Tortoise racing

Tortoise racing

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's speed is 850 feet per hour. How long will it take B to catch A?

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

The result will be an array [hour, min, sec] which is the time needed in hours, minutes and seconds (round down to the nearest second) or a string in some languages.

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

Examples:

(form of the result depends on the language)

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

** Note:

• See other examples in "Your test cases".

• In Fortran - as in any other language - the returned string is not permitted to contain any redundant trailing whitespace: you can use dynamically allocated character strings.

** Hints for people who don't know how to convert to hours, minutes, seconds:

• Tortoises don't care about fractions of seconds

• Think of calculation by hand using only integers (in your code use or simulate integer division)

• or Google: "convert decimal time to hours minutes seconds"

Solutions

👴 C

int* race(int v1, int v2, int g)
{
  int * res = (int*)malloc(sizeof(int)*3);
  if (v1 >= v2)
  {
    res[0]=-1;res[1]=-1;res[2]=-1;
    return res;
  }

  double delta_time = 1.0,
  S = g,
  how_long_to_get = 0; //in hours
  while(delta_time > 0.0000000001) // 3.6e-07 sec
  {
    delta_time =  S / (double)v2;
    S = v1 * delta_time;
    how_long_to_get+=delta_time;
  }

  how_long_to_get+=0.00000001;
  res[0] = (int)how_long_to_get; //hours
  res[1] = (how_long_to_get - res[0]) * 60; // mins
  res[2] = (how_long_to_get - res[0] - res[1]/60.0) * 3600; //secs

  return res;
}