sicp每日一题[2.41]
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]Exercise 2.41
Write a procedure to find all ordered triples of distinct positive integers i, j, and k less than or equal to a given integer n that sum to a given integer s.
这道题有一点难度，主要是要使用两次 flatmap 函数，仿照2元组的构造方式构造出3元组来，然后再筛选出和为s的。
; 获取不大于正整数n的1/3的最大整数
(define (one-third-factor n)
(cond ((= (remainder n 3) 0) (/ n 3))
((= (remainder n 3) 1) (/ (- n 1) 3))
(else (/ (- n 2) 3))))
(define (ordered-triple-sum n s)
(if (or (< s 6) (> n (- s 3))) ; n是不同的3个正整数之和，所以s至少是6,且n至少比s小3
nil
(filter (lambda (seq) (= (accumulate + 0 seq) s))
(flatmap (lambda (i)
(flatmap (lambda (j)
(map (lambda (k) (list i j k))
(enumerate-interval (+ j 1) n))) ; 第三个数比第二个大
(enumerate-interval (+ i 1) (- n 1)))) ; 第二个数比第一个大
(enumerate-interval 1 (one-third-factor n)))))) ; 需要的是有序3元组，所以第一个数最大也比n/3小
(ordered-triple-sum 15 20)
; 执行结果
'((1 4 15) (1 5 14) (1 6 13) (1 7 12) (1 8 11) (1 9 10) (2 3 15) (2 4 14) (2 5 13) (2 6 12) (2 7 11) (2 8 10) (3 4 13) (3 5 12) (3 6 11) (3 7 10) (3 8 9) (4 5 11) (4 6 10) (4 7 9) (5 6 9) (5 7 8))
【推荐】国内首个AI IDE，深度理解中文开发场景，立即下载体验Trae
【推荐】编程新体验，更懂你的AI，立即体验豆包MarsCode编程助手
【推荐】抖音旗下AI助手豆包，你的智能百科全书，全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell：AI 加持，快人一步
· 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？