Lisp的整除

Question

`（disivible「（1 2 3 4 5 6 7））這是我到目前爲止已經取得：

(defun divisible(n) 
    (cond ((eq n 0) nill) (eq(rem n 3) 0) t)('else 0))) 

但我在口齒不清新的，我不知道如何讓它不顯示我可以被3整除的數字，但只需添加數字並顯示結果。有人能幫助我嗎？

Answer 1

過濾remove-if-not和應用+

(defun divisible (num-list) 
    (apply #'+ (remove-if-not #'(lambda(x)(zerop (rem x 3))) num-list))) 

Answer 2

由於apply可能不長名單的工作，這裏有一個loop版本：

(defun divisible (lst) 
    (loop for i in lst when (zerop (rem i 3)) sum i)) 

，或者使用reduce

(defun divisible (lst) 
    (reduce '+ lst :key (lambda (i) (if (zerop (rem i 3)) i 0)))) 

遞歸版本

(defun divisible (lst) 
    (if (null lst) 
    0 
    (let ((i (car lst)))  
     (if (zerop (rem i 3)) 
     (+ i (divisible (cdr lst))) 
     (divisible (cdr lst)))))) 

或更笨拙尾遞歸版本

(defun divisible (lst) 
    (labels 
     ((sub (lst res) 
     (if (null lst) 
      res 
      (let ((i (car lst))) 
      (sub 
       (cdr lst) 
       (if (zerop (rem i 3)) (+ i res) res)))))) 
    (sub lst 0))) 

Answer 3

(defun divisible (num-list) 
    (reduce #'+ 
      (remove-if-not #'zerop 
         num-list 
         :key (lambda (x) 
           (mod x 3))))) 

Answer 4

(defun sum3s (x) (loop for i in x :if (= (mod i 3) 0) :sum i)) 

這是上述答案的另一變型。