sicp习题试解 (1.33)

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; ======================================================================
;
; Structure and Interpretation of Computer Programs
; (trial answer to excercises)
;
; 计算机程序的构造和解释(习题试解)
;
; created: code17 03/05/05
; modified:
; (保持内容完整不变前提下,可以任意转载)
; ======================================================================

;; SICP No.1.33

;; 递归版本
(define (filtered-accumulate-r filter combiner null-value term a next b)
(if (> a b)
null-value
(combiner (if (filter a) (term a) null-value)
(filtered-accumulate-r filter combiner null-value
term (next a) next b))))

;; 迭代版本
(define (filtered-accumulate-i filter combiner null-value term a next b)
(if (> a b)
null-value
(filtered-accumulate-i filter combiner
(if (filter a)
(combiner null-value (term a))
null-value)
term (next a) next b)))

;; 素数的平方和
(define (sum-of-primes-square a b)
(define (next i) (+ i 1))
(filtered-accumulate-r prime? + 0 square a next b))

;; 互质数的积
(define (product-of-relative-primes n)
(define (relative-prime? i) (= (gcd i n) 1))
(define (id i) i)
(define (next i) (+ i 1))
(filtered-accumulate-i relative-prime? * 1 id 2 next (- n 1)))


;; Test-it
;; Welcome to MzScheme version 209, Copyright (c) 2004 PLT Scheme, Inc.
;; > (sum-of-primes-square 3 10)
;; 83
;; > (product-of-relative-primes 10)
;; 189

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