[Redis] Redis Event Library

Document: http://redis.io/topics/internals-rediseventlib

Source: https://github.com/antirez/redis/

[ProjectEuler] Problem 473

大原則：碰到 Fibonacci 的問題都不會太困難，一定會有什麼遞迴。

如果問的數字很大，可以用 Q-matrix 把 complexity time O(N) 降為 O(logN)



不過還是有幾題卡住就是了，喵的。

[Redis] Hacking Strings

Link: http://redis.io/topics/internals-sds

-----------
|5|0|redis|
-----------
^   ^
sh  sh->buf

關鍵就是這個。

[GIT] Changing git remote origin

git remote rm origin 

git remote add origin https://username@bitbucket.org/your.new.repo 

[ProjectEuler] Problem #378

http://stackoverflow.com/questions/16402854/number-of-increasing-subsequences-of-length-k

[ProjectEuler] Problem #242

沒有什麼特別想法，先暴力算算看 f(n, k) 好了。

[ProjectEuler] Notes on Problem #435

Lemma: F_n(x) = (f_{n+1} x^{n+1} + f_{n} x^{n+2} - x) / (x^2 + x - 1)

F_n(x) 

= sum_{0 <= i <= n} f_i x^i 

= sum_{1 <= i <= n} f_i x^i ...... (A)

x F_n(x) 

= sum_{0 <= i <= n} f_{i} x^{i+1} 

= sum_{1 <= i+1 <= n+1} f_{i+1 - 1} x^{i+1}

= sum_{1 <= j <= n+1} f_{j-1} x^{j}

= sum_{1 <= i <= n} f_{i-1} x^{i} + f_n x^{n+1} ...... (B)

                     

(A) + (B) implies that (x+1) F_n(x) = sum_{1 <= i <= n} f_{i+1} x^i + f_n x^{n+1}

Multiply x on both sides:

x(x+1) F_n(x) = sum_{1 <= i <= n} f_{i+1} x^{i+1} + f_n x^{n+2}

= sum_{2 <= i+1 <= n+1} f_{i+1} x^{i+1} + f_n x^{n+2}

= sum_{2 <= i <= n+1} f_i x^i + f_n x^{n+2}

= sum_{1 <= i <= n} f_i x^i - f_1 x + f_{n+1} x^{n+1} + f_n x^{n+2}

= F_n(x) + f_{n+1} x^{n+1} + f_n x^{n+2} - x

(x^2 + x - 1) F_n(x) = f_{n+1} x^{n+1} + f_n x^{n+2} - x

So, F_n(x) = (f_{n+1} x^{n+1} + f_n x^{n+2} - x) / (x^2 + x - 1)

不過這離解題還有一大段的距離。