TA的每日心情 | 擦汗 2016-4-17 12:18 |
|---|
签到天数: 1 天 [LV.1]炼气
|
本帖最后由 tingsanguo 于 2014-5-2 14:01 编辑
- W) T* b9 r1 y! A/ H# t d+ J水风 发表于 2014-5-2 01:57 ![]()
* l) q# C: O: A! x再次重温自己与数学天才们之间的差距有多大了。解出来了,我还看不懂思路 ...
6 B7 R+ `6 f {- W8 {9 _ H3 J2 ^* l" v T9 d$ j3 ^8 `4 B7 k
fixed point f(X) = X = (1-n), where n is the number of people who divide the coconut.6 G5 G: ^/ O( X# v+ N5 [, g
9 T! X- B. A$ L/ f, I0 D. s( U
The least number of coconut before dividing is Z = [n^(n+1) + X], i.e., always divide (n+1) times.0 a4 ^6 w( Y6 p( c9 ?
6 G0 q/ o- C" Y$ w* @# F1 zAfter being divided (n+1) times, Z becomes 1, since X is fixed point.
6 Y) J- H/ Z% U& z5 l/ _- z7 Q
- w, M2 ~4 Z8 t, b, n4 k$ s2 nn n^(n+1)+X, n^(n)+X, n^(n-1)+X, n^(n-2)+X ( S8 i1 p) O P |! V. D2 v
1 1- i4 \2 o2 n9 c
2 7 3 1 ! v7 z( [, N$ J0 l0 b# r5 P$ a: n
3 79 25 7 1
2 }4 }5 m! b J' p7 @$ P B+ h4 1021 253 61 13 1
, B. _3 V* Q" |/ n! o5 15621 3121 621 121 21 1
8 I) s( q# C- E; R& t, u: F: t0 b' L# H1 s1 I# M
; y* \( O' s! B2 [5 B t7 P0 S9 L |
|