莫比乌斯反演

求：

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

先提一个 \(k\)

\[\sum \limits_{i=1}^{\lfloor \frac{n}{k} \rfloor} {\sum \limits_{j=1}^{\lfloor \frac{m}{k} \rfloor} {[\gcd(i,j)=1]}} \]

\[\sum \limits_{i=1}^{\lfloor \frac{n}{k} \rfloor} {\sum \limits_{j=1}^{\lfloor \frac{m}{k} \rfloor} {\sum \limits_{d \mid \gcd(i,j)}{\mu(d)}}} \]

\[\sum \limits_{d=1}^n {\sum \limits_{i=1}^{\lfloor \frac{n}{kd} \rfloor} {\sum \limits_{j=1}^{\lfloor \frac{m}{kd} \rfloor} {\mu(d)}}} \]

把 \(\mu(d)\) 提到前面，得到

\[\sum \limits_{d=1}^{n}{\mu(d )\sum \limits_{i=1}^{\lfloor \frac{n}{kd} \rfloor } \sum \limits_{j=1}^{\lfloor \frac{m}{kd} \rfloor}{1}} \]

\[\sum \limits_{d=1}^{n} {\mu(d) \lfloor \frac{n}{kd} \rfloor \lfloor \frac{m}{kd} \rfloor} \]

先预处理出\(\mu(d)\) ，再数论分块即可。

P2257 YY的GCD

求：

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

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

\[\sum \limits_{k=1,k\in prime}{\sum \limits_{d=1}^{\lfloor \frac{n}{k} \rfloor}{\mu(d) \sum \limits_{i=1}^{\lfloor \frac{n}{kd} \rfloor}{\sum \limits_{j=1}^{\lfloor \frac{m}{kd}\rfloor}{1}}}} \]

\[\sum \limits_{k=1,k\in prime}{\sum \limits_{d=1}^{\lfloor \frac{n}{k}\rfloor}{\mu(d) \times \lfloor \frac{n}{kd} \rfloor \times \lfloor \frac{m}{kd} \rfloor}} \]

令 \(T=kd\) ，得：

\[\sum \limits_{k=1,k\in prime}{\sum \limits_{d=1}^{\lfloor \frac{n}{k}\rfloor}{\mu(d) \times \lfloor \frac{n}{T} \rfloor \times \lfloor \frac{m}{T} \rfloor}} \]

\[\sum \limits_{T=1}^{n}{\lfloor \frac{n}{T} \rfloor \times \lfloor \frac{m}{T} \rfloor \sum \limits_{k \mid T,k\in prime}{\mu(\frac{T}{k})}} \]

对于最右边那个\(\sum\)，可以通过预处理得到，再数论分块，时间复杂度\(O(n+T\sqrt{n})\)