鏼爷的冬令营课件
核猩公式
\[\ln (1+x)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+...
\]
版本1
\[\begin{aligned}
\prod_{i=1}^{n}(1+x^i+x^{2i}+...)^{a_i}=&\prod_{i=1}^{n}(\frac{1}{1-x^i})^{a_i}\\
=&\exp(\sum_{i=1}^{n}-a_i\ln(1-x^i))\\
=&\exp(\sum_{i=1}^{n}a_i\sum_{j=1}^{+\infty}\frac{x^{ij}}{j})\ (其实感觉推到这步就能做了的说)\\
=&\exp(\sum_{j=1}^{+\infty}\frac{1}{j}\sum_{i=1}^{n}a_ix^{ij})\\
=&\exp(\sum_{j=1}^{+\infty}\frac{1}{j}A(x^j))
\end{aligned}
\]
可以\(A(x)\)有用的项一共只有\(\sum_{i=1}^{n}\lfloor\frac{n}{i}\rfloor=O(n \log n)\)项。
版本2
\[\begin{aligned}
\prod_{i=1}^{n}(1+x^i)^{a_i}=&\exp(\sum_{i=1}^{n}a_i\ln(1+x^i))\\
=&\exp(\sum_{i=1}^{n}a_i\sum_{j=1}^{+\infty}(-1)^{j+1}\frac{x^{ij}}{j})\\
=&\exp(\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{j}\sum_{i=1}^{n}a_ix^{ij})\\
=&\exp(\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{j}A(x^j))
\end{aligned}
\]
版本3
\[\begin{aligned}
\prod_{i=1}^{n}(1+x^i+x^{2i}+...+x^{a_ii})=&\prod_{i=1}^{n}\frac{1-x^{a_ii}}{1-x^i}\\
=&\exp(\sum_{i=1}^{n}(\ln(1-x^{a_ii})-\ln(1-x^i)))\\
=&\exp(\sum_{i=1}^{n}(\sum_{j=1}^{+\infty}\frac{x^{ij}}{j})-(\sum_{j=1}^{+\infty}\frac{x^{a_iij}}{j}))
\end{aligned}
\]