JavaScript fun: The tortoise race

Two turtles, A and B, are racing. A is moving at an hourly rate of 720 feet. Young B knows that A is slower than it, so he is still eating vegetables in A hurry. Two turtles, A and B, are racing.

A is moving at an hourly rate of 720 feet.

Young B knows that A is slower than it, so he is still eating vegetables in A hurry.

When B starts to run, it finds that A is 70 feet ahead, but B's speed is 850 feet per hour, so it will certainly catch up.

How long does B catch up with?

More common cases: Given two speeds: v1 (A's speed,> an integer of 0), v2 (B's speed,> an integer of 0 ), there is also A leading GAP g (g> 0). How long does B have to catch up with?

The result should be an array. [h, mn, s], h, mn, s indicates the hour, minute, and second.

If an exception occurs, such as v1> = v2, B will never catch up with A, then null is returned directly.

For example:

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

There are two key points for this question:

First, you have to figure out the relationship. In the process of B catching up with A, A has never been idle and has not stopped!

So if we want B to catch up with A, we must satisfy this equation:

V1 * time + g = v2 * time

In this way, time is easy, but the most important thing is how to split the time into minutes and seconds.

My approach is to calculate the clock first, and then the minute based on the remainder, and then the second based on the remainder.

function race(v1, v2, g) {      var h = -1;      var mn = -1;      var s = -1;      var remainder;      var speedGap = v2 - v1;      if(speedGap > 0){          remainder = g % speedGap;          h = parseInt(g / speedGap);          mn = parseInt(60 * remainder / speedGap);          remainder = remainder * 60 % speedGap;          s = parseInt(remainder * 60 / speedGap);          return [h, mn, s];      }      return null;  }

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