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refactor: upload
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@Tiphereth-A
Tiphereth-A committed May 23, 2024
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14 changes: 14 additions & 0 deletions ICPCnotebook.cls
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Expand Up @@ -79,6 +79,20 @@
\newcommand{\dstirlingII}[2]{\genstirlingII{0}{#1}{#2}}
\newcommand{\tstirlingII}[2]{\genstirlingII{1}{#1}{#2}}

\newcommand{\sech}{\operatorname{sech}}
\newcommand{\csch}{\operatorname{csch}}
\newcommand{\arccot}{\operatorname{arccot}}
\newcommand{\arcsec}{\operatorname{arcsec}}
\newcommand{\arccsc}{\operatorname{arccsc}}
\newcommand{\arcsinh}{\operatorname{arcsinh}}
\newcommand{\arccosh}{\operatorname{arccosh}}
\newcommand{\arctanh}{\operatorname{arctanh}}
\newcommand{\arccoth}{\operatorname{arccoth}}
\newcommand{\arcsech}{\operatorname{arcsech}}
\newcommand{\arccsch}{\operatorname{arccsch}}

\newcommand{\md}{\mathrm{d}}

\newcommand{\fullref}[1]{\ref{#1}.\nameref{#1} (\pageref{#1} 页)}


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5 changes: 1 addition & 4 deletions config.yml
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Expand Up @@ -1343,9 +1343,6 @@ notebook:
- sudoku: 数独求解
code_ext: hpp
test_ext: cpp
- texas_holdem: 德州扑克
code_ext: hpp
test_ext: cpp
- npuzzle_data: N puzzle 数据类
code_ext: hpp
test_ext: cpp
Expand Down Expand Up @@ -1376,7 +1373,7 @@ cheatsheets:
decimal: Python Decimal
bash: Bash 简单用法
vim: Vim 简单用法
lgv: lgv 定理
integral: 积分表
default_code_style: common
code_styles:
C: cpp
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230 changes: 224 additions & 6 deletions src/cheatsheet/formula.tex
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Expand Up @@ -25,6 +25,44 @@ \subsubsection{积和式}

\(\operatorname{haf}(A)\) 的定义参见 \fullref{sec:图的完美匹配计数}

\subsubsection{邻接矩阵行列式的意义}

在无向图中取若干个环, 一种取法权值就是边权的乘积, 对行列式的贡献是 \((-1)^{e}\), 其中 \(e\) 是偶环的个数


\subsubsection{LGV 引理}

\paragraph{定义}

\(\omega(P)\) 表示 \(P\) 这条路径上所有边的边权之积. (路径计数时, 可以将边权都设为 \(1\)) (事实上, 边权可以为生成函数)

\(e(u, v)\) 表示 \(u\) 到 \(v\) 的 \textbf{每一条} 路径 \(P\) 的 \(\omega(P)\) 之和, 即 \(e(u, v)=\sum\limits_{P:u\rightarrow v}\omega(P)\).

起点集合 \(A\), 是有向无环图点集的一个子集, 大小为 \(n\).

终点集合 \(B\), 也是有向无环图点集的一个子集, 大小也为 \(n\).

一组 \(A\rightarrow B\) 的不相交路径 \(S\): \(S_i\) 是一条从 \(A_i\) 到 \(B_{\sigma(S)_i}\) 的路径(\(\sigma(S)\) 是一个排列), 对于任何 \(i\ne j\), \(S_i\) 和 \(S_j\) 没有公共顶点.

\(t(\sigma)\) 表示排列 \(\sigma\) 的逆序对个数.

\paragraph{引理}

\[
M = \begin{bmatrix}
e(A_1,B_1) & e(A_1,B_2) & \cdots & e(A_1,B_n) \\
e(A_2,B_1) & e(A_2,B_2) & \cdots & e(A_2,B_n) \\
\vdots & \vdots & \ddots & \vdots \\
e(A_n,B_1) & e(A_n,B_2) & \cdots & e(A_n,B_n)
\end{bmatrix}
\]

\[
\det(M)=\sum\limits_{S:A\rightarrow B}(-1)^{t(\sigma(S))}\prod\limits_{i=1}^n \omega(S_i)
\]

其中 \(\sum\limits_{S:A\rightarrow B}\) 表示满足上文要求的 \(A\rightarrow B\) 的每一组不相交路径 \(S\).

\subsection{组合数学}

\subsubsection{组合数相关公式}
Expand Down Expand Up @@ -129,12 +167,34 @@ \subsubsection{Burnside 引理}
如: 长为 \(n\) 的项链, 珠子可以染 \(m\) 种颜色, 则总方案数为

\[
\frac{1}{n}\sum_{k=0}^{n-1} m^{\gcd(n,k)}
\frac{1}{n}\sum_{k=0}^{n-1} m^{(n,k)}
\]

\subsubsection{自然数幂次和关于次数的EGF}

\begin{equation}
\begin{aligned}
F(x)= & \sum_{k=0}^\infty \frac{\sum_{i=0}^n i^k}{k!}x^k \\
= & \sum_{i=0}^n \mathrm{e}^{ix} \\
= & \frac{\mathrm{e}^{(n+1)x-1}}{\mathrm{e}^x-1}
\end{aligned}
\end{equation}


\subsection{初等数论}

\subsubsection{Dirichlet 卷积相关公式}
\subsubsection{数论函数相关}

\begin{multicols}{3}
\(\displaystyle \sum_{\delta\mid n}\varphi(\delta)d\left(\frac{n}{\delta}\right) = \sigma(n)\) \\
\(\displaystyle \sum_{\delta\mid n}\left|\mu(\delta)\right| = 2^{\omega(n)}\) \\
\(\displaystyle \sum_{\delta\mid n}2^{\omega(\delta)} = d(n^2)\) \\
\(\displaystyle \sum_{\delta\mid n}d(\delta^2) = d^2(n)\) \\
\(\displaystyle \sum_{\delta\mid n}d\left(\frac{n}{\delta}\right)2^{\omega(\delta)} = d^2(n)\) \\
\(\displaystyle \sum_{\delta\mid n}\frac{\mu(\delta)}{\delta} = \frac{\varphi(n)}{n}\) \\
\(\displaystyle \sum_{\delta\mid n}\frac{\mu(\delta)}{\varphi(\delta)} = d(n)\) \\
\(\displaystyle \sum_{\delta\mid n}\frac{\mu^2(\delta)}{\varphi(\delta)} = \frac{n}{\varphi(n)}\)
\end{multicols}

\begin{equation}
\sum_{i=1}^n\sum_{j=1}^m(i,j)^k=\sum_{D=1}^{\min\{n,m\}}\left\lfloor\frac{n}{D}\right\rfloor\left\lfloor\frac{m}{D}\right\rfloor\{\operatorname{id}_k*\mu\}(D)
Expand All @@ -144,6 +204,96 @@ \subsubsection{Dirichlet 卷积相关公式}
\{\operatorname{id}_k*\mu\}(p^s)=p^{ks}-p^{k(s-1)}
\end{equation}

\begin{equation}
\sum_{i = 1} ^ n \left[(i, n) = 1\right] i = n \frac {\varphi(n) + e(n)} 2
\end{equation}

\begin{equation}
\sum_{i = 1} ^ n \sum_{j = 1} ^ i \left[(i, j) = d\right] = S_\varphi \left( \left\lfloor \frac n d \right\rfloor \right)
\end{equation}

\begin{equation}
\sum_{i = 1} ^ n \sum_{j = 1} ^ m \left[(i, j) = d\right] = \sum_{d \mid k} \mu\left( \frac k d \right) \left\lfloor \frac n k \right\rfloor \left\lfloor \frac m k \right\rfloor
\end{equation}

\begin{equation}
\sum_{i = 1} ^ n f(i) \sum_{j = 1} ^ {\left\lfloor \frac n i \right\rfloor} g(j) = \sum_{i = 1} ^ n g(i) \sum_{j = 1} ^ {\left\lfloor \frac n i \right\rfloor} f(j)
\end{equation}

\begin{equation}
\mu^2(n) = \sum_{d^2 \mid n} \mu (d)
\end{equation}

\begin{equation}
\prod_{k=1,(k, m) = 1}^{m} k \equiv
\begin{cases}
-1 \pmod m, & m = 4, p^q, 2p^q \\
1 \pmod m, & \text{otherwise} \\
\end{cases}
\end{equation}

\begin{equation}
\sigma_k(n) = \sum_{d\mid n}d^k = \prod_{i=1}^{\omega(n)}\frac{p_i^{(a_i+1)k}-1}{p_i^k-1}
\end{equation}

\begin{eqnarray}
&&J_k(n) = n^k\prod_{p\mid n}\left(1-\frac{1}{p^k}\right) \\
&&\sum_{\delta\mid n}J_k(\delta) = n^k \\
&&\sum_{\delta\mid n}\delta^sJ_r(\delta)J_s\left(\frac{n}{\delta}\right) = J_{r+s}(n)
\end{eqnarray}

其中 \(J_k(n)\) 是不大于 \(n\) 的正整数构成的 \(k\) 元组中, 与 \(n\) 构成互素 \((k + 1)\) 元组的个数

\begin{equation}
\sum_{i=1}^{n} \sum_{j=1}^{n} [(i,j)=1]ij = \sum_{i=1}^{n} i^2\varphi(i)
\end{equation}

\begin{equation}
n\mid \varphi(a^n-1)
\end{equation}

\begin{equation}
\sum_{\substack{1 \leq k \leq n,~(k, n) = 1}}f(\gcd(k-1, n)) = \varphi(n)\sum_{d\mid n}\frac{(\mu*f)(d)}{\varphi(d)}
\end{equation}

\begin{equation}
\varphi([m, n])\varphi(\gcd(m,n)) = \varphi(m)\varphi(n)
\end{equation}

\begin{equation}
\sum_{\delta\mid n}d^3(\delta) = \left(\sum_{\delta\mid n}d(\delta)\right)^2
\end{equation}

\begin{equation}
d(uv) = \sum_{\delta\mid (u, v)}\mu(\delta)d\left(\frac{u}{\delta}\right)d\left(\frac{v}{\delta}\right)
\end{equation}

\begin{equation}
\sigma_k(u)\sigma_k(v) = \sum_{\delta\mid (u, v)}\delta^k\sigma_k(\frac{uv}{\delta^2})
\end{equation}

\begin{equation}
\mu(n) = \sum_{k=1}^n[(k, n)=1]\cos{2\pi \frac{k}{n}}
\end{equation}

\begin{equation}
\varphi(n) = \sum_{k=1}^n[(k, n)=1] = \sum_{k=1}^n(k, n)\cos{2\pi \frac{k}{n}}
\end{equation}

\begin{equation}
\begin{cases}
S(n) = \sum_{k=1}^n(f * g)(k) \\
\sum_{k=1}^nS(\lfloor \frac n k \rfloor) = \sum_{i=1}^nf(i)\sum_{j=1}^{\lfloor n/i \rfloor}(g * 1)(j)
\end{cases}
\end{equation}

\begin{equation}
\begin{cases}
S(n) = \sum_{k=1}^n(f \cdot g)(k), g \text{ completely multiplicative} \\
\sum_{k=1}^nS\left(\left\lfloor \frac n k \right\rfloor\right)g(k) = \sum_{k=1}^n(f * 1)(k)g(k)
\end{cases}
\end{equation}

\subsubsection{升幂引理}

对 \(n>0\), \(p\in\mathbb{P}\), \(x,y\in\mathbb{Z}\), \(p\nmid x,y\) 且 \(p\mid x-y\):
Expand All @@ -163,19 +313,45 @@ \subsubsection{Pythagorean 三元组}

其中 \(0<n<m\), \(k>0\), \(m\perp n\), 且 \(m\), \(n\) 中恰有一个奇数

\subsection{概率论与统计}

\begin{equation}
D(X)=E(X-E(X))^2=E\left(X^2\right)-(E(X))^2
\end{equation}

\begin{equation}
D(X+Y)=D(X)+D(Y)D(aX)=a^2D(X)
\end{equation}

\begin{equation}
E[x]=\sum_{i=1}^{\infty}P(X\geq i)
\end{equation}

\(m\) 个数的方差: \(\displaystyle s^2=\frac{\sum_{i=1}^m x_i^2}m-\overline x^2\)

\subsection{kMAX-MIN 反演}

\begin{equation}
\operatorname{k-max} S=\sum_{T\subset S, T\neq \emptyset}(-1)^{|T|-k}\binom{|T|-1}{k-1}\min T
\end{equation}

\begin{equation}
\max S=\sum_{T\subset S, T\neq \emptyset}(-1)^{|T|-1}\min T
\end{equation}

\subsection{杂项}

\begin{equation}
a>b,(a,b)=1 \implies (a^m-b^m,a^n-b^n)=a^{(m,n)}-b^{(m,n)}
\end{equation}

\begin{equation}
\lfloor x\rfloor=\begin{cases}
\displaystyle x+\frac{\arctan\cot\pi x}{\pi}-\frac{1}{2}, & x\notin\mathrm{Z} \\
x, & x\in\mathrm{Z}
\end{cases}
\end{equation}

\begin{equation}
\varphi(xy)=\varphi(x)\varphi(y)\frac{(x,y)}{\varphi((x,y))}
\end{equation}

\begin{equation}
\mathrm{e}(\mathrm{e}+\pi)^\mathrm{e}(2e + \pi)^\mathrm{e}+\frac{\mathrm{e}}{(\pi^\mathrm{e}-\mathrm{e})(\mathrm{e}+\mathrm{e}^\pi-\pi^\mathrm{e})}\approx 114514.1919810
\end{equation}
Expand All @@ -192,3 +368,45 @@ \subsection{杂项}
\end{cases}\qquad(n\to+\infty)
\end{aligned}
\end{equation}

\begin{equation}
S_j = \sum_{k=1}^nx_k^j
\end{equation}

\begin{eqnarray}
&h_m = \sum_{1\leq j_1 < \cdots < j_m \leq n} x_{j_1}\cdots x_{j_m}\\
&H_m = \sum_{1\leq j_1 \leq \cdots \leq j_m \leq n} x_{j_1}\cdots x_{j_m}
\end{eqnarray}

\begin{eqnarray}
h_n = \frac{1}{n}\sum_{k=1}^n(-1)^{k+1}S_kh_{n-k}
H_n = \frac{1}{n}\sum_{k=1}^nS_kH_{n-k}
\end{eqnarray}

\begin{equation}
\sum_{k=0}^nkc^k = \frac{nc^{n+2}-(n+1)c^{n+1}+c}{(c-1)^2}
\end{equation}

\begin{equation}
\sum_{i=1}^n\frac{1}{n}\approx\ln\left(n + \frac 1 2\right) + \frac{1}{24(n+0.5)^2}+\Gamma,\qquad \Gamma\approx0.5772156649015328606065
\end{equation}

\begin{equation}
n! = \sqrt{2\pi n}\left(\frac{n}{\mathrm{e}}\right)^n\left(1+\frac{1}{12n}+\frac{1}{288n^2}+O\left(\frac{1}{n^3}\right)\right)
\end{equation}

\begin{equation}
(\max-\min){\{x_a-x_b, y_a-y_b, z_a-z_b\}} = \frac{1}{2}\sum_{cyc}\left| (x_a-y_a)-(x_b-y_b) \right|
\end{equation}

\begin{equation}
\sum_{cyc}(a+b) = \frac{(a+b+c)^3 - a^3 - b^3 - c^3}{3}
\end{equation}

划分问题: \(n\) 个 \(k-1\) 维向量最多把 \(k\) 维空间分为 \(\sum_{i=0}^{k}\binom{n}{i}\) 份.

设 \(U\), \(V\), \(W\), \(u\), \(v\), \(w\) 是四面体的边长 (\(U\), \(V\), \(W\) 构成三角形; \(u\) 是 \(U\) 的对边), 则该四面体的面积为

\begin{equation}
V = \frac{\sqrt{ 4u^2v^2w^2 - \sum_{cyc}{u^2(v^2+w^2-U^2)^2} + \prod_{cyc}{(v^2+w^2-U^2)} }}{12}
\end{equation}
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