Scheme:FizzBuzz
Fizz-Buzz問題
(define-syntax aa
(syntax-rules ()
((_ current v next)
(define-syntax current
(syntax-rules ()
((_)
#f)
((_ elem . rest)
(begin
(print v)
(next . rest))))))
((_ current next)
(define-syntax current
(syntax-rules ()
((_)
#f)
((_ elem . rest)
(begin
(print elem)
(next . rest))))))))
(define-syntax ab
(syntax-rules (!)
((_ v)
#f)
((_ current ! msg next rest ...)
(begin
(aa current msg next)
(ab next rest ...)))
((_ current next rest ...)
(begin
(aa current next)
(ab next rest ...)))))
(ab a1 a2 a3 ! "Fizz" a4 a5 ! "Buzz"
a6 ! "Fizz" a7 a8 a9 ! "Fizz" a10 ! "Buzz"
a11 a12 ! "Fizz" a13 a14 a15 ! "FizzBuzz"
a1)
(a1 1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18 19
20 21 22 23 24 25 26 27 28 29
30 31 32 33 34 35 36 37 38 39
40 41 42 43 44 45 46 47 48 49
50 51 52 53 54 55 56 57 58 59
60 61 62 63 64 65 66 67 68 69
70 71 72 73 74 75 76 77 78 79
80 81 82 83 84 85 86 87 88 89
90 91 92 93 94 95 96 97 98 99
100)
- nobsun(2007/05/08 11:13:53 PDT)
(use srfi-1) (use srfi-13) (use util.match)
(define (foo m s) (match-lambda ((n . t) (if (= 0 (modulo n m)) (cons n (string-append s t)) (cons n t)))))
(define bar (match-lambda ((n . s) (if (string-null? s) (x->string n) s))))
(define fizz (foo 3 "Fizz"))
(define buzz (foo 5 "Buzz"))
(define (fizz-buzz xs) (map (compose bar (compose fizz (compose buzz (cut cons <> "")))) xs))
- 齊藤(2007/05/08 18:30:21 PDT): Scheme的に素直に抽象化したつもり。
(use srfi-1)
(define-syntax fizz-buzz
(syntax-rules ()
((_ ((num word)) x)
(if (zero? (modulo x num)) word x))
((_ ((num word) (num1 word1) ...) x)
(if (zero? (modulo x num))
(string-append
word
(let ((a (fizz-buzz ((num1 word1) ...) x)))
(if (number? a) "" a)))
x))))
(for-each
(lambda(x)
(display (fizz-buzz ((3 "Fizz")(5 "Buzz")) x))
(newline))
(iota 100 1 1))
- gemma(2007/05/14 06:26:22 PDT):剰余を使わずに、循環リストを使う方法。
(use srfi-1)
(define (fizzbuzz)
(define fizz (circular-list #f #f "fizz"))
(define buzz (circular-list #f #f #f #f "buzz"))
(map (lambda (f b i)
(or (and f b (string-append f b)) f b i))
fizz buzz (iota 100 1)))
(print (fizzbuzz))
- shinya(2007/05/19 01:48:51 PDT):unfoldを使ってみた
(use srfi-1)
(define (dividable? x y) (= (modulo x y) 0))
(define fizzbuzz-spec
`((,(lambda (x) (and (dividable? x 3) (dividable? x 5))) . "FizzBuzz")
(,(cut dividable? <> 3) . "Fizz")
(,(cut dividable? <> 5) . "Buzz")))
(define (fizzbuzz spec)
(lambda (x)
(cond
((find (lambda (pair) ((car pair) x)) spec) => cdr)
(else x))))
(for-each print (unfold (pa$ < 100) (fizzbuzz fizzbuzz-spec) (pa$ + 1) 1))
- 普通だけど。 Common Lisp Loop macro で。
(loop for i from 1 to 100
collect
(cond
((= 0 (mod i 15)) "FizzBuzz")
((= 0 (mod i 3)) "Fizz")
((= 0 (mod i 5)) "Buzz")
(t i)))
(use srfi-1)
(use srfi-11)
(use text.tree)
(use gauche.sequence)
;; (fizz 0 '(3 "Fizz" 5 "Buzz") #f)
;; => ("fizz") or ("fizz" "Buzz") or 0
(define (fizz n lst has)
(cond ((null? lst)
(if has '() n))
((= 0 (remainder n (car lst)))
(cons (cadr lst) (fizz n (cddr lst) #t)))
(else
(fizz n (cddr lst) has))))
;; input: (fizzbuzz 1 100 '(3 "fizz" 5 "buzz" ...))
(define (fizzbuzz start end preds)
(map (cut fizz <> preds #f) (iota (- (+ end 1) start ) start)))
;; (parse-args '("10" "300" "3" "buz" "5" "foo"))
;; => (10 300 (3 "buz" 5 "foo"))
(define (parse-args args)
(values (string->number (car args)) (string->number (cadr args))
(map-with-index
(lambda (i x)
(if (= 0 (remainder i 2))
(string->number x)
x))
(cddr args))))
;; input: 1 100 3 Fizz 5 Buzz ...
(define (main args)
(if (> 4 (length args))
(begin (print "ex: 1 100 3 Fizz 5 Buzz")(exit 1))
(let-values (((start end preds) (parse-args (cdr args))))
(for-each (lambda (x)
(if (pair? x) (begin (write-tree x) (newline))
(print x)))
(fizzbuzz start end preds)))))