sicp每日一题[1.43]

合集 - sicp(77)
1.sicp每日一题[1.42]
2.sicp每日一题[1.43]
3.sicp每日一题[1.44]4.sicp每日一题[1.45]5.sicp每日一题[1.46]6.sicp每日一题[2.1]7.sicp每日一题[2.2]8.sicp每日一题[2.3]9.sicp每日一题[2.4]10.sicp每日一题[2.5]11.sicp每日一题[2.6]12.sicp每日一题[2.7]13.sicp每日一题[2.8]14.sicp每日一题[2.9]15.sicp每日一题[2.11]16.sicp每日一题[2.10]17.sicp每日一题[2.12]18.sicp每日一题[2.13-2.16]19.sicp每日一题[2.17]20.sicp每日一题[2.18]21.sicp每日一题[2.19]22.sicp每日一题[2.20]23.sicp每日一题[2.21]24.sicp每日一题[2.22-2.23]25.sicp每日一题[2.24-2.27]26.sicp每日一题[2.28]27.sicp每日一题[2.29]28.sicp每日一题[2.30]29.sicp每日一题[2.31]30.sicp每日一题[2.32]31.sicp每日一题[2.33]32.sicp每日一题[2.34]33.sicp每日一题[2.35]34.sicp每日一题[2.36-2.37]35.sicp每日一题[2.38-2.39]36.sicp每日一题[2.40]37.sicp每日一题[2.41]38.sicp每日一题[2.42]39.sicp每日一题[2.43]40.sicp每日一题[2.44]41.sicp每日一题[2.45]42.sicp每日一题[2.46]43.sicp每日一题[2.47]44.sicp每日一题[2.48]45.sicp每日一题[2.49]46.sicp每日一题[2.50]47.sicp每日一题[2.51]48.sicp每日一题[2.52]49.sicp每日一题[2.53-2.55]50.sicp每日一题[2.56]51.sicp每日一题[2.57]52.sicp每日一题[2.58]53.sicp每日一题[2.59]54.sicp每日一题[2.60]55.sicp每日一题[2.61]56.sicp每日一题[2.62]57.sicp每日一题[2.63-2.64]58.sicp每日一题[2.65]59.sicp每日一题[2.66]60.sicp每日一题[2.67-2.68]61.sicp每日一题[2.69]62.sicp每日一题[2.70]63.sicp每日一题[2.71]64.sicp每日一题[2.72]65.sicp每日一题[2.73]66.sicp每日一题[2.74]67.sicp每日一题[2.75]68.sicp每日一题[2.76]69.sicp每日一题[2.77]70.sicp每日一题[2.78]71.sicp每日一题[2.79]72.sicp每日一题[2.80]73.sicp每日一题[2.81]74.sicp每日一题[2.82]75.sicp每日一题[2.83]76.sicp每日一题[2.84]77.sicp每日一题[2.85-2.86]
收起

Exercise 1.43

If f is a numerical function and n is a positive integer, then we can form the nth repeated application of f, which is defined to be the function whose value at x is f (f (...(f (x)). . .)). For example, if f is the function x -> x + 1, then the nth repeated application of f is the function x -> x + n. If f is the operation of squaring a number, then the nth repeated application of f is the function that raises its argument to the 2^n-th power. Write a procedure that takes as inputs a procedure that computes f and a positive integer n and returns the procedure that computes the nth repeated application of f. Your procedure should be able to be used as follows:

((repeated square 2) 5)
625

Hint: You may find it convenient to use compose from Exercise 1.42.

这道题稍微复杂一点，要用一个循环来实现重复调用目标函数 n 次。

; f 表示函数，参数只有一个且为数值；n 表示要嵌套执行多少次 f 函数
(define (repeated f n)
  (if (= n 1)
      (lambda (x) (f x))
      (compose f (repeated f (- n 1)))))
      
      
((repeated square 2) 5)
((repeated inc 10) 5)

; 执行结果
625
15
posted @   再思即可  阅读(6)  评论(0)  编辑  收藏  举报
相关博文：
· sicp每日一题[1.44]
· sicp每日一题[1.42]
· 第四章程序段的重复执行
· 练习: 实现一个recursive函数 F(n), 使满足下述结果:
· 20240321每日一题题解
阅读排行：
· DeepSeek 开源周回顾「GitHub 热点速览」
· 物流快递公司核心技术能力-地址解析分单基础技术分享
· .NET 10首个预览版发布：重大改进与新特性概览！
· AI与.NET技术实操系列（二）：开始使用ML.NET
· 单线程的Redis速度为什么快？
点击右上角即可分享
微信分享提示
AI FOR CODE 大赛