Uva 10387 Billiard-优快云博客

本文探讨了一个关于台球桌的问题，通过输入五个关键参数：台球桌的长、宽、运动时间、碰撞竖直面次数和水平面次数，计算出球的发射角度和初始速度。通过应用三角函数和物理定律，我们能够得出准确的角度和速度值。

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C - Billiard

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Description

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, 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

题目大意：

给你5个数字，分别是台球桌的长，台球桌的宽，台球运动的时间，台球碰到竖直面的次数，台球碰到水平面的次数。

注意：在碰撞的过程中能量不损耗。

求夹角和速度

题目解析：

就是根据长宽，时间和碰的次数，算出水平运动了多远，垂直运动了多久，再用三角函数求角度和速度。

#include <stdio.h>
#include <math.h>
const double PI = acos(-1.0);
int main() {
	double a,b,s,m,n;

	while(scanf("%lf%lf%lf%lf%lf",&a,&b,&s,&m,&n) != EOF && (a||b||s||m||n)) {

		double ang = atan( (n*b)/(m*a) ) * 180 /PI;
		double v = sqrt((a*m)*(a*m) + (b*n)*(b*n)) / s;
		
		printf("%.2lf %.2lf\n",ang,v);
	}
	return 0;
}