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hsjoihs / famicom_long_sequence_32767
Created September 4, 2020 09:29
Famicom / NES Long sequence pseudorandom bits
0000000000000010000000000000110000000000001010000000000011110000000000100010000000001100110000000010101010000000111111110000001000000010000011000000110000101000001010001111000011110010001000100010110011001100111010101010101001111111111111010000000000001110000000000010010000000000110110000000001011010000000011101110000000100110010000001101010110000010111111010000111000001110001001000010010011011000110110101101001011011110111011101100011001100110100101010101011101111111111100110000000000101010000000001111110000000010000010000000110000110000001010001010000011110011110000100010100010001100111100110010101000101010111111001111111000001010000001000011110000011000100010000101001100110001111010101010010001111111110110010000000011010110000000101111010000001110001110000010010010010000110110110110001011011011010011101101101110100110110110011101011011010100111101101111101000110110000111001011010001001011101110011011100110010101100101010111110101111111000011110000001000100010000011001100110000101010101010001111111111110010
\frac{d}{dt} (x, p) = \left.\left(\frac{\partial H(a,b)}{\partial b}, -\frac{\partial H(a,b)}{\partial a} \right)\right|_{a=x, b=p}
\int_b^c a(x-b)(x-c) dx = -\frac{a}{6}(c-b)^3
\frac{4}{\pi}\int _0^{\pi/2} \sin((2n-1)x) (\sin x)^{b} dx
\frac{2}{ab}\frac{d^3}{d\theta^3} S_{true} = -2ab\left(1+m^2\right) \left( \frac{ 2\sin \theta\cos\theta}{L}-\frac{ 2abm \sin^3 \theta}{L^2} \right) + \frac{dm}{d\theta} \left(-\frac{4abm \sin^2 \theta}{L}+\cos\theta\right)-m\sin\theta -\cos \theta
faire aefir
comme cemmo
était aitté
votre eortv
pourquoi ioopqruu
cette ceett
quand adnqu
alors alors
comment cemmnot
merci ceimr
@hsjoihs
hsjoihs / !POST to wandbox
Last active June 17, 2019 01:49
POST to wandbox
POST to wandbox
@hsjoihs
hsjoihs / dijkstra_like.js
Created March 26, 2018 12:08
「友達がなんかオリジナルのDijkstra法を改良できたっぽいのでコード見てもらえますか?」というDMが来た、嘘解法っぽさあるけど反例分からん
/* DMで相談がきたソースコード */
/*
id:{image:'path',cost:INF,done:[],route:"",positionX:"latitude",positionY:"elevation",positionZ:"longitude",links:["link_1","link_0"]}
costに頂点に到達するまでのコスト、doneで最短距離を調べ終わったやつ、
routeに最短距離でどの頂点から来たか、positionXに緯度、
positionYに高度、positionZに経度、
linksに配列でその頂点につながってる頂点
@require: stdjabook
@require: hdecoset
@require: vdecoset
@require: code
@require: list
let-block ctx +centering it =
line-break true true ctx (inline-fil ++ read-inline ctx it ++ inline-fil)