数学-公式

鲜花

数论方向

取模相关

\(p\) 是质数。

费马小定理:

\[\large a^{p-1}\equiv 1\pmod p \]

\[\large a^{-1}\equiv {a^{p-2}}\pmod p \]

威尔逊定理:

\[\large (p-1)!\equiv -1\pmod p \]

卢卡斯定理:

\[\large \tbinom{n}{m} \equiv \tbinom{\lfloor n/p \rfloor}{\lfloor m/p \rfloor}\tbinom{n\bmod p}{m\bmod p} \pmod p \]

同余方程相关

线性丢番图方程有解问题:

如果 \(\large a_1,a_2,\cdots,a_n\) 是整数,方程 \(\large a_1x_1+a_2x_2+\cdots+a_nx_n=c\) 有整数解,当且仅当 \(\large \gcd(a_1,a_2,\cdots,a_n)|c\)

整除方向

鸽巢原理:

\(\large n\) 个物品分为 \(\large k\) 组,至少存在一个分组含有 \(\large \ge \left\lceil\frac{n}{k}\right\rceil\) 个物品。

卷积相关

\[\large\sum\limits_{d|n}\mu(d)=[n=1] \]

\[\large\sum\limits_{d|n}\phi(d)=n \]

\[\large\sum\limits_{d|n}\mu(d)\frac{n}{d}=\phi(n) \]

计数方向

排列相关

圆排列:

\[\large (n-1)! \]

多重集排列数:

\[\large \frac{n!}{n_1!n_2!\cdots n_k!} \]

组合相关

杨辉恒等式:

\[\large\binom{n}{m}=\binom{n-1}{m}+\binom{n-1}{m-1} \]

单行和:

\[\large\sum\limits_{i=0}^n\binom{n}{i}=2^n \]

单行平方和:

\[\large\sum\limits_{i=0}^n\binom{n}{i}^2=\binom{2n}{n} \]

二项式定理:

\[\large(a+b)^n=\sum\limits_{i=0}^n \binom{n}{i}a^ib^{n-i} \]

广义组合数:

\[\large \binom{n} {m}=\frac{\prod_{n-m+1}^n}{m!}=(-1)^m\binom{-n+m-1}{m} \]

上指标翻转:

\[\large \binom{r}{k}=(-1)^k\binom{k-r-1}{k} \]

卡特兰数公式:

\[\large Catlan(n)=\frac{(2n)!}{(n+1)!n!}=\frac{1}{n+1}\binom{2n}{n}=\binom{2n}{n}-\binom{2n}{n-1} \]

第二类斯特林数:

\[\large {n \brace m}={n-1 \brace m-1}+m{n-1 \brace m}=\frac{1}{m!}\sum\limits_{k=0}^m(-1)^k\binom{m}{k}(m-k)^n \]

普通幂转化:

\[\large x^n=\sum\limits_{i=0}^n \binom{x}{i}i!{n \brace k} \]

容斥相关

基本形式:

\[\large f_S=\sum\limits_{T\subseteq S}^n g_T \iff g_S=\sum\limits_{T\subseteq S} (-1)^{|S|-|T|}f_T \]

\[\large f_S=\sum\limits_{S\subseteq T \subseteq U}^n g_T \iff g_S=\sum\limits_{S\subseteq T \subseteq U} (-1)^{|T|-|S|}f_T \]

二项式反演:

\[\large f_n=\sum\limits_{i=0}^n\binom{n}{i}g_i \iff g_n=\sum\limits_{i=0}^n(-1)^{n-i}\binom{n}{i}f_i \]

\[\large f_k=\sum\limits_{i=k}^n\binom{i}{k}g_i \iff g_k=\sum\limits_{i=k}^n(-1)^{i-k}\binom{i}{k}f_i \]

Min-Max 容斥:

\[\large \max_{i\in S}(x_i)=\sum\limits_{T\subseteq S}(-1)^{|T|-1}\min_{j\in T}(x_j) \]

\[\large \min_{i\in S}(x_i)=\sum\limits_{T\subseteq S}(-1)^{|T|-1}\max_{j\in T}(x_j) \]

扩展 Min-Max 容斥:

\[\large \text{kth} \max_{i\in S}(x_i)=\sum\limits_{T\subseteq S}(-1)^{|T|-k}\binom{|T|-1}{k-1}\min_{j\in T}(x_j) \]

\[\large \text{kth} \min_{i\in S}(x_i)=\sum\limits_{T\subseteq S}(-1)^{|T|-k}\binom{|T|-1}{k-1}\max_{j\in T}(x_j) \]

对期望仍然成立:

\[\large E\left(\text{kth} \max_{i\in S}(x_i)\right)=\sum\limits_{T\subseteq S}(-1)^{|T|-k}\binom{|T|-1}{k-1}E\left(\min_{j\in T}(x_j)\right) \]

\[\large E\left(\text{kth} \min_{i\in S}(x_i)\right)=\sum\limits_{T\subseteq S}(-1)^{|T|-k}\binom{|T|-1}{k-1}E\left(\max_{j\in T}(x_j)\right) \]

生成函数相关

普通生成函数:

\[\large f(x)=\sum\limits_{i=0}^{+\infty} a_ix^i \]

指数生成函数

\[\large f(x)=\sum\limits_{i=0}^{+\infty} a_i\frac{x^i}{i!} \]

求前缀和:

\[\large \frac{1}{1-x} f(x)=\sum\limits_{i=0}^{+\infty} \left(\sum\limits_{j=0}^i a_j \right) x^i \]

求差分:

\[\large (1-x) f(x)=a_0+\sum\limits_{i=1}^{+\infty} \left(a_i-a_{i-1} \right) x^i \]

常见普通生成函数:

\[\large \frac{1}{1-x}=\sum\limits_{i=0}^{+\infty} p^ix=\left\{1,1,1,\cdots,1\right\} \]

\[\large \frac{1}{1-px}=\sum\limits_{i=0}^{+\infty} p^ix=\left\{1,p,p^2,\cdots,p^i\right\} \]

\[\large \frac{1}{(1-x)^k}=\sum\limits_{i=0}^{+\infty} \binom{k}{i}x^i=\left\{\binom{k}{0},\binom{k}{1},\binom{k}{2},\cdots,\binom{k}{i} \right\} \]

\[\large e^x=\sum\limits_{i=0}^{+\infty} \frac{1}{i!}x^i=\left\{1,1,\frac{1}{2!},\frac{1}{3!},\cdots,\frac{1}{n!}\right\} \]

\[\large -\ln(1-x)=\sum\limits_{i=0}^{+\infty} \frac{1}{i}x^i=\left\{1,1,\frac{1}{2},\frac{1}{3},\cdots,\frac{1}{n}\right\} \]

常见指数生成函数:

\[\large e^x=\sum\limits_{i=0}^{+\infty} \frac{x^i}{i!}=\left\{1,1,1,\cdots,1\right\} \]

\[\large e^{px}=\sum\limits_{i=0}^{+\infty} \frac{x^i}{i!}=\left\{1,p,p^2,\cdots,p^i\right\} \]

\[\large \frac{e^{x}+e^{-x}}{2}=\frac{1}{2}\sum\limits_{i=0}^{+\infty} \frac{x^i+x^{-i}}{i!}=\left\{1,0,1,0,\cdots\right\} \]

\[\large \frac{e^{x}-e^{-x}}{2}=\frac{1}{2}\sum\limits_{i=0}^{+\infty} \frac{x^i-x^{-i}}{i!}=\left\{0,1,0,1,\cdots\right\} \]

常见求和式:

\[\large \sum\limits_{i=0}^n 1=n \]

\[\large \sum\limits_{i=0}^n i=\frac{n(n+1)}{2} \]

\[\large \sum\limits_{i=0}^n i^2=\frac{n(n+1)(2n+1)}{6} \]

\[\large \sum\limits_{i=0}^n i^3=\frac{n^2(n+1)^22}{4} \]

\[\large \sum\limits_{i=0}^n a_0+id=\frac{(n+1)(a_0+a_0+nd)}{2} \]

\[\large \sum\limits_{i=0}^n a_0q^i=a_0\frac{1-q^{n+1}}{1-q} \]

多项式相关

多项式乘法:

\[\large H(x)=F(x)G(x) \]

\[\large h_k=\sum\limits_{i=0}^kf_i\times g_{k-i} \]

多项式逆元:

\[\large F(x)G(x)\equiv 1 \pmod n \]

\[\large F(x)G_0(x)\equiv 1\pmod {\left\lceil\frac{n}{2}\right\rceil} \]

\[\large G(x)\equiv2G_0(x)-F(x)(G_0)^2(x)\pmod n \]

多项式除法:

\[\large F(x)=Q(x)G(x)+R(x) \]

\[\large \bar{F}(x)=x^nF(\frac{1}{x}) \]

\[\large \bar{Q}(x)\equiv\bar F(x)\bar G^{-1}(x)\pmod x^{n-m+1} \]

\[\large R(x)=F(x)-G(x)R(x) \]

多项式求导:

\[\large f(x)=\sum\limits_{i=0}^n a_ix^n \]

\[\large f'(x)=\sum\limits_{i=0}^{n-1} (i+1)a_{i+1}x^i \]

\[\large (f(g(x)))'=f'(g(x))g'(x) \]

多项式积分:

\[\large f(x)=\sum\limits_{i=0}^n a_ix^n \]

\[\large \int f(x) \,dx =\sum\limits_{i=0}^{n-1} \frac{a_{i-1}x^i}{i} \]

多项式 \(\ln\)

\[\large G(x)\equiv \ln F(x) \pmod {x^n} \]

\[\large G'(x)\equiv \frac{F'(x)}{F(x)} \pmod{x^n} \]

\[\large G(x)\equiv \int \frac{F'(x)}{F(x)} \, dx \pmod{x^n} \]

多项式牛顿迭代;

\[\large F(G(x))\equiv0 \pmod{x^n} \]

\[\large F_0(G(x))=0\pmod {x^{2^{m-1}}} \]

\[\large G(x)\equiv G_0(x)-\frac{F(G_0(x))}{F'(G_0(x))} \pmod{x^{2^m}} \]

多项式 \(\exp\)

\[\large e^{F(x)}\equiv G(x)\pmod{x^n} \]

\[\large e^{F(x)}\equiv G_0(x)\pmod {x^{2^{m-1}}} \]

\[G(x)\equiv G_0(x)(1-\ln G_0(x)+F(x)) \pmod {x^{2^m}} \]

多项式快速幂:

\[\large F^k(x)\equiv G(x) \pmod{x^n} \]

\[\large G(x)\equiv e^{ k\ln F(x)} \pmod{x^n} \]

分治 \(FFT\)

\[\large f_n=\sum\limits_{i=1}^n f_{n-i}g_{i},f_0=1 \]

\[\large F(x)\equiv (1-G^{-1}(x)) \pmod (x^n) \]

期望方向

基本公式:

\[\large E (aX+Y)=aE(x)+E(Y) \]

\[\large E (X+Y)=aE(x)+E(Y) \]

随机变量期望:

\[\large P(x=k)=(1-p)^{k-1}p\Longrightarrow E(x=k)=\frac{1}{p} \]

统计方向

平均数:

\[\large \bar{x}=\frac{1}{n}\sum\limits_{i=1}^n x_i \]

方差:

\[\large s^2=\frac{1}{n}\sum\limits_{i=1}^n(x_i-\bar{x})^2=\frac{1}{n}\sum\limits_{i=1}^n x_i^2-\bar{x}^2 \]

线代方向

几何方向

posted @ 2024-06-30 11:58  yshpdyt  阅读(34)  评论(0)    收藏  举报