P5518 [MtOI2019] 幽灵乐团 / 莫比乌斯反演基础练习题

P5518 [MtOI2019] 幽灵乐团 / 莫比乌斯反演基础练习题

官解
这个题几乎包含了所有套路技巧。
下述算式中的部分下取整省略(基本上如果除法没写成负数次方的都是下取整)

type=0

\(\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C\frac{\rm{lcm}(i,j)}{\gcd(i,k)}\)

\(=\prod_{i=1}^A\prod_{j=1}^B(\rm{lcm}(i,j))^C\prod_{k=1}^C\frac{1}{\gcd(i,k)}\)

\(=\prod_{i=1}^A\dfrac{\prod_{j=1}^B(\rm{lcm}(i,j))^C}{\prod_{k=1}^C(\gcd(i,k))^B}\)

\(=\prod_{i=1}^A\dfrac{\prod_{j=1}^B(\frac{i\times j}{\gcd(i,j)})^C}{\prod_{k=1}^C(\gcd(i,k))^B}\)

\(=\prod_{i=1}^Ai^{BC}\prod_{j=1}^Bj^{AC}\prod_{i=1}^A\prod_{j=1}^B\gcd(i,j)^{-C}\prod_{i=1}^A\prod_{k=1}^C\gcd(i,k)^{-B}\)

\(=\prod_{i=1}^Ai^{BC}\prod_{j=1}^Bj^{AC}(\prod_{g=1}^{\min(A,B)}\prod_{i=1}^A\prod_{j=1}^Bg^{-[\gcd(i,j)=g]})^C(\prod_{g=1}^{\min(A,C)}\prod_{i=1}^A\prod_{k=1}^Cg^{-[\gcd(i,k)=g]})^B\)

令 \(fac_i=i!,S(A,B)=\prod_{g=1}^{\min(A,B)}\prod_{i=1}^A\prod_{j=1}^Bg^{[\gcd(i,j)=g]}\)

\((*)=fac_A^{BC}fac_B^{AC}S^{-C}(A,B)S^{-B}(A,C)\)

\(S(A,B)=\prod_{g=1}^{\min(A,B)}\prod_{i=1}^A\prod_{j=1}^Bg^{[\gcd(i,j)=g]}\)

\(=\prod_{g=1}^{\min(A,B)}g^{\sum_{i=1}^{\frac{A}{g}}\sum_{j=1}^{\frac{B}{g}}\sum_{d|i,d|j}\mu(d)}\)

\(=\prod_{g=1}^{\min(A,B)}g^{\sum_{d=1}^{\min(\frac{A}{g},\frac{B}{g})}\mu(d)\frac{A}{gd}\frac{B}{gd}}]\)

\(=\prod_{T=1}^{\min(A,B)}(\prod_{d|T}d^{\mu(\frac{T}{d})})^{\lfloor\frac{A}{T}\rfloor\times \lfloor\frac{B}{T}\rfloor},(T=gd)\)

type=1

\(\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C(\frac{\rm{lcm}(i,j)}{\gcd(i,k)})^{i\times j\times k}\)

\(=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C(\frac{i\times j}{\gcd(i,j)\times \gcd(i,k)})^{i\times j\times k}\)

\(=\prod_{i=1}^A\prod_{j=1}^B(i\times j)^{i\times j\times \sum_{k=1}^Ck}\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C(\gcd(i,j)\times \gcd(i,k))^{-i\times j\times k}\)

\(=\prod_{i=1}^Ai^{i\times \sum_{j=1}^Bj\times \sum_{k=1}^Ck}\prod_{j=1}^Bj^{j\times \sum_{i=1}^Ai\times \sum_{k=1}^Ck}\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C(\gcd(i,j)\times \gcd(i,k))^{-i\times j\times k}\)

令 \(sum(x)=\sum_{i=1}^xi=\frac{x\times (x+1)}{2},pw_x=\prod_{i=1}^x(i^i)\)

\((*)=pw_A^{sum(B)\times sum(C)}pw_B^{sum(A)\times sum(C)}(\prod_{i=1}^A\prod_{j=1}^B\gcd(i,j)^{i\times j})^{-sum(C)}(\prod_{i=1}^A\prod_{k=1}^C\gcd(i,k)^{i\times k})^{-sum(B)}\)

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

\((*)=pw_A^{sum(B)\times sum(C)}pw_B^{sum(A)\times sum(C)}F^{-sum(C)}(A,B)F^{-sum(B)}(A,C)\)

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

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

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

\(=\prod_{g=1}^{\min(A,B)}g^{g^2\sum_{i=1}^{\frac{A}{g}}\sum_{j=1}^{\frac{B}{g}}i\times j\times [\gcd(i,j)=1]}\)

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

\(=\prod_{g=1}^{\min(A,B)}g^{g^2\sum_{d=1}^{\min(\frac{A}{g},\frac{B}{g})}\mu(d)d^2\sum_{i=1}^{\frac{A}{gd}}\sum_{j=1}^{\frac{B}{gd}}i\times j}\)

\(=\prod_{g=1}^{\min(A,B)}g^{g^2\sum_{d=1}^{\min(\frac{A}{g},\frac{B}{g})}\mu(d)d^2sum(\frac{A}{gd})sum(\frac{B}{gd})}\)

\(=\prod_{T=1}^{\min(A,B)}(\prod_{d|T}d^{\mu(\frac{T}{d})})^{T^2sum(\frac{A}{T})sum(\frac{B}{T})},(T=gd)\)

type=2

前置： \(\sum_{d|T}d\mu(\frac{T}{d})=\varphi(T)\)

\(id=\varphi*I,(\sum_{d|n}\varphi(d)=n)\)

\(\therefore id*\mu=\varphi*I*\mu=\varphi*\varepsilon=\varphi\)

\(\therefore \sum_{d|T}d\mu(\frac{T}{d})=\varphi(T)\)

\(\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C(\frac{\rm{lcm}(i,j)}{\gcd(i,k)})^{\gcd(i,j,k)}\)

\(=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C(\frac{i\times j}{\gcd(i,j)\times \gcd(i,k)})^{\gcd(i,j,k)}\)

令 \(f1(A,B,C)=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^Ci^{\gcd(i,j,k)},f2(A,B,C)=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C\gcd(i,j)^{\gcd(i,j,k)}\)

\((*)=f1(A,B,C)f1(B,A,C)f2^{-1}(A,B,C)f2^{-1}(A,C,B)\)

\(f1(A,B,C)=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^Ci^{\gcd(i,j,k)}\)

\(=\prod_{g=1}^{\min(A,B,C)}\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^Ci^{g[\gcd(i,j,k)=g]}\)

\(=\prod_{g=1}^{\min(A,B,C)}\prod_{i=1}^{\frac{A}{g}}\prod_{j=1}^{\frac{B}{g}}\prod_{k=1}^{\frac{C}{g}}(ig)^{g[\gcd(i,j,k)=1]}\)

\(=\prod_{g=1}^{\min(A,B,C)}\prod_{i=1}^{\frac{A}{g}}(ig)^{g\sum_{j=1}^{\frac{B}{g}}\sum_{k=1}^{\frac{C}{g}}[\gcd(i,j,k)=1]}\)

\(=\prod_{g=1}^{\min(A,B,C)}\prod_{i=1}^{\frac{A}{g}}(ig)^{g\sum_{d=1}^{\min(\frac{A}{g},\frac{B}{g},\frac{C}{g})}[d|i]\mu(d)\frac{B}{gd}\frac{C}{gd}}\)

\(=\prod_{g=1}^{\min(A,B,C)}\prod_{i=1}^{\frac{A}{g}}(ig)^{g\sum_{d=1}^{\min(\frac{A}{g},\frac{B}{g},\frac{C}{g})}[d|i]\mu(d)\frac{B}{gd}\frac{C}{gd}}\)

\(=\prod_{g=1}^{\min(A,B,C)}\prod_{d=1}^{\min(\frac{A}{g},\frac{B}{g},\frac{C}{g})}\prod_{i=1}^{\frac{A}{gd}}(idg)^{g\mu(d)\frac{B}{gd}\frac{C}{gd}}\)

\(=\prod_{T=1}^{\min(A,B,C)}\prod_{d|T}\prod_{i=1}^{\frac{A}{T}}(i\times T)^{d\mu(\frac{T}{d})\frac{B}{T}\frac{C}{T}},(T=gd)\)

\(=\prod_{T=1}^{\min(A,B,C)}(\prod_{d|T}(fac(\frac{A}{T})\times T^{\frac{A}{T}})^{d\mu(\frac{T}{d})})^{\frac{B}{T}\frac{C}{T}}\)

\(=\prod_{T=1}^{\min(A,B,C)}((fac(\frac{A}{T})\times T^{\frac{A}{T}})^{\sum_{d|T}d\mu(\frac{T}{d})})^{\frac{B}{T}\frac{C}{T}}\)

\(=\prod_{T=1}^{\min(A,B,C)}(fac(\frac{A}{T})\times T^{\frac{A}{T}})^{\varphi(T)\frac{B}{T}\frac{C}{T}}\)

\(=\prod_{T=1}^{\min(A,B,C)}fac(\frac{A}{T})\times \prod_{T=1}^{\min(A,B,C)} T^{\varphi(T)\frac{A}{T}\frac{B}{T}\frac{C}{T}}\)

枚举 \(\\gcd(i,j,k)\)

\(f2(A,B,C)=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C\gcd(i,j)^{\gcd(i,j,k)}\)

\(=\prod_{d=1}^{\min(A,B,C)}\prod_{i=1}^{\frac{A}{d}}\prod_{j=1}^{\frac{B}{d}}\prod_{k=1}^{\frac{C}{d}}(d\times \gcd(i,j))^{d[\gcd(i,j,k)=1]}\)

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

\(=\prod_{d=1}^{\min(A,B,C)}\prod_{t=1}^{\min(\frac{A}{d},\frac{B}{d},\frac{C}{d})}\prod_{i=1}^{\frac{A}{dt}}\prod_{j=1}^{\frac{B}{dt}}(dt\times \gcd(i,j))^{d\mu(t)\frac{C}{dt}}\)

\(=\prod_{T=1}^{\min(A,B,C)}\prod_{i=1}^{\frac{A}{T}}\prod_{j=1}^{\frac{B}{T}}(T\times \gcd(i,j))^{\frac{C}{T}\sum_{d|T}d\mu(\frac{T}{d})},(T=dt)\)

\(=\prod_{T=1}^{\min(A,B,C)}T^{\varphi(T)\frac{A}{T}\frac{B}{T}\frac{C}{T}}\times\prod_{T=1}^{\min(A,B,C)}\prod_{i=1}^{\frac{A}{T}}\prod_{j=1}^{\frac{B}{T}}\gcd(i,j)^{\frac{C}{T}\varphi(T)}\)

\(=\prod_{T=1}^{\min(A,B,C)}T^{\varphi(T)\frac{A}{T}\frac{B}{T}\frac{C}{T}}\times\prod_{T=1}^{\min(A,B,C)}(\prod_{T'=1}^{\min(\frac{A}{T},\frac{B}{T},\frac{C}{T})}\prod_{d|T'}d^{\mu(\frac{T'}{d})\frac{A}{TT'}\frac{B}{TT'}})^{\frac{C}{T}\varphi(T)}\)

发现 \(f1,f2\) 有一部分可以消。消完之后有

\(f1(A,B,C)=\prod_{T=1}^{\min(A,B,C)}fac(\frac{A}{T})\)

\(f2(A,B,C)=\prod_{T=1}^{\min(A,B,C)}(\prod_{T'=1}^{\min(\frac{A}{T},\frac{B}{T},\frac{C}{T})}\prod_{d|T'}d^{\mu(\frac{T'}{d})\frac{A}{TT'}\frac{B}{TT'}})^{\frac{C}{T}\varphi(T)}\)

\((*)=f1(A,B,C)f1(B,A,C)f2^{-1}(A,B,C)f2^{-1}(A,C,B)\)
至此你可以选择两遍数论分块过掉它。
或者再把这个暴力拆开和 \(type=1\) 对比发现 \((*)=\prod_{i=1}^{\min(A,B,C)}F_{type=0}(\frac{A}{d},\frac{B}{d},\frac{C}{d})^{\varphi(d)}\)
然后一遍数论分块过掉它。