Re: [轉錄][閒聊] 兩個8 兩個3

看板ACMCLUB作者時間18年前 (2005/12/15 02:00), 編輯推噓0(0推 0噓 13→)
留言13則, 4人參與, 最新討論串1/4 (看更多)
手邊類似的老梗還有幾題，不過我猜也是很多人看過了罷。 :-) (據說這是 google 的，尚未考證...) 1 11 21 1211 111221 312211
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12/14 21:50, , 1F
這題我寫過程是讓他看字串增長速度有多快XD到了第4x項
12/14 21:50, 1F
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12/14 21:51, , 2F
系上工作站就爆掉了
12/14 21:51, 2F
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12/14 21:55, , 3F
[bonus] plz prove the number in each digit can't be
12/14 21:55, 3F
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12/14 21:58, , 4F
larger than 3.
12/14 21:58, 4F
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12/14 22:08, , 5F
John Conway's Look and Say Sequence
12/14 22:08, 5F
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12/14 22:23, , 6F
小時候在 "捍衛網路" 裡看到的 :P
12/14 22:23, 6F
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12/14 22:30, , 7F
樓上們您真內行 ! :-)
12/14 22:30, 7F
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12/14 23:36, , 8F
if a digit is larger than 4 then the previous num
12/14 23:36, 8F
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12/14 23:37, , 9F
must have at least xxxx in a row, which means the
12/14 23:37, 9F
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12/14 23:38, , 10F
previous-two num have 2*x's x in a row, resulting
12/14 23:38, 10F
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12/14 23:38, , 11F
in contradiction (這一列有41->前一列有1111->再前一
12/14 23:38, 11F
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12/14 23:39, , 12F
列是11->則應該是21不是1111) 這樣講可以嗎@@?
12/14 23:39, 12F
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12/15 00:11, , 13F
Good job ! 不過底下 F 網友那篇更接近系統化表述。:-)
12/15 00:11, 13F
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文章代碼(AID): #13e5t200 (ACMCLUB)
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文章代碼(AID): #13e5t200 (ACMCLUB)