一个有趣的表变换，请各位网友不吝赐教，谢谢

mahuan1279

夏生生 发表于 2022-2-27 16:17
谢谢提醒，您提供的思路已经很清晰了，我顺着这个思路向下走就是

偶数一样有遗漏。
_$ (f2 '("a" "b" "c" "d" "e" "f") 6 2)
(("a" "a") ("a" "a" "a") ("a" "a" "a" "a") ("a" "a" "a" "a" "a") ("a" "a" "a" "a" "a" "a") ("b" "b") ("b" "a" "b") ("b" "a" "a" "b") ("b" "a" "a" "a" "b") ("b" "a" "a" "a" "a" "b") ("b" "b" "b") ("b" "b" "b" "b") ("b" "b" "b" "b" "b") ("b" "b" "b" "b" "b" "b") ("c" "c") ("c" "a" "c") ("c" "a" "a" "c") ("c" "a" "a" "a" "c") ("c" "a" "a" "a" "a" "c") ("c" "b" "c") ("c" "b" "b" "c") ("c" "b" "b" "b" "c") ("c" "b" "b" "b" "b" "c") ("c" "c" "c") ("c" "c" "c" "c") ("c" "c" "c" "c" "c") ("c" "c" "c" "c" "c" "c") ("d" "d") ("d" "b" "d") ("d" "b" "b" "d") ("d" "b" "b" "b" "d") ("d" "b" "b" "b" "b" "d") ("d" "c" "d") ("d" "c" "c" "d") ("d" "c" "c" "c" "d") ("d" "c" "c" "c" "c" "d") ("d" "d" "d") ("d" "d" "d" "d") ("d" "d" "d" "d" "d") ("d" "d" "d" "d" "d" "d") ("e" "e") ("e" "c" "e") ("e" "c" "c" "e") ("e" "c" "c" "c" "e") ("e" "c" "c" "c" "c" "e") ("e" "d" "e") ("e" "d" "d" "e") ("e" "d" "d" "d" "e") ("e" "d" "d" "d" "d" "e") ("e" "e" "e") ("e" "e" "e" "e") ("e" "e" "e" "e" "e") ("e" "e" "e" "e" "e" "e") ("f" "f") ("f" "d" "f") ("f" "d" "d" "f") ("f" "d" "d" "d" "f") ("f" "d" "d" "d" "d" "f") ("f" "e" "f") ("f" "e" "e" "f") ("f" "e" "e" "e" "f") ("f" "e" "e" "e" "e" "f") ("f" "f" "f") ("f" "f" "f" "f") ("f" "f" "f" "f" "f") ("f" "f" "f" "f" "f" "f"))
_$

("f" "f" "d" "d" "f" "f")("f" "f" "e" "e" "f" "f")遗漏了。

Bao_lai

不知道楼主这个何用呢？

夏生生

Bao_lai 发表于 2022-2-27 16:53
不知道楼主这个何用呢？

不是蛮好解释，有序对称排点一类的应用

mahuan1279

夏生生 发表于 2022-2-27 15:17
应该向下面截图的这种逻辑

;;;前提m>=2
(defun f(a m n)
(defun f1(a m n)   
(setq flst (list a a) i 0 j (max 1 (- a n)) vlst nil)
(while (<= j a)
(while (< i (- m 2))
  (setq fflst (list j))
  (repeat i
       (setq fflst (cons j fflst)))
  (setq vlst (cons fflst vlst))
  (setq i (+ i 1))
)
(setq j (+ j 1) i 0)
)
(setq vlst (mapcar '(lambda (x) (cons a (reverse (cons a x)))) vlst))
(setq vlst (cons flst (reverse vlst)))
)
(defun f2(a m n)
(setq k (max 1 (- a n)) vvlst nil)
(while (< k a)
   (setq plst (list a a k) qlst (list a a k k))
   (while (<= (length plst) (/ (+ m 1) 2))
     (setq ptlst (append plst (cdr (reverse plst))))
     (setq vvlst (cons ptlst vvlst))
     (if (<= (length plst) (/ m 2))
          (progn
        (setq ptlst (append plst (reverse plst)))
        (setq vvlst (cons ptlst vvlst))
           )
         )
     (setq plst (cons a plst))
        )
   (while (<= (length qlst) (/ (+ m 1) 2))
     (setq qtlst (append qlst (cdr (reverse qlst))))
     (setq vvlst (cons qtlst vvlst))
         (if (<= (length qlst) (/ m 2))
           (progn
         (setq qtlst (append qlst (reverse qlst)))
         (setq vvlst (cons qtlst vvlst))
                )
          )
     (setq qlst (reverse (cons k (reverse qlst))))
  )
  (setq k (+ k 1))
)
vvlst
)
(setq ii 1 lst nil)
(while (<= ii a)
  (setq lst (append (f1 ii m n) lst))
  (setq ii (+ ii 1))
)
(setq lst (append (f2 a m n) lst))
)
F
_$ (f 4 3 2)
((4 4) (4 2 4) (4 3 4) (4 4 4) (3 3) (3 1 3) (3 2 3) (3 3 3) (2 2) (2 1 2) (2 2 2) (1 1) (1 1 1))
_$ (f 4 5 2)
((4 4 3 4 4) (4 4 2 4 4) (4 4) (4 2 4) (4 2 2 4) (4 2 2 2 4) (4 3 4) (4 3 3 4) (4 3 3 3 4) (4 4 4) (4 4 4 4) (4 4 4 4 4) (3 3) (3 1 3) (3 1 1 3) (3 1 1 1 3) (3 2 3) (3 2 2 3) (3 2 2 2 3) (3 3 3) (3 3 3 3) (3 3 3 3 3) (2 2) (2 1 2) (2 1 1 2) (2 1 1 1 2) (2 2 2) (2 2 2 2) (2 2 2 2 2) (1 1) (1 1 1) (1 1 1 1) (1 1 1 1 1))
_$ (f 4 4 2)
((4 4) (4 2 4) (4 2 2 4) (4 3 4) (4 3 3 4) (4 4 4) (4 4 4 4) (3 3) (3 1 3) (3 1 1 3) (3 2 3) (3 2 2 3) (3 3 3) (3 3 3 3) (2 2) (2 1 2) (2 1 1 2) (2 2 2) (2 2 2 2) (1 1) (1 1 1) (1 1 1 1))
_$ (f 5 5 2)
((5 5 4 5 5) (5 5 3 5 5) (5 5) (5 3 5) (5 3 3 5) (5 3 3 3 5) (5 4 5) (5 4 4 5) (5 4 4 4 5) (5 5 5) (5 5 5 5) (5 5 5 5 5) (4 4) (4 2 4) (4 2 2 4) (4 2 2 2 4) (4 3 4) (4 3 3 4) (4 3 3 3 4) (4 4 4) (4 4 4 4) (4 4 4 4 4) (3 3) (3 1 3) (3 1 1 3) (3 1 1 1 3) (3 2 3) (3 2 2 3) (3 2 2 2 3) (3 3 3) (3 3 3 3) (3 3 3 3 3) (2 2) (2 1 2) (2 1 1 2) (2 1 1 1 2) (2 2 2) (2 2 2 2) (2 2 2 2 2) (1 1) (1 1 1) (1 1 1 1) (1 1 1 1 1))
_$ (f 6 6 2)
((6 6 5 5 6 6) (6 6 5 6 6) (6 6 4 4 6 6) (6 6 4 6 6) (6 6) (6 4 6) (6 4 4 6) (6 4 4 4 6) (6 4 4 4 4 6) (6 5 6) (6 5 5 6) (6 5 5 5 6) (6 5 5 5 5 6) (6 6 6) (6 6 6 6) (6 6 6 6 6) (6 6 6 6 6 6) (5 5) (5 3 5) (5 3 3 5) (5 3 3 3 5) (5 3 3 3 3 5) (5 4 5) (5 4 4 5) (5 4 4 4 5) (5 4 4 4 4 5) (5 5 5) (5 5 5 5) (5 5 5 5 5) (5 5 5 5 5 5) (4 4) (4 2 4) (4 2 2 4) (4 2 2 2 4) (4 2 2 2 2 4) (4 3 4) (4 3 3 4) (4 3 3 3 4) (4 3 3 3 3 4) (4 4 4) (4 4 4 4) (4 4 4 4 4) (4 4 4 4 4 4) (3 3) (3 1 3) (3 1 1 3) (3 1 1 1 3) (3 1 1 1 1 3) (3 2 3) (3 2 2 3) (3 2 2 2 3) (3 2 2 2 2 3) (3 3 3) (3 3 3 3) (3 3 3 3 3) (3 3 3 3 3 3) (2 2) (2 1 2) (2 1 1 2) (2 1 1 1 2) (2 1 1 1 1 2) (2 2 2) (2 2 2 2) (2 2 2 2 2) (2 2 2 2 2 2) (1 1) (1 1 1) (1 1 1 1) (1 1 1 1 1) (1 1 1 1 1 1))
_$

x_s_s_1

mahuan1279 发表于 2022-2-27 18:07
;;;前提m>=2
(defun f(a m n)
(defun f1(a m n)   

谢谢了，我再仔细研读一下

mahuan1279

mahuan1279 发表于 2022-2-27 18:07
;;;前提m>=2
(defun f(a m n)
(defun f1(a m n)   

由两个表拼凑而成总显得不够完美，直接一步到位。
_$ (defun f(a m n)
  (setq i 1 vvlst nil)
  (while (<= i a)
    (setq k (max 1 (- i n)))
        (while (<= k i)       
            (setq vlst nil)
                (setq plst (list (list i k i)))
        (setq qlst (list (list i k k i)))
                (while (<= (length (car plst)) m)
               (setq vlst (append plst vlst))
               (if (= (length plst) 1)
                               (setq plst (list (list i k k k i) (list i i k i i)))
                   (setq plst (cons (cons i (cons k (cdr (reverse (cons i (cons k (cdr (car plst)))))))) (mapcar '(lambda (x) (cons i (reverse (cons i x)))) plst)))
                                )
             )
                (while (<= (length (car qlst)) m)
               (setq vlst (append qlst vlst))
               (if (= (length plst) 1)
                               (setq qlst (list (list i k k k k i) (list i i k k i i)))
                   (setq qlst (cons (cons i (cons k (cdr (reverse (cons i (cons k (cdr (car qlst)))))))) (mapcar '(lambda (x) (cons i (reverse (cons i x)))) qlst)))
                                )
             )
                (setq vvlst (append vlst vvlst))
        (setq k (+ k 1))
        )
   (setq vvlst (cons (list i i) vvlst))
   (setq i (+ i 1))
  )
  vvlst
)
F
_$ (f 4 3 2)
((4 4) (4 4 4) (4 3 4) (4 2 4) (3 3) (3 3 3) (3 2 3) (3 1 3) (2 2) (2 2 2) (2 1 2) (1 1) (1 1 1))
_$ (f 5 4 3)
((5 5) (5 5 5 5) (5 5 5) (5 4 4 5) (5 4 5) (5 3 3 5) (5 3 5) (5 2 2 5) (5 2 5) (4 4) (4 4 4 4) (4 4 4) (4 3 3 4) (4 3 4) (4 2 2 4) (4 2 4) (4 1 1 4) (4 1 4) (3 3) (3 3 3 3) (3 3 3) (3 2 2 3) (3 2 3) (3 1 1 3) (3 1 3) (2 2) (2 2 2 2) (2 2 2) (2 1 1 2) (2 1 2) (1 1) (1 1 1 1) (1 1 1))
_$

mahuan1279

mahuan1279 发表于 2022-2-28 23:42
由两个表拼凑而成总显得不够完美，直接一步到位。
_$ (defun f(a m n)
  (setq i 1 vvlst nil)

当K=i时，有重复情况，需单独处理。
_$ (defun f(a m n)
  (setq i 1 vvlst nil)
  (while (<= i a)
    (setq k (max 1 (- i n)))
        (while (< k i)      
            (setq vlst nil)
                (setq plst (list (list i k i)))
        (setq qlst (list (list i k k i)))
                (while (<= (length (car plst)) m)
               (setq vlst (append plst vlst))
               (if (= (length plst) 1)
                               (setq plst (list (list i k k k i) (list i i k i i)))
                   (setq plst (cons (cons i (cons k (cdr (reverse (cons i (cons k (cdr (car plst)))))))) (mapcar '(lambda (x) (cons i (reverse (cons i x)))) plst)))
                                )
             )
                (while (<= (length (car qlst)) m)
               (setq vlst (append qlst vlst))
               (if (= (length plst) 1)
                               (setq qlst (list (list i k k k k i) (list i i k k i i)))
                   (setq qlst (cons (cons i (cons k (cdr (reverse (cons i (cons k (cdr (car qlst)))))))) (mapcar '(lambda (x) (cons i (reverse (cons i x)))) qlst)))
                                )
             )
                (setq vvlst (append vlst vvlst))
        (setq k (+ k 1))
        )
   (setq vvlst (cons (list i i) vvlst))
   (if (= k i)
       (progn
              (setq vlst nil)
              (setq plst (list i i i))
                  (while (<= (length plst) m)
                       (setq vlst (append (list plst) vlst))
                           (setq plst (cons i plst))
                  )
           )
   )
   (setq vvlst (append vlst vvlst))
   (setq i (+ i 1))
  )
  vvlst
)
F
_$ (f 3 2 2)
((3 3) (2 2) (1 1))
_$ (f 4 3 2)
((4 4 4) (4 4) (4 3 4) (4 2 4) (3 3 3) (3 3) (3 2 3) (3 1 3) (2 2 2) (2 2) (2 1 2) (1 1 1) (1 1))
_$ (f 4 4 2)
((4 4 4 4) (4 4 4) (4 4) (4 3 3 4) (4 3 4) (4 2 2 4) (4 2 4) (3 3 3 3) (3 3 3) (3 3) (3 2 2 3) (3 2 3) (3 1 1 3) (3 1 3) (2 2 2 2) (2 2 2) (2 2) (2 1 1 2) (2 1 2) (1 1 1 1) (1 1 1) (1 1))
_$ (f 6 4 2)
((6 6 6 6) (6 6 6) (6 6) (6 5 5 6) (6 5 6) (6 4 4 6) (6 4 6) (5 5 5 5) (5 5 5) (5 5) (5 4 4 5) (5 4 5) (5 3 3 5) (5 3 5) (4 4 4 4) (4 4 4) (4 4) (4 3 3 4) (4 3 4) (4 2 2 4) (4 2 4) (3 3 3 3) (3 3 3) (3 3) (3 2 2 3) (3 2 3) (3 1 1 3) (3 1 3) (2 2 2 2) (2 2 2) (2 2) (2 1 1 2) (2 1 2) (1 1 1 1) (1 1 1) (1 1))
_$ (f 6 5 2)
((6 6 6 6 6) (6 6 6 6) (6 6 6) (6 6) (6 5 5 6) (6 5 5 5 6) (6 6 5 6 6) (6 5 6) (6 4 4 6) (6 4 4 4 6) (6 6 4 6 6) (6 4 6) (5 5 5 5 5) (5 5 5 5) (5 5 5) (5 5) (5 4 4 5) (5 4 4 4 5) (5 5 4 5 5) (5 4 5) (5 3 3 5) (5 3 3 3 5) (5 5 3 5 5) (5 3 5) (4 4 4 4 4) (4 4 4 4) (4 4 4) (4 4) (4 3 3 4) (4 3 3 3 4) (4 4 3 4 4) (4 3 4) (4 2 2 4) (4 2 2 2 4) (4 4 2 4 4) (4 2 4) (3 3 3 3 3) (3 3 3 3) (3 3 3) (3 3) (3 2 2 3) (3 2 2 2 3) (3 3 2 3 3) (3 2 3) (3 1 1 3) (3 1 1 1 3) (3 3 1 3 3) (3 1 3) (2 2 2 2 2) (2 2 2 2) (2 2 2) (2 2) (2 1 1 2) (2 1 1 1 2) (2 2 1 2 2) (2 1 2) (1 1 1 1 1) (1 1 1 1) (1 1 1) (1 1))
_$ (f 6 6 3)
((6 6 6 6 6 6) (6 6 6 6 6) (6 6 6 6) (6 6 6) (6 6) (6 5 5 5 5 6) (6 6 5 5 6 6) (6 5 5 6) (6 5 5 5 6) (6 6 5 6 6) (6 5 6) (6 4 4 4 4 6) (6 6 4 4 6 6) (6 4 4 6) (6 4 4 4 6) (6 6 4 6 6) (6 4 6) (6 3 3 3 3 6) (6 6 3 3 6 6) (6 3 3 6) (6 3 3 3 6) (6 6 3 6 6) (6 3 6) (5 5 5 5 5 5) (5 5 5 5 5) (5 5 5 5) (5 5 5) (5 5) (5 4 4 4 4 5) (5 5 4 4 5 5) (5 4 4 5) (5 4 4 4 5) (5 5 4 5 5) (5 4 5) (5 3 3 3 3 5) (5 5 3 3 5 5) (5 3 3 5) (5 3 3 3 5) (5 5 3 5 5) (5 3 5) (5 2 2 2 2 5) (5 5 2 2 5 5) (5 2 2 5) (5 2 2 2 5) (5 5 2 5 5) (5 2 5) (4 4 4 4 4 4) (4 4 4 4 4) (4 4 4 4) (4 4 4) (4 4) (4 3 3 3 3 4) (4 4 3 3 4 4) (4 3 3 4) (4 3 3 3 4) (4 4 3 4 4) (4 3 4) (4 2 2 2 2 4) (4 4 2 2 4 4) (4 2 2 4) (4 2 2 2 4) (4 4 2 4 4) (4 2 4) (4 1 1 1 1 4) (4 4 1 1 4 4) (4 1 1 4) (4 1 1 1 4) (4 4 1 4 4) (4 1 4) (3 3 3 3 3 3) (3 3 3 3 3) (3 3 3 3) (3 3 3) (3 3) (3 2 2 2 2 3) (3 3 2 2 3 3) (3 2 2 3) (3 2 2 2 3) (3 3 2 3 3) (3 2 3) (3 1 1 1 1 3) (3 3 1 1 3 3) (3 1 1 3) (3 1 1 1 3) (3 3 1 3 3) (3 1 3) (2 2 2 2 2 2) (2 2 2 2 2) (2 2 2 2) (2 2 2) (2 2) (2 1 1 1 1 2) (2 2 1 1 2 2) (2 1 1 2) (2 1 1 1 2) (2 2 1 2 2) (2 1 2) (1 1 1 1 1 1) (1 1 1 1 1) (1 1 1 1) (1 1 1) (1 1))
_$

xyp1964

1. ;; (tt lst表 n1中间最大元素数量)
   
2. (defun tt (lst n1 / i lst1 j k d)
   
3.   (defun abc (x n / b) (repeat n (setq b (cons x b))) b)
   
4.   (setq        lst1 '()
   
5.         j    0
   
6.   )
   
7.   (foreach a lst
   
8.     (setq i 0)
   
9.     (repeat (1+ n1)
   
10.       (setq k 0)
    
11.       (foreach b lst
    
12.         (if (<= k j)
    
13.           (setq        d    (append (list a) (abc b i) (list a))
    
14.                 lst1 (if (not (member d lst1))
    
15.                        (cons d lst1)
    
16.                        lst1
    
17.                      )
    
18.           )
    
19.         )
    
20.         (setq k (1+ k))
    
21.       )
    
22.       (setq i (1+ i))
    
23.     )
    
24.     (setq j (1+ j))
    
25.   )
    
26.   (reverse lst1)
    
27. )
    
28. 
    
29. ;; (tt '("a" "b" "c" "d" "e") 3)
    
30. ;; '(("a" "a") ("a" "a" "a") ("a" "a" "a" "a") ("a" "a" "a" "a" "a") ("b" "b") ("b" "a" "b") ("b" "b" "b") ("b" "a" "a" "b") ("b" "b" "b" "b") ("b" "a" "a" "a" "b") ("b" "b" "b" "b" "b") ("c" "c") ("c" "a" "c") ("c" "b" "c") ("c" "c" "c") ("c" "a" "a" "c") ("c" "b" "b" "c") ("c" "c" "c" "c") ("c" "a" "a" "a" "c") ("c" "b" "b" "b" "c") ("c" "c" "c" "c" "c") ("d" "d") ("d" "a" "d") ("d" "b" "d") ("d" "c" "d") ("d" "d" "d") ("d" "a" "a" "d") ("d" "b" "b" "d") ("d" "c" "c" "d") ("d" "d" "d" "d") ("d" "a" "a" "a" "d") ("d" "b" "b" "b" "d") ("d" "c" "c" "c" "d") ("d" "d" "d" "d" "d") ("e" "e") ("e" "a" "e") ("e" "b" "e") ("e" "c" "e") ("e" "d" "e") ("e" "e" "e") ("e" "a" "a" "e") ("e" "b" "b" "e") ("e" "c" "c" "e") ("e" "d" "d" "e") ("e" "e" "e" "e") ("e" "a" "a" "a" "e") ("e" "b" "b" "b" "e") ("e" "c" "c" "c" "e") ("e" "d" "d" "d" "e") ("e" "e" "e" "e" "e"))
    

;; (tt lst表 n1中间最大元素数量)<br /> (defun tt (lst n1 / i lst1 j k d)<br />   (defun abc (x n / b) (repeat n (setq b (cons x b))) b)<br />   (setq        lst1 '()<br />         j    0<br />   )<br />   (foreach a lst<br />     (setq i 0)<br />     (repeat (1+ n1)<br />       (setq k 0)<br />       (foreach b lst<br />         (if (<= k j)<br />           (setq        d    (append (list a) (abc b i) (list a))<br />                 lst1 (if (not (member d lst1))<br />                        (cons d lst1)<br />                        lst1<br />                      )<br />           )<br />         )<br />         (setq k (1+ k))<br />       )<br />       (setq i (1+ i))<br />     )<br />     (setq j (1+ j))<br />   )<br />   (reverse lst1)<br /> )<br /> <br /> ;; (tt '("a" "b" "c" "d" "e") 3)<br /> ;; '(("a" "a") ("a" "a" "a") ("a" "a" "a" "a") ("a" "a" "a" "a" "a") ("b" "b") ("b" "a" "b") ("b" "b" "b") ("b" "a" "a" "b") ("b" "b" "b" "b") ("b" "a" "a" "a" "b") ("b" "b" "b" "b" "b") ("c" "c") ("c" "a" "c") ("c" "b" "c") ("c" "c" "c") ("c" "a" "a" "c") ("c" "b" "b" "c") ("c" "c" "c" "c") ("c" "a" "a" "a" "c") ("c" "b" "b" "b" "c") ("c" "c" "c" "c" "c") ("d" "d") ("d" "a" "d") ("d" "b" "d") ("d" "c" "d") ("d" "d" "d") ("d" "a" "a" "d") ("d" "b" "b" "d") ("d" "c" "c" "d") ("d" "d" "d" "d") ("d" "a" "a" "a" "d") ("d" "b" "b" "b" "d") ("d" "c" "c" "c" "d") ("d" "d" "d" "d" "d") ("e" "e") ("e" "a" "e") ("e" "b" "e") ("e" "c" "e") ("e" "d" "e") ("e" "e" "e") ("e" "a" "a" "e") ("e" "b" "b" "e") ("e" "c" "c" "e") ("e" "d" "d" "e") ("e" "e" "e" "e") ("e" "a" "a" "a" "e") ("e" "b" "b" "b" "e") ("e" "c" "c" "c" "e") ("e" "d" "d" "d" "e") ("e" "e" "e" "e" "e")) <br />

mahuan1279

xyp1964 发表于 2022-3-1 22:01

_$ (tt '("a" "b" "c" "d" "e") 4)
(("a" "a") ("a" "a" "a") ("a" "a" "a" "a") ("a" "a" "a" "a" "a") ("a" "a" "a" "a" "a" "a") ("b" "b") ("b" "a" "b") ("b" "b" "b") ("b" "a" "a" "b") ("b" "b" "b" "b") ("b" "a" "a" "a" "b") ("b" "b" "b" "b" "b") ("b" "a" "a" "a" "a" "b") ("b" "b" "b" "b" "b" "b") ("c" "c") ("c" "a" "c") ("c" "b" "c") ("c" "c" "c") ("c" "a" "a" "c") ("c" "b" "b" "c") ("c" "c" "c" "c") ("c" "a" "a" "a" "c") ("c" "b" "b" "b" "c") ("c" "c" "c" "c" "c") ("c" "a" "a" "a" "a" "c") ("c" "b" "b" "b" "b" "c") ("c" "c" "c" "c" "c" "c") ("d" "d") ("d" "a" "d") ("d" "b" "d") ("d" "c" "d") ("d" "d" "d") ("d" "a" "a" "d") ("d" "b" "b" "d") ("d" "c" "c" "d") ("d" "d" "d" "d") ("d" "a" "a" "a" "d") ("d" "b" "b" "b" "d") ("d" "c" "c" "c" "d") ("d" "d" "d" "d" "d") ("d" "a" "a" "a" "a" "d") ("d" "b" "b" "b" "b" "d") ("d" "c" "c" "c" "c" "d") ("d" "d" "d" "d" "d" "d") ("e" "e") ("e" "a" "e") ("e" "b" "e") ("e" "c" "e") ("e" "d" "e") ("e" "e" "e") ("e" "a" "a" "e") ("e" "b" "b" "e") ("e" "c" "c" "e") ("e" "d" "d" "e") ("e" "e" "e" "e") ("e" "a" "a" "a" "e") ("e" "b" "b" "b" "e") ("e" "c" "c" "c" "e") ("e" "d" "d" "d" "e") ("e" "e" "e" "e" "e") ("e" "a" "a" "a" "a" "e") ("e" "b" "b" "b" "b" "e") ("e" "c" "c" "c" "c" "e") ("e" "d" "d" "d" "d" "e") ("e" "e" "e" "e" "e" "e"))
_$

似乎少了("b" "b" "a" "b" "b")、("b" "b" "a" "a" "b" "b") ……等等情形。

xyp1964

mahuan1279 发表于 2022-3-1 22:54
_$ (tt '("a" "b" "c" "d" "e") 4)
(("a" "a") ("a" "a" "a") ("a" "a" "a" "a") ("a" "a" "a" "a" "a") ...

实在是看不太懂楼主的要求