组合数学一些结论

Posted 2020-11-26 rimuru-tempest

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

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

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

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

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$f(n)=sum_{k=0}^{n}left(egin{array}{l}{n} \ {k}end{array} ight) g(k)$，则有$g(n)=sum_{k=0}^{n}(-1)^{n-k}left(egin{array}{l}{n} \ {k}end{array} ight) f(k)$

$Mobius$反演:
若满足$g(n)=sum_{d | n} f(d)$，则有$f(n)=sum_{d | n} g(d) muleft(frac{n}{d} ight)$

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: $[k | n]=frac{sum_{i=0}^{n-1} omega_{k}^{in}}{n}$