UVa 10387- Billiard

UVa 10387- Billiard

Table of Contents

1 题目

2 思路

3 代码

4 参考

1 题目

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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

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2 思路

题目的关键在于两点。一是要明白反射的过程中角度的对称性，导致小球的轨迹中所有的线与水平方向的夹角都是一样的。 二是需要根据一这个性质，体会出a*m就是水平方向总路径长，b*n就是竖直方向总路径长，而小球总的路径长就是由总水平 长与总竖直长组成的三角形的斜边的长度。明白了这两点，代码就很容易写出来了。

另外，说句题外话。这题目想了好几个小时才体会出来这两点。不断地画图，做小例子，才体会到。可能是我智商太低， 那么久才想出来，不过由自己亲自想出来一个结论，并且得到验证，那种感觉实在太美妙了！

3 代码

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

#define PI acos(-1)

int main() {
double a, b, s, m, n;
double angle, velocity;

while (scanf ("%lf%lf%lf%lf%lf", &a, &b, &s, &m, &n) != EOF) {
if (a == 0 && b==0 && s==0 && m==0 && n==0)
break;
angle = atan( (b*n)/(a*m) ) * 180 / PI;
velocity = sqrt(b*n*b*n+a*m*a*m) / s;
printf ("%.2lf %.2lf\n", angle, velocity);
}

return 0;
}

4 参考

1
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=101&page=show_problem&problem=1328