uva 10387 - Billiard

Problem A: Billiard

In a billiard table with horizontal side a inches and vertical sideb inches, a ball is launched from the middle of the table. Afters > 0 seconds the ball returns to the point from which it was launched, after
having madem bounces off the vertical sides and n bounces off the horizontal sides of the table. Find the launching angleA (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, andn, 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 angleA 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

水平长度a,垂直高度b；经过m次垂直边上的反弹和n次水平边上的反弹后历时s秒回到出发点。一个球从桌的正中间发射，求发射时的角度（0-90度之间）和球的初始速度。运动期间无能量损失，把球看成一个质点。

所有可能的情况如上图所示：所有问题的求解可归结为求解菱形的边长和内角；

单个菱形的垂直长度为b/(2m);水平宽度为a/(2n )

tan（angle）=（bn）/（am）； angle=arctan（bn/am);

总的路程就是菱形的边长总和，这样的菱形有m*n个。

边长l=sqrt（（b/4M)^2+(a/4N)^2);

总路程L=4*m*n*l；

有了上述基础还不能AC，题目里有句话between 0 and 90 degrees inclusive,0（m=0）和90（n=0）是特殊情况计算路程特殊对待

还有对于第三组数据我们发现撞击次数比边长大很多时，由于精度限制l=(b*b/(4*m*m)+a*a/(4*n*n)); l会等于0；

为了避免这个问题，计算总路程L时把根号外的4*m*n化入根号内得 L=sqrt（（bn）^2+（an）^2）；

#include<stdio.h>

#include<math.h>

void main()

{double a,b,m,n,s,l,angle;

while (scanf("%lf%lf%lf%lf%lf",&a,&b,&s,&m,&n)&& a+b+s+m+n)

{angle=atan2(n*b,(m*a));

if (m==0) l=n*b;

else

if (n==0) l=m*a;

else l=sqrt(b*b*n*n+a*a*m*m);

printf("%.2lf %.2lf\n",angle*180/3.1415926,l/s);

}

}