Tortoise-racing

題目：

Two tortoises named A and B must run a race. A starts with an average speed of 720 feet per hour. Young B knows she runs faster than A and furthermore has not finished her cabbage.

When she starts, at last, she can see that A has a 70 feet lead but B speed is 850 feet per hour. How long will it take B to catch A?

More generally: given two speeds v1 (A speed, integer > 0) and v2 (B speed, integer > 0) and a lead g (integer > 0) how long will it take B to catch A?

The result will be an array [h, mn, s] where h, mn, s is the time needed in hours, minutes and seconds (don't worry for fractions of second).

If v1 >= v2 then return nil, nothing, null, None or {-1, -1, -1} for C++, C, Go, Nim.

Examples:

race(720, 850, 70) => [0, 32, 18]
race(80, 91, 37) => [3, 21, 49]

Note: you can see some other examples in "Your test cases".

思路：

物理計算問題，重點在計算的時候要先把數字轉為浮點數後再轉回整數回傳

使用到的語法：

if和數學計算式

程式碼：

def race(v1, v2, g)
    if v1 >= v2
      return nil
    else
      t = ((g.fdiv(v2 - v1)) * 3600).to_i
      result = [t / 3600,(t / 60) % 60, t % 60]
    end
end