Tortoise.rb

=begin

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++.

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".

=end

# My Solution

def race(v1, v2, g)
  return nil if v2 < v1
  t = (g.to_f/(v2.to_f - v1.to_f))
  s = ((t.to_f*3600))%60
  m = ((t.to_f)*60.00)%60
  h = (t.to_f)%60
  [h.floor, m.floor, s.floor]
end

# Better Solution
def race(v1, v2, g)
   sec = (g * 3600 / (v2 - v1))
   [sec / 3600,  (sec / 60) % 60, sec % 60] if v2 > v1
end