UVa 10387 - Billiard

最新推荐文章于 2017-10-26 08:01:00 发布

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最新推荐文章于 2017-10-26 08:01:00 发布

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分类专栏： UVa 文章标签： velocity numbers output each parallel input

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本文探讨了一个关于台球在特定尺寸的台球桌上反弹的问题，通过数学模型计算发射角度和初始速度，确保球能在指定次数的垂直和水平碰撞后返回起点。文中提供了一段C语言实现代码作为解决方案。

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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

看着题目给出的数据猜猜就行

#include <stdio.h>
#include <math.h>
#define PI 3.141592654;
int main()
{
    double a,b,s,m,n;
    double s1,s2,l1,l2;
    while(scanf("%lf %lf %lf %lf %lf",&a,&b,&s,&m,&n)!=EOF)
    {
        if(!a&&!b&&!s&&!m&&!n)
        {
            break;
        }
        l1=a*m;
        l2=b*n;
        s1=atan(l2/l1)*180/PI;
        s2=sqrt(l1*l1+l2*l2)/s;
        printf("%.2lf %.2lf\n",s1,s2);
    }
    return 0;
}