Problem 6

本帖于 2009-07-28 16:38:03 时间, 由普通用户 康MM 编辑

通常第6题会比较难,今年的不一样?试着解,不对的话请指正。

There are n! different ways to arrange the jumps.
Given a point p between 0 and s = a_1+a_2+...a_n, it
has at most (n-1)! ways to reach. So the total ways to be matched is (n-1)(n-1)! which is less than n!. Also it
means there is at least one way that the grasshopper never lands on any point in M.











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通常第六题不是最难的 -累死算了- 给 累死算了 发送悄悄话 (148 bytes) () 07/26/2009 postreply 01:05:48

通常第六题是最难的 -botong- 给 botong 发送悄悄话 botong 的博客首页 (408 bytes) () 07/26/2009 postreply 09:25:21

有点问题吧 -dynamic- 给 dynamic 发送悄悄话 (226 bytes) () 07/26/2009 postreply 22:14:34

回复:有点问题吧 -botong- 给 botong 发送悄悄话 botong 的博客首页 (209 bytes) () 07/27/2009 postreply 11:01:10

回复:回复:有点问题吧 -dynamic- 给 dynamic 发送悄悄话 (79 bytes) () 07/27/2009 postreply 18:12:36

回复:回复:回复:有点问题吧 -botong- 给 botong 发送悄悄话 botong 的博客首页 (91 bytes) () 07/28/2009 postreply 10:22:54

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