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 ll long long
    #define rep(i,a,b) for(i=a;i<b;i++)
    #define pb push_back
    #define all(v) v.begin(),v.end()
    #define F first
    #define S second
    #define vi vector < ll  int >
    #define CLR(a) memset((a),0,sizeof(a))
    #define mp make_pair
    #define hell 1000000007
    using namespace std;

    #define MAXN 1000005
    int spf[MAXN];
    vector < ll > vp;
    void sieve()
    {
        spf[1] = 1;
        for (int i=2; i<MAXN; i++)
            spf[i] = i;
        for (int i=4; i<MAXN; i+=2)
            spf[i] = 2;

        for (int i=3; i*i<MAXN; i++)
        {
            if (spf[i] == i)
            {
                for (int j=i*i; j<MAXN; j+=i)

                    if (spf[j]==j)
                        spf[j] = i;
            }
        }
        for(int i=2;i<=100000;i++)
        {
            if(spf[i]==i) vp.pb(i);
        }
    }


    vector<int> gf(int x)
    {
        vector<int> ret;
        while (x != 1)
        {
            ret.push_back(spf[x]);
            x = x / spf[x];
        }
        return ret;
    }
    int BIT[100050];
    void upd(int x) { ++x;  while(x<=MAXN)  {  BIT[x]++;  x+=(x&-x);  } }
    int cal(int x) {  ++x;  int res=0;  while(x>0)  {  res+=BIT[x];  x-=(x&-x);  } return res; }

    ll mod=hell;
    ll pr( ll x , ll y)
    {
        ll ans=1;
        if(y==0) return ans;
        while(y>1)
        {
            if(y&1) ans*=x;
            x*=x;
            y/=2;
            x%=mod;
            ans%=mod;
        }
        ans*=x;
        ans%=mod;
        return ans;
    }
    ll inv(ll n)
    {
        return pr(n,mod-2);
    }
    ll stoi(string s)
    {
        int n=s.size();
        ll k=0; int i;
        rep(i,0,n)
        {
            k+=s[i]-'0';
            k*=10;
        }
        return k/10;
    }
    string to_string(ll n)
    {  if(!n) return "0";
        string s=""; int a,b;
        while(n>0)
        {
            b=n%10;
            s+=(b+'0');
            n/=10;
        }
        reverse(all(s));
        return s;
    }
  ll invvf[1000500],fact[1000500];
  void pre()
  {
      ll i,j,n=1000050;
      fact[0]=1; fact[1]=1;
      rep(i,2,n) {fact[i]=fact[i-1]*i; fact[i]%=hell;}
      invvf[1]=1; invvf[0]=1;
      rep(i,2,n)
      {
          invvf[i]=inv(fact[i]);
      }

  }
 ll ncr(ll n , ll r)
 {
     ll y=fact[n];
     y*=invvf[r];
     y%=hell;
     y*=invvf[n-r];
     y%=hell;
     return y;
 }


int phi(int n) {
    int result = n;
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            while (n % i == 0)
                n /= i;
            result -= result / i;
        }
    }
    if (n > 1)
        result -= result / n;
    return result;
}

 bool fun(string s){
    int one = 0, zero = 0;
    for(int i = s.length() - 1; i >= 0; i--){
		if(s[i] == '0')
			zero++;
		else
			one++;
		if(zero > one)
			return false;
    }
    return true;
}
  ll ck(string s)
  {
      int n=s.size(); int i;
      char ch='9'; char ch2='0';
      rep(i,0,n)
      {
          ch=min(ch,s[i]);
          ch2=max(ch2,s[i]);
      }
      ll a1=ch-'0';
      ll a2=ch2-'0';
      return a1*a2;
  }
  bool okk(string &s,ll val)
{
	map<char,ll>M;

	int i,n=s.size();
	rep(i,0,val)
	{
		M[s[i]]++;
	}
	if(M.size()==3) return true;


	rep(i,val,n)
	{
		M[s[i-val]]--;

		if(M[s[i-val]]==0)
			M.erase(s[i-val]);
		if(M.find(s[i])==M.end())
			M[s[i]]=1;
		else
			M[s[i]]++;

		if(M.size()==3)
			return true;
	}

	if(M.size()==3)
		return true;
		else
	return false;
}

    void solve()
    {
    long long  i,j=0,r=200,p,q,k=0,d,y=0,n,h=100000000000000,flag=0,m=0,a=hell,b=hell,id,h1,h2,a1,b1,a2,w,x,u,f,l,c=0,e,ans=1e9;
    cout<<setprecision(15);
     double dh,dm,d1,d2;
     cin>>d1>>d2>>dh>>dm;
     double ang1=dm*6;
     double ang2=dh*30+(ang1)/12;
     double an;
     if(ang1>=180) ang1-=180;
     else if(ang2>=180) ang2-=180;
     else ang1+=180;
     if(ang1>ang2) an=ang1-ang2;
     else an=ang2-ang1;

     if(an>180) an=360-an; double ass;
      //cout<<ang1<<" "<<ang2<<" "<<an<<endl;
      double  pi=2*acos(0);
      double dd=an;
     an=(an*pi)/180;
      ass=d1*d1+d2*d2+2*d1*d2*cos(an);
     ass=sqrt(ass);
     cout<<ass;

}

    signed main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);
        int t=1;
        //sieve();
       // pre();
        //cin>>t;
        while(t--)
        solve();
    }