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(defvar *prog-input*)
(let ((ui (read-line)))
(if (equal ui "")
(setq *prog-input* *standard-input*)
(setq *prog-input* (open ui))))
(let
((cal (read-line *prog-input* NIL)) (total-score 0)
(your-rps '(#\0 #\X #\Y #\Z)) (enemy-rps '(#\0 #\A #\B #\C))
(cy-rps 0) (ce-rps 0))
(loop until (or (equal cal "end") (not cal)) do
;; Rock is 1, paper is 2, scissors is 3
(setq ce-rps (position (char cal 0) enemy-rps))
(setq cy-rps (position (char cal 2) your-rps))
;; We want, when we lose to add 3 * 0, when we draw to add 3 * 1 and when we win to add 3 * 2
;; To get numbers 0, 1, and 2, we'll use `3 mod N`
;; Writing down all possible combinations
;; (from left to right, columns are "Your choice", "Enemy choice", "Result"):
;; 1 1 | 1 = 3 % 1 (Draw)
;; 1 2 | 0 = 3 % 3 (Lose)
;; 1 3 | 2 = 3 % 2 (Win)
;; ----+-----------------
;; 2 1 | 2 = 3 % 2 (Win)
;; 2 2 | 1 = 3 % 1 (Draw)
;; 2 3 | 0 = 3 % 3 (Lose)
;; ----+-----------------
;; 3 1 | 0 = 3 % 3 (Lose)
;; 3 2 | 2 = 3 % 2 (Win)
;; 3 3 | 1 = 3 % 1 (Draw)
;; We notice, that N (in "3 % N") is a rotation of the numbers 3, 2, 1, where if your choice
;; is 3, we don't rotate them, if it's 2 we rotate by one (backwards) and so on.
;; We can get 3, 2, 1 from the enemie's 1, 2, 3 by subtracting them from 4. Then we can use
;; your choice to do the "rotation" (since we're doing mod, 3 % 3 = 3 % 6).
(setq total-score (+ total-score (* 3 (mod (+ (- 4 ce-rps) cy-rps) 3)) cy-rps))
(setq cal (read-line *prog-input* NIL)))
(print total-score))
(if (not (eq *prog-input* *standard-input*))
(close *prog-input*))
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