找到方案中数字的数字的平方和(find sum of the squares of the digits of a number in scheme)

找到方案中数字的数字的平方和(find sum of the squares of the digits of a number in scheme)

我需要在scheme中编写一个函数来计算平方数字的总和。

ex - (sum-of-digits 130) > 10

这是我的功能。

(define (sum-of-digits x) (if (= x 0) 0 (+ (modulo x 10) (sum-of-digits (/ (- x (modulo x 10)) 10)))))

它对某些数字不起作用。 当我输入（数字和130）时，它返回4.我该如何解决这个问题？

另外我需要用这个函数来找到0,1,4,16,20,37,42,58,89,145的停止数字

ex :- (stop? 42)  #t (stop? 31)  #f

如何使用上面的和位函数和来做到这一点？

I need to write a function in scheme which calculates the sum of square digits.

ex - (sum-of-digits 130) > 10

This is my function.

(define (sum-of-digits x) (if (= x 0) 0 (+ (modulo x 10) (sum-of-digits (/ (- x (modulo x 10)) 10)))))

it doesn't work for some numbers. When I entered (sum-of-digits 130) , it returns 4. How can i fix this ?

Also I need to use this function to find the stop numbers which are 0,1,4,16,20,37,42,58,89,145

ex :- (stop? 42)  #t (stop? 31)  #f

How can I do this using the function sum of sum-of-digits above?

最满意答案

你忘了实际对每个数字进行平方 ，并且有一种更简单的方法来获得商：

(define (sum-of-digits x) (if (= x 0) 0 (+ (sqr (modulo x 10)) (sum-of-digits (quotient x 10)))))

对于问题的第二部分：

(define (stop? x) (let ((sum (sum-of-digits x))) (if (member sum '(0 1 4 16 20 37 42 58 89 145)) #t #f)))

You forgot to actually square each digit, and there's a simpler way to obtain the quotient:

(define (sum-of-digits x) (if (= x 0) 0 (+ (sqr (modulo x 10)) (sum-of-digits (quotient x 10)))))

For the second part of the question:

(define (stop? x) (let ((sum (sum-of-digits x))) (if (member sum '(0 1 4 16 20 37 42 58 89 145)) #t #f)))