组合数学一些结论

seasongirl 2019-12-05


$C(n, m)=\frac{m !}{n !(m-n) !}$
$\left(C_{n}^{0}\right)^{2}+\left(C_{n}^{1}\right)^{2}+\left(C_{n}^{2}\right)^{2}+\cdots+\left(C_{n}^{n}\right)^{2}=C_{2 n}^{n}$
$(1+x)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l}{n} \\ {k}\end{array}\right) x^{k}$

斯特林公式: $n ! \approx \sqrt{2 \pi n}\left(\frac{n}{e}\right)^{n}$,即$lim_{n\rightarrow \infty}\frac{n!}{\sqrt{2 \pi n}\left(\frac{n}{e}\right)^{n}}=1$

$Catalan$数:
$C_{n+1}=\sum_{i=0}^{n} C_{i} \cdot C_{n-i}=C_{n-1} \cdot \frac{4 n-2}{n+1}$
$C_{n+1}=\left(\begin{array}{c}{2 n} \\ {n}\end{array}\right)-\left(\begin{array}{c}{2 n} \\ {n-1}\end{array}\right)$

$Lucas$定理: 当$p$为素数时,$C_{n}^{m} \equiv C_{n \bmod p}^{m \bmod p} * C_{n / p}^{m / p}(\bmod p)$

在$DAG$中有,最长反链=最小链覆盖,最长链=最小反链覆盖

二项式反演:
若满足$f(n)=\sum_{k=0}^{n}\left(\begin{array}{l}{n} \\ {k}\end{array}\right) g(k)$,则有$g(n)=\sum_{k=0}^{n}(-1)^{n-k}\left(\begin{array}{l}{n} \\ {k}\end{array}\right) f(k)$

$Mobius$反演:
若满足$g(n)=\sum_{d | n} f(d)$,则有$f(n)=\sum_{d | n} g(d) \mu\left(\frac{n}{d}\right)$

子集反演:
若满足$f(S)=\sum_{T \subseteq S} g(T)$,则有$g(S)=\sum_{S \subseteq T}(-1)^{|T|-|S|} S(T)$

单位根反演: $[k | n]=\frac{\sum_{i=0}^{n-1} \omega_{k}^{in}}{n}$

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