Language Scheme: find the sum of proper divisors

问题

I am wondering how to write a function calculating the sum of proper divisors of a integer greater than 1.

(define (sum-of-proper-divisors n)
          (cond
           [(= n 1) 1]
           [(= 0 (remainder n (sub1 n)))
            (+ (remainder n (sub1 n)) (sum-of-proper-divisors (sub1 (sub1 n))))]
           [else (sum-of-proper-divisors (sub1 n))]))

This is the code that I wrote, however, it does not work. It will never stop evaluating because it will always do n-1. And I don't know how to fix this. Also, there might be other problems. How to put the restriction that makes the function stop evaluating when the divisor becomes 1?

回答1:

You're confusing the number n whose divisors you want to find, with said divisors. Notice that n never changes, what must be modified at each step is the current integer being tested (a possible divisor). For that you'll need to pass around two parameters:

(define (sum-of-proper-divisors n i)
  (cond
    [(= i 1) 1]
    [(= (remainder n i) 0)
     (+ i (sum-of-proper-divisors n (sub1 i)))]
    [else (sum-of-proper-divisors n (sub1 i))]))

Call it like this, at the beginning i must be one unit less than n:

(sum-of-proper-divisors 10 9)
=> 8

If having two parameters bothers you there are several ways to pass a single parameter, for instance using a named let:

(define (sum-of-proper-divisors n)
  (let loop ((i (sub1 n)))
    (cond
      [(= i 1) 1]
      [(= (remainder n i) 0)
       (+ i (loop (sub1 i)))]
      [else (loop (sub1 i))])))

来源：https://stackoverflow.com/questions/19760269/language-scheme-find-the-sum-of-proper-divisors