浮点数和整数的计算，Ruby 30 天刷题修行篇第十三话

大家好，我是 A Fei，大天要来解甚麽题目呢？让我们看下去：
（题目来源：Codewars）

A child is playing with a ball on the nth floor of a tall building. The height of this floor, h, is known.

He drops the ball out of the window. The ball bounces (for example), to two-thirds of its height (a bounce of 0.66).

His mother looks out of a window 1.5 meters from the ground.

How many times will the mother see the ball pass in front of her window (including when it's falling and bouncing?

Three conditions must be met for a valid experiment:
Float parameter "h" in meters must be greater than 0
Float parameter "bounce" must be greater than 0 and less than 1
Float parameter "window" must be less than h.
If all three conditions above are fulfilled, return a positive integer, otherwise return -1.

Note:
The ball can only be seen if the height of the rebounding ball is strictly greater than the window parameter.

Examples:

- h = 3, bounce = 0.66, window = 1.5, result is 3

- h = 3, bounce = 1, window = 1.5, result is -1 

(Condition 2) not fulfilled).

题目给了一个情境：一个小孩正 n 层楼玩球。该楼层的高度为 h，他把球扔出窗外，球反弹到其三分之二的高度（bounce 等於 0.66）。 他的母亲从离地 1.5 公尺的窗户往外看。请问，妈妈会看到球在窗外经过多少次（包括球掉落和弹跳的时候）。此外，题目给了三个条件：

1. h（公尺）是浮点数且必须大於 0
2. bounce 必须大於 0 且小於 1
3. 窗户高度必须小於 h

如果以上三个条件都满足，则回传看到球的次数（整数），反之则回传 -1

我的答案：

def bouncingBall(h, bounce, window)
  return -1 if h <= 0 || (bounce <= 0 && bounce >= 1) || window >= h 
  i = 1
  while h * bounce > window
    h = (h * bounce)
    i += 2
  end 
  i
end