WD与积木

题目：
https://www.luogu.com.cn/problem/P5162

思路：
设总的方案数为$g_n$，所有方案的层数的总数为$f_n$，则答案就是$\frac{f_n}{g_n}$。先算$g_n$
$$g_n=\sum _{i=1}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)g_{n-i}$$
含义就是我们枚举第一层里都有哪些积木，再乘上剩下的积木贡献的方案数$g_{n-i}$，化简。
$$\begin{array}{rl}{g}_n& =\frac{n!}{i!(n-i)!}{g}_{n-i}\\ \frac{g_n}{n!}& =\sum _{i=1}^n\frac{1}{i!}\frac{g_{n-i}}{(n-i)!}\end{array}$$
设$G_n=\frac{g_n}{n!},H_n=\frac{1}{n!}$，则
$$G_n=\sum _{i=1}^n{H}_i{G}_{n-i}$$
$g_0=G_0=H_0=1$
类似卷积，考虑生成函数$G={\sum }_{i=0}^{\infty }{G}_i{x}^i,H={\sum }_{i=0}^{\infty }{H}_i{x}^i$
$$H\ast G=\sum _{i=0}^{\infty }\sum _{j=0}^i{H}_i{G}_{i-j}{x}^i$$
当$i>0$
$$\begin{array}{rl}\sum _{j=0}^i{H}_i{G}_{i-j}{x}^i& =H_0{G}_i{x}^i+\sum _{j=1}^i{H}_i{G}_{i-j}{x}^i\\ & =(H_0{G}_i+G_i)x^i\\ & =2G_i{x}^i\end{array}$$
当$i=0$
$$H_0{G}_0{x}^0=x^0=(2G_0-1)x^0$$
所以
$$\begin{array}{rl}H\ast G=\sum _{i=0}^{\infty }\sum _{j=0}^i{H}_i{G}_{i-j}{x}^i& =H_0{G}_0{x}^0+\sum _{i=1}^{\infty }\sum _{j=0}^i{H}_i{G}_{i-j}{x}^i\\ & =(2G_0-1)x^0+\sum _{i=1}^{\infty }2G_i{x}^i\\ & =-1+2\sum _{i=0}^{\infty }{G}_i{x}^i\\ & =2G-1\\ G=(2-H{)}^{-1}& \end{array}$$
用多项式求逆即可。
再来算$f_n$
$$\begin{array}{rl}{f}_n& =\sum _{i=1}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)(f_{n-i}+g_{n-i})\\ \frac{f_n}{n!}& =\sum _{i=1}^n\frac{1}{i!}[\frac{f_{n-i}}{(n-i)!}+\frac{g_{n-i}}{(n-i)!}]\end{array}$$
解释就是还是枚举第一层，然后总层数就是剩下$n-i$个积木贡献的总层数，加上第一个盒子的贡献。还是考虑生成函数。
设$F_n=\frac{f_n}{n!},F_0=f_0=0$
$$F_n=\sum _{i=1}^n{H}_i(F_{n-i}+G_{n-i})$$
$F={\sum }_{i=0}^{\infty }{F}_i{x}^i$
$$H\ast (F+G)=\sum _{i=0}^{\infty }\sum _{j=0}^i{H}_i(F_{i-j}+G_{i-j})x^i$$
当$i>0$
$$\begin{array}{rl}\sum _{j=0}^i{H}_i(F_{i-j}+G_{n-i})x^i& =H_0(F_i+G_i)x^i+\sum _{j=1}^i{H}_i(F_{i-j}+G_{n-i})x^i\\ & =(2F_i+G_i)x^i\end{array}$$
当$i=0$
$$H_0(F_0+G_0)x^0=x^0=(2F_0+G_0)x^0$$
所以
$$\begin{array}{rl}H\ast (F+G)& =2F+G\\ F& =(1-G)\ast (H-2{)}^{-1}\end{array}$$
最后答案就是$\frac{[x^n]F}{[x^n]G}$