scheme(gauche)で世界のナベアツ問題

3がつく数と3の倍数の時だけアホになる(片仮名になる)それ以外は普通に数字を表示
peercastの大会で課題だったので、っていうか就活の課題やれ俺www

(use srfi-1)
(use srfi-13)

(define (unpack x y)
  (if (= x 0)
      y
      (unpack (quotient x 10) (append  (list (modulo x 10)) y))))

(define (3baisu? x) (= (modulo x 3) 0))

(define (3gatuku? x)
  (any (lambda (i) (= i 3)) (unpack x '())))

(define (senkurai x)
  "ｾﾝ")

(define (hyakukurai x)
  (ref '( ""　 "ﾋｬｸ"  "ﾆﾋｬｸ" "ｻﾝﾋﾞｬｸ" "ﾖﾝﾋｬｸ" "ｺﾞﾋｬｸ" "ﾛｯﾋﾟｬｸ" "ﾅﾅﾋｬｸ" "ﾊｯﾋﾟｬｸ" "ｷｭｳﾋｬｸ") x))

(define (jukurai x)
  (ref '( "" "ｼﾞｭｳ" "ﾆｼﾞｭｳ" "ｻﾝｼﾞｭｳ" "ﾖﾝｼﾞｭｳ" "ｺﾞｼﾞｭｳ" "ﾛｸｼﾞｭｳ" "ﾅﾅｼﾞｭｳ" "ﾊﾁｼﾞｭｳ" "ｷｭｳｼﾞｭｳ") x))

(define (ichikurai x)
  (ref '( "" "ｲﾁ" "ﾆ" "ｻﾝ" "ﾖﾝ" "ｺﾞ" "ﾛｸ" "ﾅﾅ" "ﾊﾁ" "ｷｭｳ") x))

(define func  `(() ,ichikurai ,jukurai ,hyakukurai ,senkurai))

(define (tokatakana x)
  (let loop ((lis (unpack x '())) (kana ""))
    (if (= (length lis) 0)
	kana
	(loop (cdr lis) (string-append kana ((ref func (length lis)) (car lis)))))))
		    

(for-each print (map (lambda (x)
		       (if (or (3baisu? x) (3gatuku? x))
			   (string-append (tokatakana x) "!")
			   x))
		       (iota 1000 1 1)))

出力の最後のほう

950
ｷｭｳﾋｬｸｺﾞｼﾞｭｳｲﾁ!
952
ｷｭｳﾋｬｸｺﾞｼﾞｭｳｻﾝ!
ｷｭｳﾋｬｸｺﾞｼﾞｭｳﾖﾝ!
955
956
ｷｭｳﾋｬｸｺﾞｼﾞｭｳﾅﾅ!
958
959
ｷｭｳﾋｬｸﾛｸｼﾞｭｳ!
961
962
ｷｭｳﾋｬｸﾛｸｼﾞｭｳｻﾝ!
964
965
ｷｭｳﾋｬｸﾛｸｼﾞｭｳﾛｸ!
967
968
ｷｭｳﾋｬｸﾛｸｼﾞｭｳｷｭｳ!
970
971
ｷｭｳﾋｬｸﾅﾅｼﾞｭｳﾆ!
ｷｭｳﾋｬｸﾅﾅｼﾞｭｳｻﾝ!
974
ｷｭｳﾋｬｸﾅﾅｼﾞｭｳｺﾞ!
976
977
ｷｭｳﾋｬｸﾅﾅｼﾞｭｳﾊﾁ!
979
980
ｷｭｳﾋｬｸﾊﾁｼﾞｭｳｲﾁ!
982
ｷｭｳﾋｬｸﾊﾁｼﾞｭｳｻﾝ!
ｷｭｳﾋｬｸﾊﾁｼﾞｭｳﾖﾝ!
985
986
ｷｭｳﾋｬｸﾊﾁｼﾞｭｳﾅﾅ!
988
989
ｷｭｳﾋｬｸｷｭｳｼﾞｭｳ!
991
992
ｷｭｳﾋｬｸｷｭｳｼﾞｭｳｻﾝ!
994
995
ｷｭｳﾋｬｸｷｭｳｼﾞｭｳﾛｸ!
997
998
ｷｭｳﾋｬｸｷｭｳｼﾞｭｳｷｭｳ!
1000