P5518 [MtOI2019] 幽灵乐团 / 莫比乌斯反演基础练习题

\[\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}\left(\frac{\operatorname{lcm}(i,j)}{\gcd(i,k)} \right)^{f(\text{Type})} \]

\[\begin{cases} f(0)=1\\ f(1)=i\times j\times k\\ f(2)=\gcd(i,j,k) \end{cases} \]

\[A,B,C\le 10^5 \]

对一个大质数 \(p\) 取模，多组数据。

---

首先有

\[\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}\left(\frac{\operatorname{lcm}(i,j)}{\gcd(i,j)} \right)^{f(\text{Type})} \]

\[=\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}\left(\frac{i\times j}{\gcd(i,j)\times \gcd(i,k)} \right)^{f(\text{Type})} \]

根据连乘的性质，我们只需要关心下面两个东西：

\[\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}i^{f(\text{Type})} \]

\[\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}\gcd(i,j)^{f(\text{Type})} \]

记 \(S_1(A,B,C),S_2(A,B,C)\) 是上面那个东西，则原式等于

\[\frac{\left( \prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}i^{f(\text{Type})} \right) \left( \prod_{j=1}^{B}\prod_{i=1}^{A}\prod_{k=1}^{C}i^{f(\text{Type})} \right)}{\left( \prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}\gcd(i,j)^{f(\text{Type})} \right)\left( \prod_{i=1}^{A}\prod_{k=1}^{C}\prod_{j=1}^{B}\gcd(i,j)^{f(\text{Type})} \right)} \]

\[=\frac{S_1(A,B,C)\times S_1(B,A,C)}{S_2(A,B,C)\times S_2(A,C,B)} \]

\(\text{Type}=0,F(0)=1\)

第一部分

\[\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}i=\prod_{i=1}^{A}\prod_{j=1}^{B}i^C=\prod_{i=1}^{A}i^{BC}=\left( \prod_{i=1}^{A}i \right)^{BC} \]

第二部分

\[\left( \prod_{i=1}^{A}\prod_{j=1}^{B}\gcd(i,j) \right)^{C} \]

只考虑里面的，为

\[\prod_{d=1}d^{\sum_{i=1}^{\frac{A}{d}}\sum_{j=1}^{\frac{B}{d}}[\gcd(i,j)=1]} \]

\[=\prod_{d=1}d^{\sum_{i=1}^{\frac{A}{d}}\sum_{j=1}^{\frac{B}{d}}\sum_{c|i,c|j}\mu(c)} \]

\[=\prod_{d=1}d^{\sum_{c=1}\mu(c)\left\lfloor \frac{A}{cd} \right\rfloor\left\lfloor \frac{B}{cd} \right\rfloor} \]

令 \(F(T)=\left\lfloor \frac{A}{T} \right\rfloor\times \left\lfloor \frac{B}{T} \right\rfloor\)，则

\[=\prod_{d=1}\prod_{c=1}d^{\mu(c)F(cd)} \]

\[=\prod_{T=1}\left( \prod_{d|T}d^{\mu\left( \frac{T}{d} \right)} \right) ^{F(T)} \]

里面的东西一开始预处理直接枚举是 \(O(n\log n)\) 的，然后因为不存在 \(0\) 的项所以维护前缀积之后整除分块。

\(\text{Type}=1,F(1)=i\times j\times k\)

令 \(S(x)=\sum_{i=1}^x i=\frac{x(x+1)}{2}\)。

第一部分

\[\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}i^{i\times j\times k}=\prod_{i=1}^{A}\prod_{j=1}^{B}i^{i\times j\times \sum_{k=1}^{C}k}=\left( \prod_{i=1}^{A} i^i \right)^{\sum{j}\times \sum k} \]

\[=\left( \left( \prod_{i=1}^{A} i^i \right)^{S(B)} \right)^{S(C)} \]

第二部分

还是考虑里面的

\[\prod_{i=1}^{A}\prod_{j=1}^{B} \gcd(i,j)^{i\times j} \]

\[=\prod_{d=1}d^{d^2\sum_{i=1}^{\frac{A}{d}}\sum_{j=1}^{\frac{B}{d}}i\times j\times \sum_{c|i,c|j}\mu(c)} \]

\[=\prod_{d=1}d^{d^2\sum_{c=1}\mu(c)\sum_{i=1}^{\frac{A}{cd}}\sum_{j=1}^{\frac{B}{cd}}(i\times c\times j\times c)} \]

\[=\prod_{d=1}d^{d^2\sum_{c=1}\mu(c)c^2\sum_{i=1}^{\frac{A}{cd}}\sum_{j=1}^{\frac{B}{cd}}(ij)} \]

\[=\prod_{d=1}d^{d^2\sum_{c=1}\mu(c)c^2S(\left\lfloor \frac{A}{cd} \right\rfloor)S(\left\lfloor \frac{B}{cd} \right\rfloor)} \]

还是令 \(T=cd\)，

\[=\prod_{T=1}\left( \prod_{d|T}d^{\mu(\frac{T}{d})\times T^2} \right)^{S(\left\lfloor \frac{A}{T} \right\rfloor)\times S(\left\lfloor \frac{B}{T} \right\rfloor)} \]

那么和 \(\text{Type}=0\) 部分之后的一样了，只是维护的中间括号里的式子东西不太一样。

\(\text{Type}=2,F(2)=\gcd(i,j,k)\)

第一部分

拆不了一层了，先考虑枚举 \(d\)，则有

\[\prod_{d=1}\prod_{i=1}^{\left\lfloor \frac{A}{d} \right\rfloor}\prod_{j=1}^{\left\lfloor \frac{B}{d} \right\rfloor}\prod_{k=1}^{\left\lfloor \frac{C}{d} \right\rfloor}(id)^{d[\gcd(i,j,k)=1]} \]

\[=\prod_{d=1}\prod_{i=1}^{\left\lfloor \frac{A}{d} \right\rfloor}\left( (id)^d \right)^{\sum_{j=1}^{\left\lfloor \frac{B}{d} \right\rfloor}\sum_{k=1}^{\left\lfloor \frac{C}{d} \right\rfloor}[\gcd(i,j,k)=1]} \]

\[=\prod_{d=1}\prod_{i=1}^{\left\lfloor \frac{A}{d} \right\rfloor}\left( (id)^d \right)^{\sum_{j=1}^{\left\lfloor \frac{B}{d} \right\rfloor}\sum_{k=1}^{\left\lfloor \frac{C}{d} \right\rfloor}\sum_{c|i,c|j,c|k}\mu(c)} \]

\[=\prod_{d=1}\prod_{i=1}^{\left\lfloor \frac{A}{d} \right\rfloor}\left( (id)^d \right)^{\sum_{c|i}\mu(c)\left\lfloor \frac{B}{dc} \right\rfloor\left\lfloor \frac{C}{dc} \right\rfloor} \]

\[=\prod_{d=1}\prod_{c=1}\prod_{i=1}^{\left\lfloor \frac{A}{cd} \right\rfloor} (icd)^{\mu(c)d\left\lfloor\frac{B}{dc}\right\rfloor\left\lfloor \frac{C}{dc} \right\rfloor } \]

\[=\prod_{T=1}\prod_{d|T}\prod_{i=1}^{\left\lfloor \frac{A}{T} \right\rfloor} (iT)^{\mu(\frac{T}{d})d\left\lfloor\frac{B}{T}\right\rfloor\left\lfloor \frac{C}{T} \right\rfloor} \]

\[=\prod_{T=1}\left( \prod_{d|T}\left( \left( \left\lfloor \frac{A}{T} \right\rfloor \right)!\times T^{\left\lfloor \frac{A}{T} \right\rfloor} \right)^{\mu(\frac{T}{d})d} \right)^{\left\lfloor\frac{B}{T}\right\rfloor\left\lfloor \frac{C}{T} \right\rfloor } \]

\[=\prod_{T=1}\left( \left( \left( \left\lfloor \frac{A}{T} \right\rfloor \right)!\times T^{\left\lfloor \frac{A}{T} \right\rfloor} \right)^{\sum_{d|T}\mu(\frac{T}{d})d} \right)^{\left\lfloor\frac{B}{T}\right\rfloor\left\lfloor \frac{C}{T} \right\rfloor } \]

\[=\prod_{T=1}\left( \left( \left( \left\lfloor \frac{A}{T} \right\rfloor \right)!\times T^{\left\lfloor \frac{A}{T} \right\rfloor} \right)^{\varphi(T)} \right)^{\left\lfloor\frac{B}{T}\right\rfloor\left\lfloor \frac{C}{T} \right\rfloor } \]

最后一步是因为

\[\sum_{c|n}\varphi(c)=n=f(n) \]

\[\varphi(n)=\sum_{c|n}f(c)\mu(\frac{n}{c}) \]

\[=\sum_{c|n}c\mu(\frac{n}{c}) \]

那么对于 \(T^{\varphi(T)}\) 的部分直接预处理前缀积，考虑处理

\[\prod_{T=1}\left( \left\lfloor \frac{A}{T} \right\rfloor \right)!^{\varphi(T)} \]

考虑整除分块，令 \(a=\left\lfloor \frac{A}{T} \right\rfloor\) 变成求

\[\prod_{l=1}^{r}(a!)^{\varphi(i)}=(a!)^{\sum_{i=l}^r \varphi(i)} \]

又因为取模的质数很大，所以可以依照费马小定理指数对 \(p-1\) 取模，维护一个前缀和即可。

\[a^{p-1}\equiv 1 \pmod p,p\bot a \]

第二部分

求的是

\[\prod_{x=1}^{A} \prod_{y=1}^{B} \prod_{z=1}^{C} \gcd(x,y)^{\gcd(x,y,z)} \]

考虑换元枚举 \(d=\gcd(x,y,z)\) 也就是 \(x=id,y=jd,z=kd\) 那么有 \(\gcd(i,j,k)=1\) 则

\[\gcd(x,y)=\gcd(di,dj)=d\gcd(i,j) \]

那么有

\[\prod_{d=1}\prod_{i=1}^{\left\lfloor \frac{A}{d} \right\rfloor} \prod_{j=1}^{\left\lfloor \frac{B}{d} \right\rfloor} (d\gcd(i,j))^{d\sum_{k=1}^{\left\lfloor \frac{C}{d} \right\rfloor}[\gcd(i,j,k)=1]} \]

\[=\prod_{d=1}\prod_{i=1}^{\left\lfloor \frac{A}{d} \right\rfloor} \prod_{j=1}^{\left\lfloor \frac{B}{d} \right\rfloor} (d\gcd(i,j))^{d\sum_{k=1}^{\left\lfloor \frac{C}{d} \right\rfloor}\sum_{t|i,t|j,t|k}\mu(t)} \]

\[=\prod_{d=1}\prod_{i=1}^{\left\lfloor \frac{A}{d} \right\rfloor} \prod_{j=1}^{\left\lfloor \frac{B}{d} \right\rfloor} (d\gcd(i,j))^{d\sum_{t|i,t|j}\mu(t)\left\lfloor \frac{C}{td} \right\rfloor} \]

\[=\prod_{d=1}\prod_{t=1}\prod_{i=1}^{\left\lfloor \frac{A}{dt} \right\rfloor}\prod_{j=1}^{\left\lfloor \frac{B}{dt} \right\rfloor}(dt\gcd(i,j))^{d \mu(t)\left\lfloor \frac{C}{dt} \right\rfloor} \]

\[=\left(\prod_{d=1}\prod_{t=1}(dt)^{d \mu(t)\left\lfloor \frac{A}{dt} \right\rfloor\left\lfloor \frac{B}{dt} \right\rfloor\left\lfloor \frac{C}{dt} \right\rfloor}\right)\times \left( \prod_{d=1}\prod_{t=1}\prod_{i=1}^{\left\lfloor \frac{A}{dt} \right\rfloor}\prod_{j=1}^{\left\lfloor \frac{B}{dt} \right\rfloor}\gcd(i,j)^{d \mu(t)\left\lfloor \frac{C}{dt} \right\rfloor} \right) \]

那么前面的其实前面求过了是

\[\prod_{T=1}T^{\left\lfloor \frac{A}{T} \right\rfloor\left\lfloor \frac{B}{T} \right\rfloor\left\lfloor \frac{C}{T} \right\rfloor\textcolor{red}{\sum_{d|T}\mu(\frac{T}{d})d}} \]

\[=\prod_{T=1}T^{\textcolor{red}{\varphi(T)}\left\lfloor \frac{A}{T} \right\rfloor\left\lfloor \frac{B}{T} \right\rfloor\left\lfloor \frac{C}{T} \right\rfloor} \]

考虑后面的

\[\prod_{d=1}\prod_{t=1}\prod_{i=1}^{\left\lfloor \frac{A}{dt} \right\rfloor}\prod_{j=1}^{\left\lfloor \frac{B}{dt} \right\rfloor}\gcd(i,j)^{d \mu(t)\left\lfloor \frac{C}{dt} \right\rfloor} \]

再次换元可以得到

\[\prod_{d=1}\prod_{t=1}\prod_{d^\prime=1}\prod_{t^\prime=1}(d^\prime)^{\mu(t^\prime)\mu(t)d\left\lfloor \frac{C}{dt} \right\rfloor\left\lfloor \frac{A}{dd^\prime tt^\prime} \right\rfloor\left\lfloor \frac{B}{dd^\prime tt^\prime} \right\rfloor} \]

\[=\prod_{T=1}\left( \prod_{T^\prime=1}\left( \prod_{d|T^\prime}d^{\mu(\frac{T^\prime}{d})} \right)^{\left\lfloor \frac{A}{TT^\prime} \right\rfloor \left\lfloor \frac{B}{TT^\prime} \right\rfloor} \right)^{\left\lfloor \frac{C}{T} \right\rfloor\color{red}{\sum_{d|T}\mu(\frac{T}{d})d}} \]

\[=\prod_{T=1}\left( \prod_{T^\prime=1}\left(\textcolor{lime}{\prod_{d|T^\prime}d^{\mu(\frac{T^\prime}{d})}} \right)^{\left\lfloor \frac{A}{TT^\prime} \right\rfloor \left\lfloor \frac{B}{TT^\prime} \right\rfloor} \right)^{\left\lfloor \frac{C}{T} \right\rfloor\color{red}{\varphi(T)}} \]

那么记 \(H(T^\prime)=\color{lime}\prod_{d|T^\prime}d^{\mu(\frac{T^\prime}{d})}\)，\(f(T)=\left\lfloor \frac{C}{T} \right\rfloor\varphi(T)\)，

则变成了求

\[\prod_{T}\prod_{T^\prime}H(T^\prime)^{f(T)\left\lfloor \frac{A}{TT^\prime} \right\rfloor\left\lfloor \frac{B}{TT^\prime} \right\rfloor} \]

我们考虑枚举小的那个然后做根号分治。

\(T\le \sqrt n\)

固定 \(T\)，根号分治相同的 \(x=\left\lfloor \frac{A}{TT^\prime} \right\rfloor,y=\left\lfloor \frac{B}{TT^\prime} \right\rfloor\)，则这段相同 \(x,y\) 的区间 \([l,r]\) 贡献为

\[\left( \prod_{T^\prime=l}^{r}H(T^\prime) \right)^{f(T)xy} \]

那么在固定 \(T\) 时，整除分块的复杂度是 \(O(\sqrt{\frac{n}{T}})\) 的，所以复杂度为

\[\sum_{T\le \sqrt n} O\left( \sqrt{\frac{n}{T}} \right) \]

\[= O\left( \sqrt n\sum_{T\le \sqrt n}\frac{1}{\sqrt T} \right) \]

后面的这个东西考虑使用柯西不等式

\[\left( \sum x_i y_i \right)^2\le (\sum x_i^2)(\sum y_i^2) \]

\[\left( \sum_{T=1}^{\sqrt n} \frac{1}{\sqrt T} \right)^2\le \left( \sum_{T=1}^{\sqrt n} 1^2\right)\left( \sum_{T=1}^{\sqrt n} \left( \frac{1}{\sqrt T} \right)^2 \right) \]

第一项是 \(\sqrt n\)，第二项是调和级数 \(\le O(\log n)\)，所以分析的复杂度是 \(O(n^{\frac{3}{4}}\sqrt{\log n})\)。实际上可以通过积分的方法得到复杂度就是 \(\Theta(n^{\frac{3}{4}})\)。

\(T> \sqrt(n)\)

改为固定 \(T^\prime\)，贡献为

\[\prod_{T=l}^rH(T^\prime)^{f(T)xy} \]

\[=\left( H(T^\prime)^{xy} \right)^{\sum_{T=l}^r f(T)} \]

那么同理的复杂度，多一个快速幂。