简单的伪代码问题 [英] Simple Pseudocode Code Question

问题描述

我对伪代码有点陌生.我明白代码在说什么，但很难把这些片段放在一起.我应该如何理解这段代码在做什么:

I'm a little new to pseudocode. I understand what the code is saying, but having a hard time putting the pieces together. How should I be thinking to understand what this code is doing:

假设 a1, a2, ..., ak 是 k 个数字的数组.下面有什么作用代码片段做什么?简要说明原因.假设所有缩进的行属于循环内.

Suppose that a1, a2, . . . , ak is an array of k numbers. What does the following code fragment do? Briefly explain why. Assume that all the indented lines belong inside the loop.

1 for p := 1 to ⌊k/2⌋
2     t := ap
3     ap := ak−p+1
4     ak−p+1 := t

推荐答案

好吧，

1 for p := 1 to ⌊k/2⌋

意味着，我们要到数组的一半.

means, we're going up to the half of the array.

2 t := ap
3 ap := ak−p+1
4 ak−p+1 := t

这种模式可以被识别为与临时t交换".什么是交换?

This pattern can be recognized as a "swap with temporary t". And what is swapped?

好吧，ap 和 ak-p+1，其中一个是数组 start 中的 p-th 个元素，另一个是 p-从末尾开始的第一个.

Well, ap and ak-p+1, one being the p-th element from the array start, the other one the p-th one from the end.

总结一下:

您将第 n 的第一个与 n 的最后一个数组值交换到数组的一半.然后呢?数组颠倒了.

You swap the n-th first with the n-th last array value up to the half of the array. And afterwards? The array is reversed.

请注意，您的伪代码格式看起来很奇怪 - 而且，最重要的是 - 模棱两可.

Note that your pseudocode-format looks really weird - and, most importantly - ambiguous.

ak-p+1 等价于 a[k-p+1] 还是 a[k]-p+1还是a[kp]+1?如果不是，你是如何表达其他的.

Is ak-p+1 equivalent to a[k-p+1] or to a[k]-p+1 or a[k-p]+1? If not, how did you express the other ones.

因此，首先，我会将这段代码转换为类似于 Python 的实际源代码，这样更易​​于阅读.

So at first, I'll convert this code to an actual source like Python's, which is much easier to read.

编辑.

I) 好吧，正如您发布的那样，数组范围从 a1 到 ak.

I) Well, as you posted, the array ranges from a1 to ak.

II) 想想如何交换两个变量(a 和 b)的值:

II) Think how you could swap the values of two variables (a and b):

1 temp := a
2 a    := b
3 b    := temp

当然，由于您在 2 行中用 b 覆盖了 a，因此您必须存储旧的 a 临时值，即 temp.

Of course, since you overwrote a with b in line 2, you had to store the old a value in a temporary, which is temp.

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