模拟时钟的最短路径算法 [英] Shortest path algorithm for an analog clock

问题描述

我遇到了一个家庭作业问题，如果有人可以帮助我指出正确的方向，那将是非常有用的。

I've got a c homework problem that is doing my head in and will be greatful if anyone can help point me in the right direction.

如果我有两个在诸如t1（55分钟）和t2（7分钟）之类的模拟手表上，您需要计算两分钟之间的最短步数。

If I have two minutes points on an analog watch such as t1 (55 minutes) and t2 (7 minutes), I need to calculate the shortest amount of steps between the two points.

什么到目前为止，我想出了以下两个等式：

What I've come up with so far is these two equations:

-t1 + t2 + 60 =
    -55 + 7 + 60 
    = 12

t1 - t2 + 60 = 
    55 - 7 + 60 
    = 108

12 is lower then 108, therefore 12 steps is the shortest distance.

如果我比较两个结果并使用最低的结果，这似乎很好用。但是，如果我选择另外两个点，例如让t1 = 39和t2 = 34并将它们插入等式：

This appears to work fine if I compare the two results and use the lowest. However, if I pick out another two points for example let t1 = 39 and t2 = 34 and plug them into the equation:

-t1 + t2 + 60 = -39 + 34 + 60 = 55
t1 - t2 + 60 = 39 - 34 + 60 = 35

35 is lower then 55, therefore 35 steps is the shortest distance.

但是，35不是正确的答案。 5步是最远的距离（39-34 = 5）。

However, 35 isn't the correct answer. 5 steps is the shorest distance (39 - 34 = 5).

我的大脑有点炸了，我知道我缺少一些简单的东西。谁能帮忙？

My brain is a little fried, and I know I am missing something simple. Can anyone help?

推荐答案

您想要的是60的加法和减法。查看％操作员。确保正确处理负片。

What you want is addition and subtraction modulo 60. Check out the % operator. Make sure you handle negatives correctly.

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