Catalan
Stirling
容斥
- (状压+容斥)
- (容斥+递推)
莫比乌斯反演与筛法
\[g(n)=\sum_{d|n}f(d)\] \[f(n)=\sum_{d|n}{\mu(d)g(\frac{n}{d})}\]
blogs:题目:
- + +
三倍经验 - (莫比乌斯反演+树状数组)
- (真的毒瘤的预处理啊orz)
- (关于幂的莫比乌斯反演orz)
二项式反演
\[f(n)=\sum_{k=p}^n (\begin{matrix} n \\ k \end{matrix})g(k)\]
\[g(n)=\sum_{k=p}^n(-1)^{n-k}(\begin{matrix} n \\ k \end{matrix})f(k)\]中国剩余定理
- \[ \begin{Bmatrix} x \equiv c_1 (mod\; m_1)\\ x \equiv c_2 (mod\; m_2)\\ \cdots\\ x \equiv c_n (mod\; m_n) \end{Bmatrix} \] 其中\((m_i, m_j)==1, i\not= j\) 令\(M=\prod_{i=1}^nm_i\), 则\(x=(\sum_{i=1}^nc_i*\frac{M}{m_i}*inv(\frac{M}{m_i}, m_i))mod\;M\)
- \[ \begin{Bmatrix} x \equiv c_1 (mod\; m_1)\\ x \equiv c_2 (mod\; m_2)\\ \cdots\\ x \equiv c_n (mod\; m_n) \end{Bmatrix} \] 对于两个方程\[x \equiv c_1 (mod\; m_1)\\ x \equiv c_2 (mod\; m_2)\] 合并为一个,有解条件为\((m_1, m_2)|(c_2-c_1)\)\[d=(m_1, m_2), m=\frac{m_1m_2}{d}\\ c=(inv(\frac{m_1}{d}, \frac{m_2}{d})*\frac{c_2-c_1}{d})\% (\frac{m_2}{d})*m_1+c_1\] 最终\(x=c\%m\)
- 求\(C_n^m%p\) 唯一分解\(p=\prod_{i=1}^qp_i^{k_i}\)\[ \begin{Bmatrix} x \equiv C_n^m \%p_1^{k_1} (mod\; p_1^{k_1})\\ x \equiv C_n^m \%p_2^{k_2} (mod\; p_2^{k_2})\\ \cdots\\ x \equiv C_n^m \%p_q^{k_q} (mod\; p_q^{k_q}) \end{Bmatrix} \] 则\(x\)即为答案
公式:
- \[g(n)=\sum_{d|n}f(d)\] \[f(n)=\sum_{d|n}{\mu(d)g(\frac{n}{d})}\]
- \[f(n)=\sum_{k=p}^n (\begin{matrix} n \\ k \end{matrix})g(k)\]\[g(n)=\sum_{k=p}^n(-1)^{n-k}(\begin{matrix} n \\ k \end{matrix})f(k)\]
- \[g(n)=\sum_{n|d}{f(d)[d \leq m]}\] \[f(n)=\sum_{n|d}{\mu(\frac{d}{n})g(d)[d \leq m]}\]
- 如果\(f(n)\)是积性函数,且\((x, y) = 1\),则有\[f(xy)=f(x)f(y)\]
- \[\sum_{i=1}^{n}{i[(i, n) == 1]}= \frac{\varphi(n)*n}2\] (用到结论:\(if (i, n) == 1, then (n-i, n) = 1\))
- \[(id*\mu)(i)=\varphi(i)\]\[(\varphi*I)(i)=id(i)\]\[(\mu*I)(i)=e(i)\]
- \[[n == 1]=\sum_{d|n}{\mu(d)}\]
- \[n=\sum_{d|n}{\varphi(d)}\]
- \[\sum_{i=1}^{n}{\sum_{j=1}^{m}ij}=\frac{n^2(n+1)^2}{4}\]
- 除数函数 \[\sigma_k(n)=\sum_{d|n}{d^k}\] 约数个数函数 \[\tau(n)=\sigma_0(n)=\sum_{d|n}1\] 约数和函数\[\sigma(n)=\sigma_1(n)=\sum_{d|n}d\]
- \[\sum_{i=1}^ni=\frac{n(n+1)}{2}\]\[\sum_{i=1}^ni^2=\frac{(n+1)(2n+1)n}{6}\]\[\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}\]
- \[\varphi(ij)=\frac{\varphi(i)\varphi(j)(i,j)}{\varphi((i,j))}\]
- \[[f(x) == 1] = e(f(x))=(\mu*I)(f(x))\](常用于引进\(\mu\)以进行莫比乌斯反演,如[NOI2016]循环之美)
- \[(\begin{matrix} 0 \\ n \end{matrix})+(\begin{matrix} 1 \\ n \end{matrix})+(\begin{matrix} 2 \\ n \end{matrix})+...+(\begin{matrix} n \\ n \end{matrix})=2^n\]
- \[d(ij)=\sum_{x|i}\sum_{y|j}[gcd(x, y) == 1]\]
- \[ \begin{Bmatrix} x \equiv c_1 (mod\; m_1)\\ x \equiv c_2 (mod\; m_2)\\ \cdots\\ x \equiv c_n (mod\; m_n) \end{Bmatrix} \] 其中\((m_i, m_j)==1, i\not= j\) 令\(M=\prod_{i=1}^nm_i\), 则\(x=(\sum_{i=1}^nc_i*\frac{M}{m_i}*inv(\frac{M}{m_i}, m_i))mod\;M\)
- \[ \begin{Bmatrix} x \equiv c_1 (mod\; m_1)\\ x \equiv c_2 (mod\; m_2)\\ \cdots\\ x \equiv c_n (mod\; m_n) \end{Bmatrix} \] 对于两个方程\[x \equiv c_1 (mod\; m_1)\\ x \equiv c_2 (mod\; m_2)\] 合并为一个,有解条件为\((m_1, m_2)|(c_2-c_1)\)\[d=(m_1, m_2), m=\frac{m_1m_2}{d}\\ c=(inv(\frac{m_1}{d}, \frac{m_2}{d})*\frac{c_2-c_1}{d})\% (\frac{m_2}{d})*m_1+c_1\] 最终\(x=c\%m\)
- 求\(C_n^m%p\) 唯一分解\(p=\prod_{i=1}^qp_i^{k_i}\)\[ \begin{Bmatrix} x \equiv C_n^m \%p_1^{k_1} (mod\; p_1^{k_1})\\ x \equiv C_n^m \%p_2^{k_2} (mod\; p_2^{k_2})\\ \cdots\\ x \equiv C_n^m \%p_q^{k_q} (mod\; p_q^{k_q}) \end{Bmatrix} \] 则\(x\)即为答案