【题解】P5518 [MtOI2019] 幽灵乐团 / 莫比乌斯反演基础练习题
https://www.luogu.com.cn/problem/P5518
Part I：
\[\begin{aligned}
\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C \left(\frac{\operatorname{lcm}(i,j)}{\gcd(i,k)}\right)^1=&
\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C \frac{ij}{\gcd(i,j)\gcd(i,k)}\\
=&\prod_{i=1}^A i^{BC}
\left(\prod_{j=1}^{B} \frac{j}{\gcd(i,j)}\right)^C
\left(\prod_{k=1}^{C} \frac{1}{\gcd(i,k)}\right)^B\\
=&(A!)^{BC}(B!)^{AC}
\left(\prod_{i=1}^A\prod_{j=1}^{B} \frac{1}{\gcd(i,j)}\right)^C
\left(\prod_{i=1}^A\prod_{k=1}^{C} \frac{1}{\gcd(i,k)}\right)^B\\
\end{aligned}
\]
使用中括号和正斜杠表示取整符号。其中
\[\begin{aligned}
G(n,m)=\prod_{i=1}^{n}\prod_{j=1}^m \gcd(i,j)=&\prod_{d=1}^{n} d^{\sum_{i=1}^{n}\sum_{j=1}^m [\gcd(i,j)=d]}\\
=&\prod_{d=1}^n d^{\sum_{i=1}^{[n/d]}\sum_{j=1}^{[m/d]} [\gcd(i,j)=1]}\\
=&\prod_{d=1}^n d^{\sum_{i=1}^{[n/d]}\sum_{j=1}^{[m/d]} \sum_{o|i,o|j} \mu(o)}\\
=&\prod_{d=1}^n d^{\sum_{o=1}^n \mu(o) [n/do][m/do]}\\
=&\prod_{t=1}^{n} \left({\color{Red}\prod_{d|t} d^{\mu(\frac{t}{d})}}\right)^{[n/t][m/t]}
\end{aligned}
\]
Part II：
\[\begin{aligned}
\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C \left(\frac{\operatorname{lcm}(i,j)}{\gcd(i,k)}\right)^{ijk}=&
\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C \left(\frac{ij}{\gcd(i,j)\gcd(i,k)}\right)^{ijk}\\
=&\left(\prod_{i=1}^A i^{i\sum_{j=1}^B\sum_{k=1}^C jk}\right)
\left(\prod_{j=1}^B j^{j\sum_{i=1}^A\sum_{k=1}^C ik}\right)
\left(\prod_{i=1}^A\prod_{j=1}^B \frac{1}{\gcd^{ij\sum_{k=1}^C k}(i,j)} \right)
\left(\prod_{i=1}^A\prod_{k=1}^C \frac{1}{\gcd^{ik\sum_{j=1}^B j}(i,k)} \right)\\
=&\left( \prod_{i=1}^A i^i \right)^{\frac{B(B+1)C(C+1)}{4}}
\left( \prod_{j=1}^B j^j \right)^{\frac{A(A+1)C(C+1)}{4}}
\left(\frac{1}{H(A,B)}\right)^{\frac{C(C+1)}{2}}
\left(\frac{1}{H(A,C)}\right)^{\frac{B(B+1)}{2}}\\
H(n,m)=\prod_{i=1}^n\prod_{j=1}^m {\gcd}^{ij}(i,j)=&\prod_{d=1}^{n} d^{\sum_{i=1}^n\sum_{j=1}^n [\gcd(i,j)=d]ij}\\
=&\prod_{d=1}^n d^{d^2\sum_{i=1}^{[n/d]}\sum_{j=1}^{[m/d]} [\gcd(i,j)=1]ij}\\
=&\prod_{d=1}^n d^{d^2 \sum_{i=1}^{[n/d]}\sum_{j=1}^{[m/d]}ij\sum_{o|i,o|j} \mu(o) }\\
=&\prod_{d=1}^n d^{d^2 \sum_o \mu(o)o^2{\color{Red}(\sum_{i=1}^{[n/od]}i)(\sum_{j=1}^{[m/od]} j)}}\\
=&\prod_{t=1}^n \left({\color{Red}\prod_{d|t} d^{d^2 \mu(\frac{t}{d})(\frac{t}{d})^2}}\right)^{S([n/t])S([m/t])}
\end{aligned}
\]
Part III：
\[\begin{aligned}
\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C \left(\frac{\operatorname{lcm}(i,j)}{\gcd(i,k)}\right)^{\gcd(i,j,k)}=& \prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C \left(\frac{ij}{\gcd(i,j)\gcd(i,k)} \right)^{\gcd(i,j,k)}\\
=& \left(\prod_{i=1}^A i^{\sum_{j=1}^B\sum_{k=1}^C \gcd(i,j,k)}\right)
\left(\prod_{j=1}^B j^{\sum_{i=1}^A\sum_{k=1}^C \gcd(i,j,k)}\right)
\left(\prod_{i=1}^A\prod_{j=1}^B \left(\frac{1}{\gcd(i,j)}\right)^{\sum_{k=1}^C\gcd(i,j,k)}\right)
\left(\prod_{i=1}^A\prod_{k=1}^C \left(\frac{1}{\gcd(i,k)}\right)^{\sum_{j=1}^B\gcd(i,j,k)}\right)
\end{aligned}
\]
其中
\[\begin{aligned}
\prod_{i=1}^A i^{\sum_{j=1}^B\sum_{k=1}^C \gcd(i,j,k)}=&
\prod_{i=1}^A i^{\sum_{d|i}d\sum_{j=1}^B\sum_{k=1}^C [\gcd(i,j,k)=d]}\\
=&\prod_{i=1}^A i^{\sum_{d|i}d\sum_{j=1}^{[B/d]}\sum_{k=1}^{[C/d]} [\gcd(\frac{i}{d},j,k)=1]}\\
=&\prod_{i=1}^A i^{\sum_{d|i}d\sum_{j=1}^{[B/d]}\sum_{k=1}^{[C/d]} \sum_{o|\frac{i}{d},o|j,o|k}\mu(o)}\\
=&\prod_{i=1}^A i^{\sum_{d|i}d\sum_{o|\frac{i}{d}}\mu(o)[B/od][C/od]}\\
=&\prod_{i=1}^A i^{\sum_{t|i} [B/t][C/t]\sum_{d|t} d\mu(\frac{t}{d})}\\
=&\prod_{t=1}^A\left(\prod_{t|i,i\le A} i\right)^{[B/t][C/t]\sum_{d|t} d\mu(\frac{t}{d})}\\
=&\prod_{t=1}^A\left(\prod_{t|i,i\le A} i \right)^{[B/t][C/t]\varphi(t)}\\
=&\prod_{t=1}^A\left(\prod_{i=1}^{[A/t]} it \right)^{[B/t][C/t]\varphi(t)}\\
=&\prod_{t=1}^A\left(t^{[A/t]}([A/t])! \right)^{[B/t][C/t]\varphi(t)}\\
\end{aligned}
\]
以及
\[\begin{aligned}
\prod_{i=1}^A\prod_{j=1}^B \gcd(i,j)^{\sum_{k=1}^C\gcd(i,j,k)}=&
\prod_{d=1}^A d^{\sum_{i=1}^A\sum_{j=1}^B[\gcd(i,j)=d]\sum_{k=1}^C\gcd(d,k)}\\
=&\prod_{d=1}^A d^{\sum_{i=1}^{[A/d]}\sum_{j=1}^{[B/d]}[\gcd(i,j)=1]\sum_{k=1}^C\gcd(d,k)}\\
=&\prod_{d=1}^A d^{\sum_{i=1}^{[A/d]}\sum_{j=1}^{[B/d]}\sum_{o|i,o|j} \mu(o)\sum_{k=1}^C\gcd(d,k)}\\
=&\prod_{d=1}^A d^{\sum_{o} \mu(o) [A/od][B/od]\sum_{k=1}^C\gcd(d,k)}\\
=&\prod_{d=1}^A d^{{\sum_{o} \mu(o) [A/od][B/od]}{\sum_{t|d}[C/t]\varphi(t)}}\\
=&\prod_{t=1}^{A} \left(\prod_{t|d}^{A} d^{\sum_{o} \mu(o)[A/od][B/od] } \right)^{[C/t]\varphi(t)}\\
=&\prod_{t=1}^{A} \left(\prod_{i=1}^{[A/t]} (it)^{\sum_{o} \mu(o)[A/oit][B/oit] } \right)^{[C/t]\varphi(t)}\\
=&\prod_{t=1}^{A} \left(t^{\sum_{i=1}^{[A/t]}\sum_{o} \mu(o)[A/oit][B/oit]}\prod_{i=1}^{[A/t]} i^{\sum_{o} \mu(o)[A/oit][B/oit] } \right)^{[C/t]\varphi(t)}\\
=&\prod_{t=1}^{A} \left(t^{\sum_{j=1}^{[A/t]}\sum_{o|j} \mu(o)[A/jt][B/jt]}\prod_{i=1}^{[A/t]} i^{\sum_{o} \mu(o)[A/oit][B/oit] } \right)^{[C/t]\varphi(t)}\\
=&\prod_{t=1}^{A} \left(t^{\sum_{j=1}^{[A/t]}[j=1][A/jt][B/jt]}\prod_{i=1}^{[A/t]} i^{\sum_{o} \mu(o)[A/oit][B/oit] } \right)^{[C/t]\varphi(t)}\\
=&\prod_{t=1}^{A} \left(t^{[A/t][B/t]}\prod_{i=1}^{[A/t]} i^{\sum_{o} \mu(o)[A/oit][B/oit] } \right)^{[C/t]\varphi(t)}\\
=&\prod_{t=1}^{A} \left(t^{[A/t][B/t]}\prod_{p=1}^{[A/t]} \left(\prod_{i|p} i^{\mu(\frac{p}{i}))}\right)^{[A/pt][B/pt]}\right)^{[C/t]\varphi(t)}\\
\end{aligned}
\]
所有东西都需要整除分块，复杂度 \(O(Tn^{0.75}\log P)\)。
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