2014-02-17 21:35:25Morris
[UVA] 11880 - Ball in a Rectangle
Ball in a Rectangle
Input: Standard Input
Output: Standard Output
| Ball in a Rectangle |
� There is a rectangle on the cartesian plane, with bottom-left corner at (0,0) and top-right corner at (L, W). There is a ball centered at (x, y), with radius=R, shown below
At time 0, the ball starts to move along a ray with polar angle a (the angle from positive x-axis to the ray, rotating counter-clockwise). When hitting the rectangle boundary, the reflex angle always equals to the incidence angle. The ball's velocity is always v (i.e. it never changes when hitting the rectangle). Where is the center of the ball at time s?
Input
There will be at most 25 test cases, each contains a line with 8 integers L,W,x,y,R,a,v,s (100
L,W109,
1R5,
RxL - R,
RyW - R,
0a < 360, 1v, s109), as stated above. The input
terminates with
L = W = x = y = R = a = v = s = 0, which should not be processed.
Output
For each test case, output a line containing two floating-point numbers x, y, rounded to two decimal points, indicating that the center of ball will be at (x, y) at time s.
Sample Input
100 100 80 10 5 90 2 23 110 100 70 10 5 180 1 9999 0 0 0 0 0 0 0 0
Sample Output
80.00 56.00 71.00 10.00
Problemsetter: Rujia Liu, Special Thanks: Yiming Li, Shamim Hafiz & Sohel Hafiz
題目描述：
求球在矩形內部的彈性碰撞後 s 秒後的位置。
題目解法：
忽略外框不計，最後再用取模和奇偶性質還原。
#include <stdio.h>
#include <math.h>
int main() {
const double pi = acos(-1);
double L, W, x, y, R, a, v, s;
while(scanf("%lf %lf %lf %lf %lf %lf %lf %lf",
&L, &W, &x, &y, &R, &a, &v, &s) == 8) {
if(R == 0)
continue;
L -= 2 * R;
W -= 2 * R;
double theta = a / 180.0 * pi;
x -= R;
y -= R;
x = x + v * cos(theta) * s;
y = y + v * sin(theta) * s;
//printf("%lf %lf \n", x, y);
if((((int)ceil((x / L))) %2+2)%2 == 0)
x = L - fmod(x, L);
else
x = fmod(x, L);
if((((int)ceil((y / W))) %2+2)%2 == 0)
y = W - fmod(y, W);
else
y = fmod(y, W);
x = fmod(x+L, L);
y = fmod(y+W, W);
printf("%.2lf %.2lf\n", x + R, y + R);
}
return 0;
}