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)


#pragma warning(disable:4996)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <float.h>

#pragma region 定義
#define i_cin(X) scanf("%d", &X)
#define i_cin2(X,Y) scanf("%d %d",&X,&Y)
#define i_cin4(X,Y,Z,W) scanf("%d %d %d %d",&X,&Y,&Z,&W)
#define i_cin3(X,Y,Z) scanf("%d %d %d",&X,&Y,&Z)
#define l_cin(X) scanf("%ld",&X)
#define f_cin(X) scanf("%f",&X)
#define l_cin2(X,Y) scanf("%ld %ld",&X,&Y)
#define s_cin(X) scanf("%s",&X)
#define c_cin(X) scanf("%c",&X);
#define ll_cin(X) scanf("%lld", &X)
#define ull_cin(X) scanf("%llu", &X)
#define lf_cin(X) scanf("%lf", &X)

#define  i_cout(X) printf("%d\n",X)
#define  f_cout(X) printf("%g\n",X)
#define  i_cout2(X,Y) printf("%d %d\n",X,Y)
#define  l_cout(X) printf("%ld\n",X)
#define  s_cout(X) printf("%s\n",X)
#define  s_coutc(X) printf("%s",X)

#define  c_cout(X) printf("%c",X)


#define  ll_cout(X) printf("%lld\n",X)
#define  ull_cout(X) printf("%llu\n",X)



typedef long long  ll;
typedef unsigned long long  ull;

#define rept(x, s, n) for (int x = s; x < n; x++)

#define S_RTN(S)  s_cout(S);return 0


#define _itoa(A,N) sprintf(A, "%d", N);

int i_cins(int n, int* A);
int l_cins2(int n, long* A, long* B);
int s_dsort(const void* a, const void* b);
int s_asort(const void* a, const void* b);

int  _gcd(int a, int b);
int  _swp(int* a, int* b);
int _cknum(char* a, int n);
int _atoi(char* s, int len);


int s_asorts(const void* a, const void* b);

//昇順
int s_asorts(const void* a, const void* b) {
	return(strcmp((char*)a, (char*)b));
}
int s_dsort(const void* a, const void* b) {
	return(*(int*)b - *(int*)a);
}
int s_asort(const void* a, const void* b) {
	return(*(int*)a - *(int*)b);
}
int l_cins2(int n, long* a, long* b) {
	int i;
	rept(i, 0, n) {
		l_cin2(*(a + i), *(b + i));
	}
	return 0;
}
int i_cins(int n, int* a) {
	int i;
	for (i = 0; i < n; i++) {
		i_cin(*(a + i));
	}

	return 0;
}
int _gcd(int a, int b) {
	int r, tmp;
	if (a < b) { tmp = a; a = b; b = tmp; }
	r = a % b;
	while (r) {
		if (!r) break;
		a = b; b = r;
		r = a % b;
	}
	return(b);
}

#define pai  3.14159265      
#define deg_rad(X)  (X*(pai/180))

long _max(long a, long b) {
	return a > b ? a : b;
}
long _min(long a, long b) {
	return a < b ? a : b;
}
int _swp(int* a, int* b)
{
	int tmp;
	tmp = *b; *b = *a; *a = tmp;
	return 0;
}

int _cknum(char* a, int n) {
	int i;
	char t = '0';

	for (i = 0; i < n; i++) {
		if (a[i] < '0' || a[i]>'9') return 1;
	}
	return 0;
}


int _atoi(char* s, int len) {
	char tmp[20];
	memcpy(tmp, s, len);
	tmp[len] = 0x00;
	return (atoi(tmp));
}

#pragma endregion


int main(void) {
	double a, b, h,m;
	double ans,r;

	scanf("%lf %lf %lf %lf", &a,&b,&h,&m);


	h = 30.0*h+m*0.5;

	m = 6.0*m;
	r = h-m;
	

	ans = pow(a,2) + pow(b,2)- (cos(deg_rad(r))*a * b *2.0);
    
	printf("%.19lf\n",sqrt(ans));


	return 0;
	

}

// ARC-80 C A-adjacent
//int main(void)
//{
//	int n, i, cnt4 = 0, cnt2 = 0, cnt = 0;
//	int a;
//
//	i_cin(n);
//
//	for (i = 0; i < n; i++) {
//		i_cin(a);
//		if (a % 2) {
//			cnt++; continue;
//		}
//		if (!(a % 4)) {
//			cnt4++; continue;
//		}
//		cnt2++; cnt++;
//	}
////	printf("cnt:%d cnt2:%d\n",cnt,cnt2+1);
//	if (cnt2 == cnt) { S_RTN("Yes\n"); }
//
//	if (cnt > cnt4 + 1) { S_RTN("No\n"); }
//	//if (cnt2<=1) { S_RTN("Yes"); }
//	//if (cnt4 <= cnt2) { S_RTN("No\n"); }
//	S_RTN("Yes\n");
//}


//167-D
//static int a[200001];
//static int c[200001];
//int main(void) {
//	ll k,ps,pe,i,j=0,n,r;
//
//	ll_cin(n);
//	ll_cin(k);
//
//	for (i = 0; i < n; i++) {
//		i_cin(a[i]);
//		if ((ll)(i + 1) == k) {
//			ll_cout(c[i]); return 0;
//		}
//		if (c[i] == 1) {
//			ps = i;
//			pe = i - 1; break;
//		}
//		c[i]++;
//	}
//	k -= i;
//	r = (int)(k % (pe-ps))+ps;
//	i_cout(a[r]);
//	return 0;
//}

//
//167-C
//int main(void) {
//	int n,m,x,i,j;
//	int c[13][2], a[13][12];
//
//	i_cin3(n, m, x);
//	for (i = 0; i < n; i++) {
//		i_cin(c[i][0]); c[i][1] = i;
//		i_cins(m, a[i]);
//	}
//	
//	qsort(c[0],n, 8, s_asort);
//	for (i = 0; i < n; i++) {
//		
//	}
//	return 0;
//
//}



//#define _N 1000000000

//int main(void)
//{
//
//	int  n,i;
//	ull   tmp=1;
//
//	i_cin(n);
//
//	for (i = 1; i <= n; i++) {
//		if (!(tmp % ((ull)_N + 7)))
//			tmp = 1;
//		tmp *= (ull)i;
//	}
//	tmp %= ((ull)_N + 7);
//		
//	ull_cout(tmp);
//	return 0;
//}