基础莫比乌斯反演推导

本篇文章主要记录了一些基础的莫比乌斯反演的推导套路以及过程，如果有错误欢迎指出。

感谢：

command_block \(\text{大佬的博客}\)

pengym \(\text{大佬的博客}\)

peterwuyihong \(\text{大佬的题单}\)

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\(\text{这里是常用的反演结论}\)

\[\sum\limits_{i \mid n}^{}{\mu(i)} = [n=1] \]

\[\sum\limits_{i \mid n}^{}{\varphi(i)} = n \]

\[\mu(ij)=\mu(i)\mu(j) [i \perp j] \]

\[\varphi(ij) = \dfrac{\varphi(i)\varphi(j)(i,j)}{\varphi((i,j))} \]

（你可能需要整除分块）

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题目1

\[\sum\limits_{i=1}^{n}{\sum\limits_{j=1}^{m}{[(i,j)=1]}} \]

\(\text{令}\) \(n \leq m\)

\[\sum\limits_{i=1}^{n}{\sum\limits_{j=1}^{m}{\sum\limits_{d=1}^{n}{\mu(d)[d\mid i][d\mid j]}}} \]

\[\sum\limits_{d=1}^{n}{\mu(d) \sum\limits_{i=1}^{n}{[d\mid i] \sum\limits_{j=1}^{m}{[d\mid j]}}} \]

\[\sum\limits_{d=1}^{n}{\mu(d) \left\lfloor\dfrac{n}{d}\right\rfloor \left\lfloor\dfrac{m}{d}\right\rfloor} \]

\(O(\sqrt{n})\)

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