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 2022-2-25 08:49
谢谢您的关注，我在一楼修改并尽力明确了需求，能否麻烦您再看下，谢谢

是这么分解么？

mahuan1279

好像这样分解才对。如此分解之后，算法也就跟着出来了。

guosheyang

研究的好深奥，很训练思维逻辑 赞一个！

mahuan1279

_$ (defun f1(a m)  
(setq flst (list a a) i 0 j 1  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)))
)
F1
_$ (f1 1 5)
((1 1) (1 1 1) (1 1 1 1) (1 1 1 1 1))
_$ (f1 2 5)
((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))
_$ (f1 3 5)
((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))
_$ (f1 4 5)
((4 4) (4 1 4) (4 1 1 4) (4 1 1 1 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))
_$

夏生生

mahuan1279 发表于 2022-2-27 12:56
是这么分解么？

应该向下面截图的这种逻辑

夏生生

mahuan1279 发表于 2022-2-27 15:16
_$ (defun f1(a m)  
(setq flst (list a a) i 0 j 1  vlst nil)
(while (

非常感谢，您的逻辑能力太强了！！我试着用nth将它转成针对表看看

mahuan1279

再补上（aa*****aa）(aaa***aaa)(aaaa*aaaa)……形式答案就齐全了。（*与a不相同）

夏生生

mahuan1279 发表于 2022-2-27 15:37
再补上（aa*****aa）(aaa***aaa)(aaaa*aaaa)……形式答案就齐全了。（*与a不相同）

谢谢您，完美解决

2. (defun f2 (lst m n / i lsta)
   
3.   (defun f1 (a m n / fflst flst i j vlst)
   
4.     (setq flst (list a a)
   
5.           i    0
   
6.           j    (max 1 (- a n))
   
7.           vlst nil
   
8.           )
   
9.     (while (<= j a)
   
10.       (while (< i (- m 2))
    
11.         (setq fflst (list j))
    
12.         (repeat        i
    
13.           (setq fflst (cons j fflst))
    
14.           )
    
15.         (setq vlst (cons fflst vlst))
    
16.         (setq i (+ i 1))
    
17.         )
    
18.       (setq j (+ j 1)
    
19.             i 0
    
20.             )
    
21.       )
    
22.     (setq
    
23.       vlst (mapcar '(lambda (x) (cons a (reverse (cons a x)))) vlst)
    
24.       )
    
25.     (setq vlst (cons flst (reverse vlst)))
    
26.     )
    
27.   (setq        i    (1+ (length lst))
    
28.         lsta nil
    
29.         )
    
30.   (repeat (1- i)
    
31.     (setq lsta (cons (f1 (setq i (1- i)) m n) lsta))
    
32.     )
    
33.   (setq
    
34.     lst
    
35.      (mapcar '(lambda (x) (mapcar '(lambda (y) (nth (1- y) lst)) x))
    
36.              (apply 'append lsta)
    
37.              )
    
38.     )
    
39.   )
    
40. (f2 '("a" "b" "c" "d" "e") 4 2)(("a" "a") ("a" "a" "a") ("a" "a" "a" "a") ("b" "b") ("b" "a" "b") ("b" "a" "a" "b") ("b" "b" "b") ("b" "b" "b" "b") ("c" "c") ("c" "a" "c") ("c" "a" "a" "c") ("c" "b" "c") ("c" "b" "b" "c") ("c" "c" "c") ("c" "c" "c" "c") ("d" "d") ("d" "b" "d") ("d" "b" "b" "d") ("d" "c" "d") ("d" "c" "c" "d") ("d" "d" "d") ("d" "d" "d" "d") ("e" "e") ("e" "c" "e") ("e" "c" "c" "e") ("e" "d" "e") ("e" "d" "d" "e") ("e" "e" "e") ("e" "e" "e" "e"))
    

(defun f2 (lst m n / i lsta)<br />   (defun f1 (a m n / fflst flst i j vlst)<br />     (setq flst (list a a)<br />           i    0<br />           j    (max 1 (- a n))<br />           vlst nil<br />           )<br />     (while (<= j a)<br />       (while (< i (- m 2))<br />         (setq fflst (list j))<br />         (repeat        i<br />           (setq fflst (cons j fflst))<br />           )<br />         (setq vlst (cons fflst vlst))<br />         (setq i (+ i 1))<br />         )<br />       (setq j (+ j 1)<br />             i 0<br />             )<br />       )<br />     (setq<br />       vlst (mapcar '(lambda (x) (cons a (reverse (cons a x)))) vlst)<br />       )<br />     (setq vlst (cons flst (reverse vlst)))<br />     )<br />   (setq        i    (1+ (length lst))<br />         lsta nil<br />         )<br />   (repeat (1- i)<br />     (setq lsta (cons (f1 (setq i (1- i)) m n) lsta))<br />     )<br />   (setq<br />     lst<br />      (mapcar '(lambda (x) (mapcar '(lambda (y) (nth (1- y) lst)) x))<br />              (apply 'append lsta)<br />              )<br />     )<br />   )<br /> (f2 '("a" "b" "c" "d" "e") 4 2)(("a" "a") ("a" "a" "a") ("a" "a" "a" "a") ("b" "b") ("b" "a" "b") ("b" "a" "a" "b") ("b" "b" "b") ("b" "b" "b" "b") ("c" "c") ("c" "a" "c") ("c" "a" "a" "c") ("c" "b" "c") ("c" "b" "b" "c") ("c" "c" "c") ("c" "c" "c" "c") ("d" "d") ("d" "b" "d") ("d" "b" "b" "d") ("d" "c" "d") ("d" "c" "c" "d") ("d" "d" "d") ("d" "d" "d" "d") ("e" "e") ("e" "c" "e") ("e" "c" "c" "e") ("e" "d" "e") ("e" "d" "d" "e") ("e" "e" "e") ("e" "e" "e" "e")) <br />

mahuan1279

夏生生 发表于 2022-2-27 15:54
谢谢您，完美解决

_$ (f2 '("a" "b" "c" "d" ) 5 2)
(("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" "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" "b" "c") ("c" "b" "b" "c") ("c" "b" "b" "b" "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" "c" "d") ("d" "c" "c" "d") ("d" "c" "c" "c" "d") ("d" "d" "d") ("d" "d" "d" "d") ("d" "d" "d" "d" "d"))
_$
有遗漏，如("c" "c" "a" "c" "c")("c" "c" "b" "c" "c")等等。

夏生生

mahuan1279 发表于 2022-2-27 16:08
_$ (f2 '("a" "b" "c" "d" ) 5 2)
(("a" "a") ("a" "a" "a") ("a" "a" "a" "a") ("a" "a" "a" "a" "a")  ...

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