uva-10387 - Billiard

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

题意：输入a,b,s,m,n 表示长a 宽b的桌子。。一个球打出去 经过s秒回到原点。撞击了长m次 宽n次。。

解题思路：起点在中间，速度不变，最终回到原点。所以。。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，直接水平垂直距离相比得出tan，然后在转换为度数即可。。

代码

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cmath>

using namespace std;

const double pi = asin(1.0) * 2;

int main()
{
double a, b, s, m, n, v, angle;
while (~scanf("%lf%lf%lf%lf%lf", &a, &b, &s, &m, &n))
{
if (a == 0 && b == 0 && s == 0 && m == 0 && n == 0)
break;
v = sqrt((a * m) * (a * m) + (b * n) * (b * n)) / s;
angle = atan(b * n / (a * m)) / pi * 180;
printf("%.2lf %.2lf\n", angle, v);
}
return 0;
}