Problem F
Traffic Volume
Input: standard input
Output: standard output
Time Limit: 1 second
 

In the picture below (or above depending on HTML response :)) you can see a street. It has infinite number of cars on it. The distance between any two consecutive cars is d, length of each car is L and the velocity of each car is v. The volume of cars through a road means the number of cars passing through a road in a specific amount of time. When the velocity is constant, d must be minimum for the volume of cars passing through the road to be maximal. In our model when the velocity of all the cars is v then the minimum possible value of d is v²/(2f) (The more the car velocity the more distance you need to bring down your velocity to zero). Here f is the deceleration due to break.

Keeping this model in mind and given the value of L and f your job is to find the value of v for which the volume of traffic through the road is maximal.

Input

The input file contains several lines of input. Each line of input contains two integers L (0<L<=100) and f(0<f<=10000). The unit of L is meter and the unit of f is meter/second². The input is terminated by a single line whose value of L and f is zero.

Output

For each line of input except the last one produce one line of output. Each line contains two floating-point number v and volume separated by a single space. Here v is the velocity for which traffic flow is maximal and volume is the maximum number of vehicles (of course it is a fraction) passing through the road in an hour. These two floating points should have eight digits after the decimal. Errors less than 1e-5 will be ignored.

Sample Input                           Output for Sample Input

5 3
0 0

Problem setter: Shahriar Manzoor, Member of Elite Problemsetters' Panel

Special Thanks: Derek Kisman

#include <stdio.h>
#include <math.h>

int main() {
    double L, f;
    while(scanf("%lf %lf", &L, &f) == 2) {
        if(L == 0 && f == 0)
            break;
        double v = sqrt(2*L*f);
        double fun = 3600*v/(L+v*v/2/f);
        printf("%.8lf %.8lf\n", v, fun);
    }
    return 0;
}