所以我很难写这个程序 范围是一个需要大量的程序(比如1,000,000)并将其分成数字(即1,500,310 - > 1百万500,000 3百1 10 0一)。
#lang r5rs
(define (three_names x)
(let loop ((x x)
(myList '()))
(if (< x 10)
(cons x myList)
(loop (quotient x 10)
(cons (remainder x 10) myList)))))
我已经得到它以便循环并将这些值返回到列表中,并提供stackoverflow的帮助。
(即,这将取100并将其放入'(1 0 0),这是有帮助的,但没有足够的帮助。我从星期三起就一直在敲我的头!)
答案 0 :(得分:1)
我会选择这样的事情:
#lang r5rs
(define (three-names n)
(let loop ((n n)
(units '((10 "one") (10 "ten") (10 "hundred") (1000 "thousand") (1000 "million")))
(res ""))
(if (or (zero? n) (null? units))
res
(let* ((unit (car units)) (div (car unit)) (txt (cadr unit)))
(let ((q (quotient n div)) (r (remainder n div)))
(loop q
(cdr units)
(string-append (number->string r) " " txt " " res)))))))
测试:
> (three-names 1500310)
"1 million 500 thousand 3 hundred 1 ten 0 one "
> (three-names 1001)
"1 thousand 0 hundred 0 ten 1 one "
修改强>
的替代版本
可能是:
(define (three-names n)
(if (zero? n)
"zero"
(let loop ((n n)
(units '((10 one) (10 ten) (10 hundred) (1000 thousand) (1000 million)))
(res '()))
(display n) (display " - ") (display res) (display " - ") (display units) (newline)
(if (or (zero? n) (null? units))
res
(let* ((unit (car units)) (div (car unit)) (txt (cadr unit)))
(let ((q (quotient n div)) (r (remainder n div)))
(loop q
(cdr units)
(cons r (cons txt res)))))))))
现在:
> (display (three-names 1500310))
1500310 - () - ((10 one) (10 ten) (10 hundred) (1000 thousand) (1000 million))
150031 - (0 one) - ((10 ten) (10 hundred) (1000 thousand) (1000 million))
15003 - (1 ten 0 one) - ((10 hundred) (1000 thousand) (1000 million))
1500 - (3 hundred 1 ten 0 one) - ((1000 thousand) (1000 million))
1 - (500 thousand 3 hundred 1 ten 0 one) - ((1000 million))
0 - (1 million 500 thousand 3 hundred 1 ten 0 one) - ()
(1 million 500 thousand 3 hundred 1 ten 0 one)