list - Lisp函数接受两个列表并返回它们之间的乘积

这是家庭作业，我不知道从哪里开始。
例如：
列表一是（1 2 3），列表二是（3 4 6）
返回29，因为（1*3）+（2*4）+（3*6）

最佳答案

不破坏答案，但这里有一点你可以期待的：

(defun sum-lists% (x y z)  ...)
(defun sum-lists (x y) ... )

跟踪两个函数：

(trace sum-lists sum-lists%)

你的例子：

(sum-lists '(1 2 3) '(3 4 6))

  0: (SUM-LISTS (1 2 3) (3 4 6))
    1: (SUM-LISTS% (1 2 3) (3 4 6) 0)
      2: (SUM-LISTS% (2 3) (4 6) 3)
        3: (SUM-LISTS% (3) (6) 11)
          4: (SUM-LISTS% NIL NIL 29)
          4: SUM-LISTS% returned 29
        3: SUM-LISTS% returned 29
      2: SUM-LISTS% returned 29
    1: SUM-LISTS% returned 29
  0: SUM-LISTS returned 29

如果你处理角落的箱子，你也可以有：

 (sum-lists '(1 2 3 4) '(3 4))

  0: (SUM-LISTS (1 2 3 4) (3 4))
    1: (SUM-LISTS% (1 2 3 4) (3 4) 0)
      2: (SUM-LISTS% (2 3 4) (4) 3)
        3: (SUM-LISTS% (3 4) NIL 11)
          4: (SUM-LISTS% (4) NIL 14)
            5: (SUM-LISTS% NIL NIL 18)
            5: SUM-LISTS% returned 18
          4: SUM-LISTS% returned 18
        3: SUM-LISTS% returned 18
      2: SUM-LISTS% returned 18
    1: SUM-LISTS% returned 18
  0: SUM-LISTS returned 18

关于list - Lisp函数接受两个列表并返回它们之间的乘积，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/53772975/