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(ns repeating-decimal) | |
(defn repeating-decimal [n] | |
(loop [rem 1 rems #{} divs [] i 0] | |
(let [div (int (/ (* rem 10) n)) | |
rem (mod (* rem 10) n)] | |
(if (and (not= rem 0) (contains? rems rem)) | |
[n divs] | |
(when (not= rem 0) | |
(recur rem (conj rems rem) (conj divs div) (inc i))))))) | |
(comment | |
(repeating-decimal 7) | |
(repeating-decimal 11) | |
(repeating-decimal 12) | |
(repeating-decimal 97) | |
) | |
(defn solve [n] | |
(loop [i n, m 0, divs []] | |
(let [l (count divs)] | |
(if (> i l) | |
(let [divs' (repeating-decimal i) | |
l' (count divs')] | |
(if (> l' l) | |
(recur (dec i) i divs') | |
(recur (dec i) m divs))) | |
[m divs])))) | |
;; (solve 100000) |
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1/n
の循環節の長さはn
を超えない0
〜n-1
のn
通り10000
以下の数n
で1/n
が10000
以上の長さの循環節を持つことはない100000
,99999
,99998
, ...,その時点の循環節の最大長
の範囲の数だけを探索すればいい