乘以不带+或* [英] Multiply without + or *

问题描述

我正在独自研究如何设计程序.我还没有完全掌握复杂的线性递归，所以我需要一些帮助.

I'm working my way through How to Design Programs on my own. I haven't quite grasped complex linear recursion, so I need a little help.

问题: 定义multiply，它使用两个自然数n和x，并在不使用Scheme的*的情况下生成n * x.也从该定义中删除+.

The problem: Define multiply, which consumes two natural numbers, n and x, and produces n * x without using Scheme's *. Eliminate + from this definition, too.

直接带+号:

(define (multiply n m)
  (cond
    [(zero? m) 0]
    [else (+ n (multiply n (sub1 m)))]))

(= (multiply 3 3) 9)

我知道要使用add1，但是我不能递归.

I know to use add1, but I can't it the recursion right.

谢谢.

推荐答案

将问题分为两个功能.首先，您需要一个将加到n的函数(add m n).基本情况是什么?当n为零时，返回m.什么是递归步骤?再次调用add的结果加1，但递减n.您猜对了，add1和sub1会很有用.

Split the problem in two functions. First, you need a function (add m n) which adds m to n. What is the base case? when n is zero, return m. What is the recursive step? add one to the result of calling add again, but decrementing n. You guessed it, add1 and sub1 will be useful.

另一个功能(mul m n)与之类似.基本情况是什么?如果m或n为零，则返回0.递归步骤是什么?将m添加(使用先前定义的函数)再次调用mul的结果，但递减n.就是这样！

The other function, (mul m n) is similar. What is the base case? if either m or n are zero, return 0. What is the recursive step? add (using the previously defined function) m to the result of calling mul again, but decrementing n. And that's it!

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