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>
 
using namespace std;
 
#define rep(i, n) for(int i=0; i<(n); i++)
#define FOR(i, m, n) for(int i=(m);i<(n);i++)
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define SORT(x) sort((x).begin(),(x).end())
#define RSORT(x) sort((x).rbegin(),(x).rend())
#define REVE(x) reverse((x).begin(),(x).end())
#define mp make_pair
#define pb push_back
#define chmax(x,y) x = max(x,y)

 
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<vector<int> > VVI;
typedef pair<int, int> PII;
typedef long long LL;
typedef vector<bool> VB;
typedef vector<LL> VL;
typedef vector<vector<long long>> VVLL;
typedef vector<set<int>> VSI;
const int MOD = 1e9+7;
const int INF = 1e9;
const LL LINF = 1e18;
int gcd(int a, int b)
{
   if (a%b == 0)
   {
       return(b);
   }
   else
   {
       return(gcd(b, a%b));
   }
}

int lcm(int a, int b)
{
   return a * b / gcd(a, b);
}
struct UnionFind {
    VI d;
    UnionFind(int n = 0): d(n,-1) {}
    int find(int x) {
        if(d[x] < 0) return x;
        return d[x] = find(d[x]);
    }
    bool unite(int x,int y) {
        x = find(x); y = find(y);
        if(x == y) return false;
        if(d[x] > d[y]) swap(x,y);
        d[x] += d[y];
        d[y] = x;
        return true;
    }
    bool same(int x,int y) { return find(x) == find(y); }
    int size(int x) { return -d[find(x)]; }
};
int deg[100005];
VI to[100005];

struct mint {
  LL x; // typedef long long ll;
  mint(LL x=0):x((x%MOD+MOD)%MOD){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= MOD) x -= MOD;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += MOD-a.x) >= MOD) x -= MOD;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= MOD;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(LL t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
 
  // for prime mod
  mint inv() const {
    return pow(MOD-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};

mint f(int n) {
    if(n == 0) return 1;
    mint x = f(n/2);
    x *= x;
    if(n%2 == 1) x *= 2;
    return x;
}

mint choose(int n,int a) {
    mint x = 1, y = 1;
    rep(i,a) {
        x *= n-i;
        y *= i+1;
    }
    return x/y;
}

int main() {
    LL a,b,h,m;
    cin >> a >> b >> h >> m;
    double e,f;
    static const double pi = 3.141592653589793;

    h += (m/60.0);
    e = 30.0 * h;
    f = 6.0 * m;
    // e += 30.0 * (m/60.0);
    LL t = min(e-f,f-e);
    double n = cos((t*M_PI)/180.0);
    // if(abs(t) == 180) {
    //   printf("%.30f",0.0);
    //   return 0;
    // } else if(e == f) {
    //   printf("%.30f",abs(a-b));
    //   return 0;
    // }

    printf("%30.30f\n",sqrt(a*a + b*b -2.0 * a * b * n) );

}