我正在经历Lisp Koans，这很有趣！但我坚持得分项目（我有一个糟糕的解决方案）。在这个项目中，我们被要求实现一个名为*Greed*的简单游戏。问题描述如下：
; *Greed* is a dice game where you roll up to five dice to accumulate
; points. The following "score" function will be used to calculate the
; score of a single roll of the dice.
;
; A greed roll is scored as follows:
; * A set of three ones is 1000 points
; * A set of three numbers (other than ones) is worth 100 times the
; number. (e.g. three fives is 500 points).
; * A one (that is not part of a set of three) is worth 100 points.
; * A five (that is not part of a set of three) is worth 50 points.
; * Everything else is worth 0 points.
;
; Examples:
;
; (score '(1 1 1 5 1)) => 1150 points
; (score '(2 3 4 6 2)) => 0 points
; (score '(3 4 5 3 3)) => 350 points
; (score '(1 5 1 2 4)) => 250 points
;
; More scoring examples are given in the tests below:
;
; Your goal is to write the score method.
我的解决方案如下：
我使用occurs函数来计算数字的出现次数并在assoc-list中表示。还有formula-wrapper函数为formula函数提供正确的参数。用于计算分数的formula函数。我的解决方案非常难看！欢迎任何建议！提前谢谢。
(defun occurs (lst)
(let ((acc nil))
(dolist (obj lst)
(let ((p (assoc obj acc)))
(if p
(incf (cdr p))
(push (cons obj 1) acc))))
(sort acc #'> :key #'cdr)))
(defun formula-wrapper (lst)
(formula (car lst) (cdr lst)))
(defun formula (number times)
(cond ((= times 0) 0)
((= times 1)
(case number
(1 100)
(5 50)
(otherwise 0)))
((= times 2)
(case number
(1 200)
(5 100)
(otherwise 0)))
((= times 3)
(case number
(1 1000)
(otherwise (* 100 number))))
((= times 4)
(case number
(1 1100)
(5 550)
(otherwise 0)))
((= times 5)
(case number
(1 1200)
(5 600)
(otherwise 0)))
(times 0)))
(defun score (dice)
(let ((rolls (occurs dice)))
(if (null rolls)
0
(apply #'+ (mapcar #'formula-wrapper rolls))))))
(define-test test-score-of-an-empty-list-is-zero
(assert-equal 0 (score nil)))
(define-test test-score-of-a-single-roll-of-5-is-50
(assert-equal 50 (score '(5))))
(define-test test-score-of-a-single-roll-of-1-is-100
(assert-equal 100 (score '(1))))
(define-test test-score-of-multiple-1s-and-5s-is-the-sum-of-individual-scores
(assert-equal 300 (score '(1 5 5 1))))
(define-test test-score-of-single-2s-3s-4s-and-6s-are-zero
(assert-equal 0 (score '(2 3 4 6))))
(define-test test-score-of-a-triple-1-is-1000
(assert-equal 1000 (score '(1 1 1))))
(define-test test-score-of-other-triples-is-100x
(assert-equal 200 (score '(2 2 2)))
(assert-equal 300 (score '(3 3 3)))
(assert-equal 400 (score '(4 4 4)))
(assert-equal 500 (score '(5 5 5)))
(assert-equal 600 (score '(6 6 6))))
(define-test test-score-of-mixed-is-sum
(assert-equal 250 (score '(2 5 2 2 3)))
(assert-equal 550 (score '(5 5 5 5))))
答案 0 :(得分：2)
(defun score (dice)
(let ((freq (make-hash-table)))
(loop for x in dice do (incf (gethash x freq 0)))
(loop for x being the hash-key of freq using (hash-value c)
sum (if (<= 3 c)
(case x
(1 (+ 1000 (* 100 (- c 3))))
(5 (+ 500 (* 50 (- c 3))))
(t (* x 100)))
(case x
(1 (* c 100))
(5 (* c 50))
(t 0))))))
答案 1 :(得分：1)
写一种方法：
(defun find-set (roll)
"which number from 1 to 6 occurs at least three times in a list of five?"
(assert (= (length roll) 5))
(loop for i from 1 to 6
when (>= (count i roll) 3)
do (return i)))
(defun score-set (i)
"compute the set score for number i"
(case i
(1 1000)
(otherwise (* i 100))))
(defun score (roll &aux (s (find-set roll)) (score 0))
(when s
(setf score (score-set s)
roll (remove s roll :count 3)))
(incf score (* (count 1 roll) 100))
(incf score (* (count 5 roll) 50))
score)
(defun test ()
(assert (= (score '(1 1 1 5 1)) 1150))
(assert (= (score '(2 3 4 6 2)) 0))
(assert (= (score '(3 4 5 3 3)) 350))
(assert (= (score '(1 5 1 2 4)) 250))
t)
答案 2 :(得分：0)
尾递归版：
(defun score (dice)
(labels ((iter (left ans)
(if (not left) ans
(cond ((and (>= (length left) 3)
(= (car left) (cadr left) (caddr left)))
(cond ((= (car left) 1)
(iter (cdddr left) (+ ans 1000)))
((= (car left) 5)
(iter (cdddr left) (+ ans 500)))
(t (iter (cdddr left) (+ ans (* (car left) 100))))))
((= (car left) 1) (iter (cdr left) (+ ans 100)))
((= (car left) 5) (iter (cdr left) (+ ans 50)))
(t (iter (cdr left) ans))))))
(iter (sort dice #'<) 0)))