I was so awkward the first time to write such a thing
First %%% Koala seniors!
Part is written in front of PowerPoint -
Probably briefly learning outcomes -
1> Cattleya number
Catalan $ $ ($ H_i used herein to indicate the number $ $ $ I-bit first)
The first is a recursive formula:
$h_n=h_0 \times h_{n-1}+h_1 \times h_{n-2} + \cdots + h_{n-1} \times h_0(n \geq 2)$
Therefore, to satisfy this form can be processed by a number slightly Cattleya
Followed by a general term formula:
$\begin{array}{lcl}h_n & = & C_{2n}^n-C_{2n}^{n-1}\\ & = & \frac{C_{2n}^n}{n+1}\end{array}$
application:
1. seeking a number of binary tree form of the program
2. Drawing sequence number obtaining program
3. Program for Finding grid walking limit
And these can be converted into a problem: the $ 1 and $ $ $ -1 arranged in array manipulation is greater than $ 0 and $ prefix
1. Can the son left the right note for the $ 1 $ to $ -1 $ credited his son, then-first search left.
2. The stack may be referred to as a stack of $ $ $ referred to as $ -1
Come to prove
First set $ k $ is the last element of the stack,
Then pop up in front of the elements must be (otherwise $ k $ is not the bottom up) before pressing into the $ k $
Subsequent elements must also be popped pop-up, pop-up to the $ k $
So there is the formula:
$f_i=\sum\limits_{k=1}^{i}f_{k-1} \times f_{i-k}$
Well, and I feel like a formula ~
Yes, that recurrence formula! ! !
3. Such is the naked: "link"
Only up to the right, you can not cross the Blue Line
Well, how count program?
But it is still $ Catalan $ number.
One is referred to the right as you $ 1 + $, $ -1 $ referred to as upward so that the request can be according to the formula
A total of $ 2n $ plus and minus, it is $ h_n $
Plus folding
The first is to draw a secondary line, this line is illegal path must be off
Is this.
Then folded portion of the rear contact path red illicit
(Incidentally, the grid also turn over)
Like this, then it was the end turn to the upper right corner of the yellow rectangle
So the answer is,
$ C_ {2n} ^ n-C_ {2n} ^ {n-1} $
Namely: to end the illegal scheme reduced to the end of the program
Go to the end $ $ 2N step, which accounted for $ $ n-step to the right, is $ C_ {2n} ^ n $
Illegal focus away $ $ 2N step, which accounted for $ n-1 $ step to the right, is $ C_ {2n} ^ {n-1} $
And then there's a version upgrade!
like this
The same problem
Folding:
So easy to say
To the end: $ n + m $ walking step, representing $ n $ is the right $ C_ {n + m} ^ n $
Illegal endpoint: walking step $ n + m $, $ m-1 $ account is $ C_ {n + m} ^ {m-1} $ rightward
The answer is:
$ C_ {n + m} ^ n-C_ {n + m} ^ {m-1} (n \ geq m) $
Up here with the same, but the answer into:
$ C_ {n + m} ^ {n + m-C_ m} ^ {n + 1} (n \ geq m) $
This formula is not $ C_n ^ m = C_n ^ {nm} $ it (laughs)
2> prufer sequence
Etc. - the amount of content will be too large, unfinished