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为便于理解,给组反例,详见内.

Posted by ferret_chao at 2005-10-12 12:28:54 on Problem 1013
In Reply To:我的方案;每个硬币一个权重遇到平衡时，每一个硬币一定为真币（让其权重为“无穷小”）。每遇到一次不平衡时，每一个币都受怀疑，权重加一（见正文） Posted by:peterpan at 2005-10-12 02:26:46

1
ABCD EFGK up
EFDL ABGK up
GHIJ EFDK up

正确答案显然是:
K is the counterfeit coin and it is light.
而你的程序是:
D is the counterfeit coin and it is light.
(第1个碰到的权重最大的硬币.)

Followed by:

Re:为便于理解,给组反例,详见内. (54) peterpan 2005-10-12 12:36:59
Re:为便于理解,给组反例,详见内. (2037) peterpan 2005-10-12 14:00:43
Re:为便于理解,给组反例,详见内. (51) chgsh8089 2006-03-13 19:44:51
Re:为便于理解,给组反例,详见内. (22) infanting 2006-04-27 12:52:45
Re:为便于理解,给组反例,详见内. (3) waitandsee 2006-09-28 08:47:36
Re:为便于理解,给组反例,详见内. (138) answer_allon 2008-02-27 09:35:07
- 这个方案可以吧！ (179) answer_allon 2008-02-27 09:45:10

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