UVa 10387 Billiard

思路：可以发现走过的总的水平距离为与垂直边碰撞的次数乘以水平边的长度，即：(a * m)，同理总的垂直距离为(b * n)。所以，其弧度为 (a*m)/(b*n) 的反正切值，再转换为角度，总长度用勾股定理算出来再除以时间，就是速度。

#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;
double pi = 3.1415926;
int main()
{
int a,b,s,n,m;
while(scanf("%d%d%d%d%d",&a,&b,&s,&m,&n)!=EOF && a||b||s||m||n)
{
double aa = 1.0*n*b/(m*a);
double t = atan(aa);
double v = (1.0*n*b) / (sin(t)*s);
t = t/pi * 180;
printf("%.2lf %.2lf\n",t,v);
}
}

In a billiard table with horizontal side a inches and vertical side b inches, a ball is launched from the

middle of the table. After s > 0 seconds the ball returns to the point from which it was launched,

after having made m bounces off the vertical sides and n bounces off the horizontal sides of the table.

Find the launching angle A (measured from the horizontal), which will be between 0 and 90 degrees

inclusive, and the initial velocity of the ball.

Assume that the collisions with a side are elastic (no energy loss), and thus the velocity component of

the ball parallel to each side remains unchanged. Also, assume the ball has a radius of zero. Remember

that, unlike pool tables, billiard tables have no pockets.

Input

Input consists of a sequence of lines, each containing five nonnegative integers separated by whitespace.

The five numbers are: a, b, s, m, and n, respectively. All numbers are positive integers not greater

than 10000.

Input is terminated by a line containing five zeroes.

Output

For each input line except the last, output a line containing two real numbers (accurate to two decimal

places) separated by a single space. The first number is the measure of the angle A in degrees and the

second is the velocity of the ball measured in inches per second, according to the description above.

Sample Input

100 100 1 1 1

200 100 5 3 4

201 132 48 1900 156

0 0 0 0 0

Sample Output

45.00 141.42

33.69 144.22

3.09 7967.81