$[q^k]f$ 表示多项式 $f$中$q^k$ 的系数
n的奇拆分数目 $[q^n]\prod\limits_{k=1}^\infty\sum\limits_{i=0}^\infty (q^{(2k-1)})^i=[q^n]\prod\limits_{k=1}^\infty\frac{1}{1-q^{2k-1}}$
n的互异拆分数目 $[q^k]\prod\limits_{k=1}^\infty(1+q^k)$
而
![qwq](https://cdn.luogu.com.cn/upload/image_hosting/sn2833z3.png)
(不知道为啥markdown复制到luogu上就炸了
```latex
$$
\begin{align}
&\prod\limits_{k=1}^\infty\frac{1}{1-q^{2k-1}} \\
=&\frac{\prod\limits_{k=1}^\infty\frac{1}{1-q^{k}}}{\prod\limits_{k=1}^\infty\frac{1}{1-q^{2k}}} \\
=&\prod\limits_{k=1}^\infty\frac{1-q^{2k}}{1-q^{k}} \\
=&\prod\limits_{k=1}^\infty(1+q^k) \\
\end{align}
$$
```
by suyue1098765432 @ 2024-03-27 20:53:02