组合对象符号化小记

卷不动题，随便摆一摆😭😭😭。

组合对象的大小函数对应生成函数的次数。

笛卡尔积：

C = A \times B C (z) = A (z) \times B (z)

$\mathcal{C}=\mathcal{A}\times\mathcal{B}\\C(z)=A(z)\times B(z)$

不交并：

A = B + C = (E_{1} \times B) \cup (E_{2} \times C) C (z) = A (z) + B (z)

$\mathcal{A}=\mathcal{B}+\mathcal{C}=(\mathcal{E_1}\times\mathcal{B})\cup(\mathcal{E_2}\times C)\\C(z)=A(z)+B(z)$

Sequence 构造：

B = SEQ (A) = E + A + A^{2} + \dots B (z) = \frac{1}{1 - A (z)}

$\mathcal{B}=\operatorname{SEQ}(\mathcal{A})=\mathcal{E}+A+A^2+\cdots\\B(z)=\frac{1}{1-A(z)}$

Cycle 构造：

B = (SEQ (A) - E) / S

$\mathcal{B}=(\operatorname{SEQ}(\mathcal{A})-\mathcal{E})/S$

根据 Burnside 引理我们计数置换下不动点的平均数：

B (z) = \sum_{k} \frac{1}{k} \sum_{i = 0}^{k - 1} A (z^{\frac{k}{gcd (i, k)}})^{gcd (i, k)} = \sum_{k} \frac{1}{k} \sum_{d ∣ k} φ (d) A (z^{d})^{\frac{k}{d}} = \sum_{d} \frac{φ (d)}{d} \sum_{k} \frac{A (z^{d})^{k}}{k} = \sum_{d} \frac{φ (d)}{d} \ln (\frac{1}{1 - A (z^{d})})

$B(z)=\sum_k\frac{1}{k}\sum_{i=0}^{k-1}A(z^{\frac{k}{\gcd(i,k)}})^{\gcd(i,k)} =\sum_k\frac{1}{k}\sum_{d\mid k}\varphi(d)A(z^d)^{\frac k d}\\ =\sum_d\frac{\varphi(d)}{d}\sum_{k}\frac{A(z^d)^k}{k} =\sum_d\frac{\varphi(d)}{d}\ln(\frac{1}{1-A(z^d)})$

我们预先计算出 $\ln(\frac{1}{1-A(x)})$ ，枚举 $d$ 之后暴力累加 $\lfloor\frac{n}{d}\rfloor$ 项点值即可。

Multiset 构造：

B = SEQ (A) / R B (z) = \prod_{a \in A} \sum_{k} z^{| a | k} = \prod_{a \in A} \frac{1}{1 - z^{| a |}} = \prod_{i} (\frac{1}{1 - z^{i}})^{A_{i}} = \exp (\ln (\prod_{i} (\frac{1}{1 - z^{i}})^{A_{i}})) = \exp (\sum_{i} A_{i} \ln (\frac{1}{1 - z^{i}})) = \exp (\sum_{i} A_{i} \sum_{i ∣ k} \frac{i x^{k}}{k}) = \exp (\sum_{k} \frac{\sum_{k} A_{i} i x^{i k}}{i k}) = \exp (\sum_{k} \frac{A (z^{k})}{k})

$\mathcal{B}=\operatorname{SEQ}(\mathcal{A})/R\\ B(z)=\prod_{a\in A}\sum_k z^{|a|k}=\prod_{a\in A}\frac{1}{1-z^{|a|}} =\prod_i(\frac{1}{1-z^i})^{A_i}\\ =\exp(\ln(\prod_i(\frac{1}{1-z^i})^{A_i})) =\exp(\sum_iA_i\ln(\frac{1}{1-z^i}))\\ =\exp(\sum_iA_i\sum_{i\mid k}\frac{ix^k}{k}) =\exp(\sum_k\frac{\sum_kA_iix^{ik}}{ik})\\ =\exp(\sum_k\frac{A(z^k)}{k})$

Powerset 构造：

Pointing 构造：（选代表元）