Problem A: Billiard
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
#include<stdio.h> #include<math.h> int main() { int a, b, m, n, s; double angle, v; double pi = acos(0.0)*2; while(scanf("%d%d%d%d%d", &a, &b, &s, &m, &n) != EOF) { if(a+b+s+m+n == 0) break; angle = atan(1.0*b*n/(1.0*a*m))/pi*180; while(angle > 90) angle -= 90; v = sqrt(1.0*a*a*m*m+1.0*b*b*n*n)/s; printf("%.2lf %.2lf\n", angle, v); } return 0; }
解题思路：
起点在中间，速度没有改变，最终回到原点。那么可以得出的结论是：在水平边上撞击了多少次就代表着垂直的路程它走了多少次；在垂直边上撞击多少次就代表着水平的路程它走了多少次，根据勾股定理可以得出它走的路程总长度，再加上有时间，很快就能算出速度。至于角度，根据刚才得到的三角形，求角的运算就简单了，因为整个过程一直都没有受到其他的外力，所以不像重力作用的抛物线，什么速度形成的角的tan值是路程形成的角的tan的1/2……
