=0;i<N;i++){cin >> a[i] >> b[i];}
#define output(answer) cout << answer << endl;
#define fN(i,N)    for(int i=0; i<N; i++)
#define fSE(i,s,e) for(int i=s; i<=e; i++)
#define ll long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define FOR(i,N) for(int i=0; i<N; i++)
#define cinv(v,N) vector<int> v(N); for(int i=0; i<N; i++){ cin >> v[i]; }
#define mt make_tuple
#define vll vector<long long int>
#define vvll vector<vector<long long int>>
#define all(v) (v).begin(),(v).end()
#define show(v) FOR(i,v.size()){ cout << v[i] << " "; } cout << endl;
const ll MOD = 1e9+7;
const ll MOD2 = 998244353;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

//Combination計算
const int LIMIT = 1e6;//1e6→50ms,1e7→500ms,1e8→5000ms程度。
//const ll MOD = 1e9+7;
vector<ll> FacMod(LIMIT), InvMod(LIMIT), FacInvMod(LIMIT);
const int dpComTableLIMIT = 3e3+1;
vector<vector<ll>> Com(dpComTableLIMIT,vector<ll>(dpComTableLIMIT,0));
const int ComKTableLIMIT = 1e7;
vector<ll> ComK(ComKTableLIMIT);

void DPComInit(){
  //Com[0][j]は0で初期化されている。
  for(int i=0; i<dpComTableLIMIT; i++){ Com[i][0] = 1; }
  
  for(int i=1; i<dpComTableLIMIT; i++){
    for(int j=1; j<dpComTableLIMIT; j++){
      Com[i][j] = (Com[i-1][j-1]+Com[i-1][j])%MOD;
    }
  }
}

void ComKInit(ll N){
  for(int i=0; i<=ComKTableLIMIT; i++){
    if(0<i && i<=N/2){
      ComK[i] = (ComK[i-1]*(N-i+1)/i)%MOD;
    }else if(i==0){
      ComK[i] = 1;
    }else if(N/2<i && i<=N){
      ComK[i] = ComK[N-i];
    }else{//i>Nの時
      ComK[i] = -1;
    }
  }
}

void ComInit(){//tableを作成
  FacMod[0] = FacMod[1] = 1;
  InvMod[0] = 0; InvMod[1] = 1; //InvMod[0]は定義できない
  FacInvMod[0] = FacInvMod[1] = 1;
  
  for(int i=2; i<LIMIT; i++){
    FacMod[i] = FacMod[i-1]*i%MOD;
    InvMod[i] = -InvMod[MOD%i]*(MOD/i)%MOD + MOD;
    FacInvMod[i] = FacInvMod[i-1] * InvMod[i]%MOD;
  }
}

ll ComMod(int n, int k){
  if(n>=0 && k>=0 && n>=k){
    return FacMod[n]*(FacInvMod[k]*FacInvMod[n-k]%MOD)%MOD;
  }else{
    return 0; //戻り値を0にする方が計算結果に影響が少ない。
  }
}


//繰り返し2乗法
ll power(ll x, ll y){//繰り返し2乗法
  if(y < 0){
    return 0;
  }else if(y == 0){
    return 1LL;
  }else if(y%2 == 0){//偶数
    return power(x,y/2)*power(x,y/2);
  }else{//奇数
    return power(x,y-1)*x;
  }
}

//繰り返し2乗法
ll mpower(ll x, ll y, ll m){//繰り返し2乗法
  if(y < 0){
    return 0;
  }else if(y == 0){
    return 1LL;
  }else if(y%2 == 0){//偶数
    return mpower(x,y/2,m)*mpower(x,y/2,m)%m;
  }else{//奇数
    return mpower(x,y-1,m)*x%m;
  }
}

//pair型のソート
bool compare_by_second(pair<ll,ll> a, pair<ll,ll> b){
  return a.second != b.second ? a.second < b.second : a.first < b.first;
}

class UnionFind {
  vector<int> parent;
  vector<int> rank;
  
  public:
  
  UnionFind(int N);
  int root(int x);
  void unite(int x, int y);
  bool isUnited(int x, int y);
  int getRank(int x);
};

UnionFind::UnionFind(int N):parent(N),rank(N,1){
  for(int i=0; i<N; i++){parent[i]=i;}
}
  
int UnionFind::root(int x){
  if(parent[x] == x) return x;
  return root(parent[x]);
  /*parent[x] = parent[parent[x]];
  if(parent[x]!=parent[parent[x]]) rank[x]-=1;*/
}
  
void UnionFind::unite(int x, int y) { // xとyの木を併合
  int rx = root(x); //xの根をrx
  int ry = root(y); //yの根をry
  if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
  
  if(rank[x] >= rank[y]){parent[ry] = rx; rank[y]+=1;}//大きい木に小さい木が併合される
  else{parent[rx] = ry; rank[x]+=1;}//根に直接辺を張る
}
  
bool UnionFind::isUnited(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
  int rx = root(x);
  int ry = root(y);
  parent[x]=rx; parent[y]=ry;rank[x]=2;rank[y]=2;
  return rx == ry;
}

int UnionFind::getRank(int x){
  return rank[x];
}

int main()
{
  //input
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  ll A, B, H, M; cin >> A >> B >> H >> M;
  double ha = 360*H/12+30*M/60;
  double ma = 360*M/60;
  double angle = min(abs(ha-ma), 360-abs(ha-ma));
  
  //main 
  double ans = 0;
  double rangle = angle*2*acos(-1)/360; 
  ans = A*A+B*B-2*A*B*cos(rangle);
  ans = sqrt(ans);
  // output
  cout << fixed << setprecision(10) << ans << endl;
  
  return 0;
}