莫比乌斯反演学习笔记

莫比乌斯反演

数论函数

数论函数是指定义域为正整数的一类函数。

基本的数论函数

• 恒等函数$I(n)=1$
• 元函数$e(n)=[n=1]$
• 单位函数$id(n)=n$
• 莫比乌斯函数

$$\mu (n)=\{\begin{array}{ll}0,& n\mathrm{的}\mathrm{约}\mathrm{数}\mathrm{中}\mathrm{包}\mathrm{含}\mathrm{大}\mathrm{于}1\mathrm{的}\mathrm{完}\mathrm{全}\mathrm{平}\mathrm{方}\mathrm{数}\\ (-1{)}^k,& k\mathrm{为}x\mathrm{含}\mathrm{有}\mathrm{的}\mathrm{质}\mathrm{因}\mathrm{子}\mathrm{种}\mathrm{类}\mathrm{数}\end{array}$$

• 欧拉函数$\phi (n)=\mathrm{小}\mathrm{于}\mathrm{等}\mathrm{于}n\mathrm{且}\mathrm{与}n\mathrm{互}\mathrm{质}\mathrm{的}\mathrm{正}\mathrm{整}\mathrm{数}\mathrm{个}\mathrm{数}$

积性函数

称一个数论函数$f$为积性函数，当且仅当$\mathrm{\forall }a,b\subseteq {\mathbb{N}}^{\ast }\gcd (a,b)=1$满足$f\left(a\right)\cdot f\left(b\right)=f(a\cdot b)$。

若一个数论函数满足$\mathrm{\forall }a,b\subseteq {\mathbb{N}}^{\ast }$有$f\left(a\right)\cdot f\left(b\right)=f(a\cdot b)$则称数论函数$f$为完全积性函数。

狄利克雷卷积

狄利克雷卷积是定义在两个数论函数上的二元运算，$$(f\ast g)\left(n\right)=\sum _{d|n}f(d)\cdot g\left(\frac{n}{d}\right)$$
这样得到的新数论函数就称为$f$和$g$的狄利克雷卷积。

下面是狄利克雷卷积的一些性质：

• 狄利克雷卷积具有交换律和结合律。

• 两个积性函数的狄利克雷卷积还是积性函数

  • proof：
    $$\begin{array}{rl}& \mathrm{\forall }a,b\subseteq {\mathbb{N}}^{\ast },\gcd (a,b)=1\\ & \sum _{d|a}f\left(d\right)\cdot g\left(\frac{a}{d}\right)\times \sum _{d|b}f\left(d\right)\cdot g\left(\frac{b}{d}\right)\\ & =\sum _{d_1|a,d_2|b}f\left(d_1\right)\cdot g\left(\frac{a}{d_1}\right)\times f\left(d_2\right)\cdot g\left(\frac{b}{d_2}\right)\\ & =\sum _{d_1|a,d_2|b}f(d_1\cdot d_2)\cdot g(\frac{a}{d_1}\cdot \frac{b}{d_2})\\ & =\sum _{d|ab}f\left(d\right)\cdot g\left(\frac{ab}{d}\right)\end{array}$$

• 若两个数论函数$f,g$满足$f\ast g=e$则称$f,g$互为对方的逆。

• 任何一个满足$f(1)\ne 0$的数论函数都存在逆

  • 我们已知一个函数$f(n)$,设$g(n)=\frac{e(n)-\sum _{d|n,d\ne 1}f(d)\cdot g\left(\frac{n}{d}\right)}{f(1)}$
    $(f\ast g)(n)=\sum _{d|n}f(d)\cdot g\left(\frac{n}{d}\right)$
    $=\sum _{d|n,d\ne 1}f(d)\cdot g\left(\frac{n}{d}\right)+f(1)\cdot g(n)$
    $=\sum _{d|n,d\ne 1}f(d)\cdot g\left(\frac{n}{d}\right)+f(1)\cdot \frac{(e-\sum _{d|n,d\ne 1}f(d)\cdot g\left(\frac{n}{d}\right))}{f(1)}$
    $=\sum _{d|n,d\ne 1}f(d)\cdot g\left(\frac{n}{d}\right)+e(n)-\sum _{d|n,d\ne 1}f(d)\cdot g\left(\frac{n}{d}\right)$
    $=e(n)$

• 积性函数的逆还是积性函数

  • proof：
    $$\begin{array}{rl}& g(1)\cdot f(1)=e(1)=1\\ & \mathrm{假}\mathrm{设}\mathrm{\forall }n,m\subseteq {\mathbb{N}}^{\ast },\gcd (n,m)=1,n\cdot m\le k\mathrm{有}g(n\cdot m)=g(n)\cdot g(m)\\ & \mathrm{则}\mathrm{当}n\cdot m=k+1\mathrm{时}\\ & g(n)\cdot g(m)=(-\sum _{d|n,d\ne 1}f(d)\cdot g\left(\frac{n}{d}\right))\cdot (-\sum _{d|m,d\ne 1}f(d)\cdot g\left(\frac{m}{d}\right))\\ & =\sum _{d_1|n,d_2|m,d_1\ne 1,d_2\ne 1}f(d_1)\cdot g\left(\frac{n}{d_1}\right)\cdot f(d_2)\cdot g\left(\frac{m}{d_2}\right)\\ & =\sum _{d|n\cdot m,d\ne 1}f(d)\cdot g\left(\frac{n\cdot m}{d}\right)-f(n)\cdot \sum _{d|n,d\ne 1}f(d)\cdot g\left(\frac{n}{d}\right)-f(m)\cdot \sum _{d|m,d\ne 1}f(d)\cdot g\left(\frac{m}{d}\right)\\ & =-g(n\cdot m)+2\cdot g(n)\cdot g(m)\\ & \mathrm{因}\mathrm{此}g(n\cdot m)=g(n)\cdot g(m),\mathrm{由}\mathrm{归}\mathrm{纳}\mathrm{法}\mathrm{可}\mathrm{知}g(n)\mathrm{为}\mathrm{积}\mathrm{性}\mathrm{函}\mathrm{数}\end{array}$$

莫比乌斯函数

对于数论函数$F,f$其中$F$易求，$f$难求。如果满足$F(n)=\sum _{d|n}f(d)$我们可以利用$F$反推出$f$。

注意到，$F=f\ast I\Rightarrow F\ast I^{-1}=f$我们只需要研究$I^{-1}$的性质。事实上这个函数就是莫比乌斯函数$\mu (n)$。

由积性函数的逆还是积性函数可知，$\mu (n)$也是积性函数。其中$\mu (1)\cdot I(1)=e(1)=1\Rightarrow \mu (1)=1$

研究积性函数一般先从质数的情况入手。$\mathrm{\forall }p\subseteq prime,\mu (p)\cdot I(1)+\mu (1)\cdot I(p)=e(p)=0\Rightarrow \mu (p)=-1$

再来研究质数的幂次的情况。$\sum _{d|p^k}\mu (d)\cdot I\left(\frac{p^k}{d}\right)=e(p^k)=0\Rightarrow \mu (1)+\mu (p)+\mu (p^2)+\cdots +\mu (p^k)=0$可以归纳推出$\mu (p^k)=0(k\ge 2)$

这样我们就得到了莫比乌斯函数的完整定义$$\mu (n)=\{\begin{array}{lll}0,& n\mathrm{的}\mathrm{约}\mathrm{数}\mathrm{中}\mathrm{包}\mathrm{含}\mathrm{大}\mathrm{于}1\mathrm{的}\mathrm{完}\mathrm{全}\mathrm{平}\mathrm{方}\mathrm{数}(-1{)}^k,& k\mathrm{为}x\mathrm{含}\mathrm{有}\mathrm{的}\mathrm{质}\mathrm{因}\mathrm{子}\mathrm{种}\mathrm{类}\mathrm{数}\end{array}$$

莫比乌斯反演的几种形式

• 定义形式：$\sum _{d|n}\mu (d)=[n=1]$
  变体：$[n=m]=[m|n]\cdot [\frac{n}{m}=1]=[m|n]\cdot \sum _{d|\frac{n}{m}}\mu (d)$

• 约数形式：如果$F(n)=\sum _{d|n}f(d)$我们可以得到$f(n)=\sum _{d|n}\mu (d)\cdot F\left(\frac{n}{d}\right)$

• 倍数形式：
  如果$F(n)=\sum _{k=1}^{+\mathrm{\infty }}f(k\cdot n)$那么$f(n)=\sum _{k=1}^{+\mathrm{\infty }}\mu (k)\cdot F(k\cdot n)$

  • proof:
    $$\begin{array}{rl}& \sum _{k=1}^{+\mathrm{\infty }}\mu (k)\cdot F(k\cdot n)\\ & =\sum _{k=1}^{+\mathrm{\infty }}\mu (k)\sum _{i=1}^{+\mathrm{\infty }}f(i\cdot k\cdot n)\\ & =\sum _{i=1}^{+\mathrm{\infty }}f(i\cdot n)\sum _{k|i}\mu (k)\\ & =\sum _{i=1}^{+\mathrm{\infty }}f(i\cdot n)\cdot [i=1]\\ & =f(n)\end{array}$$

一些例子

• $\mathrm{求}\sum _{i=1}^n\sum _{j=1}^m[\gcd (i,j)=1]$

  $$\begin{array}{rl}& \sum _{i=1}^n\sum _{j=1}^m[\gcd (i,j)=1]\\ & =\sum _{i=1}^n\sum _{j=1}^m\sum _{d|\gcd (i,j)}\mu (d)\\ & =\sum _{i=1}^n\sum _{j=1}^m\sum _{d|i,d|j}\mu (d)\\ & =\sum _{d=1}^{min\{n,m\}}\sum _{d|i}\sum _{d|j}1\\ & =\sum _{d=1}^{min\{n,m\}}\mu (d)\cdot \lfloor \frac{n}{d}\rfloor \cdot \lfloor \frac{m}{d}\rfloor \\ & \mathrm{接}\mathrm{下}\mathrm{来}\mathrm{就}\mathrm{可}\mathrm{以}\mathrm{直}\mathrm{接}\mathrm{数}\mathrm{论}\mathrm{分}\mathrm{块}\mathrm{了}。\end{array}$$

• 求$\sum _{i=1}^n\sum _{j=1}^m[\gcd (i,j)=prime]n,m\le {10}^7,T\mathrm{组}\mathrm{数}\mathrm{据}T\le {10}^4$

  $$\begin{array}{rl}& \sum _{i=1}^n\sum _{j=1}^m[\gcd (i,j)=prime]\\ & =\sum _{p\subseteq prime}\sum _{i=1}^n\sum _{j=1}^m[\gcd (i,j)=p]\\ & =\sum _{p\subseteq prime}\sum _{i=1}^n\sum _{j=1}^m[p|\gcd (i,j)]\cdot [\frac{\gcd (i,j)}{p}=1]\\ & =\sum _{p\subseteq prime}\sum _{i=1}^{\lfloor \frac{n}{p}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{p}\rfloor }[\frac{\gcd (i\cdot p,j\cdot p)}{p}=1]\\ & =\sum _{p\subseteq prime}\sum _{i=1}^{\lfloor \frac{n}{p}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{p}\rfloor }[\gcd (i,j)=1]\\ & =\sum _{p\subseteq prime}\sum _{d=1}^{min\{\lfloor \frac{n}{p}\rfloor ,\lfloor \frac{m}{p}\rfloor \}}\mu (d)\cdot \lfloor \frac{\lfloor \frac{n}{p}\rfloor }{d}\rfloor \cdot \lfloor \frac{\lfloor \frac{m}{p}\rfloor }{d}\rfloor \\ & \mathrm{小}trick：\lfloor \frac{\lfloor \frac{m}{p}\rfloor }{d}\rfloor =\lfloor \frac{m}{p\cdot d}\rfloor \\ & proof：\mathrm{设}m=k_1\cdot p+r_1(r_1<p),k_1=k_2\cdot d+r_2(r_2<d)\\ & \mathrm{则}m=k_2\cdot d\cdot p+p\cdot r_2+r_1\\ & \mathrm{而}p\cdot r_2+r_1\le p\cdot (d-1)+(p-1)=p\cdot d-1\\ & \mathrm{故}\lfloor \frac{m}{p\cdot d}\rfloor =k_2=\lfloor \frac{\lfloor \frac{m}{p}\rfloor }{d}\rfloor \\ & \mathrm{枚}\mathrm{举}p\cdot d:\\ & =\sum _{k=1}^{min\{n,m\}}\lfloor \frac{n}{k}\rfloor \cdot \lfloor \frac{m}{k}\rfloor \cdot \sum _{p\subseteq prime,p|k}\mu (\frac{k}{p})\\ & \mathrm{然}\mathrm{后}\mathrm{前}\mathrm{面}\mathrm{数}\mathrm{论}\mathrm{分}\mathrm{块}，\mathrm{后}\mathrm{面}\mathrm{可}\mathrm{以}\mathrm{预}\mathrm{处}\mathrm{理}\mathrm{前}\mathrm{缀}\mathrm{和}。\end{array}$$

• $\mathrm{求}f(n)=\sum _{d=1}^n{d}^m\cdot [\gcd (n,d)=1],\mathrm{其}\mathrm{中}n=\prod _{i=1}^{\omega }{p}_i^{{\alpha }_i},\omega \le {10}^3,p_i\subseteq prime,p_i\le {10}^9,{\alpha }_i\le {10}^9,\mathrm{对}{10}^9+7\mathrm{取}\mathrm{模}$

  $$\begin{array}{rl}& \mathrm{令}g(n)=\sum _{i=1}^n{i}^m,\mathrm{那}\mathrm{么}g(n)=\sum _{d|n}f(\frac{n}{d})\cdot d^m\\ & \frac{g(n)}{n^m}=\sum _{d|n}f(\frac{n}{d})\cdot \frac{d^m}{n^m}\\ & \mathrm{根}\mathrm{据}\mathrm{反}\mathrm{演}\mathrm{公}\mathrm{式}:\\ & f(n)\cdot \frac{1}{n^m}=\sum _{d|n}\mu (d)\cdot \frac{g(\frac{n}{d})}{{\left(\frac{n}{d}\right)}^m}\\ & f(n)=\sum _{d|n}\mu (d)\cdot d^m\cdot g(\frac{n}{d})\\ & \mathrm{函}\mathrm{数}g(n)\mathrm{可}\mathrm{以}\mathrm{表}\mathrm{示}\mathrm{成}\mathrm{关}\mathrm{于}n\mathrm{的}m+1\mathrm{次}\mathrm{多}\mathrm{项}\mathrm{式}，\mathrm{设}g(n)=\sum _{i=0}^{m+1}{f}_i\cdot n^i\\ & f(n)=\sum _{d|n}\mu (d)\cdot d^m\cdot \sum _{i=0}^{m+1}{f}_i\cdot {\left(\frac{n}{d}\right)}^i\\ & =\sum _{i=0}^{m+1}{f}_i\cdot n^i\cdot \sum _{d|n}\mu (d)\cdot d^{m-i}\\ & =\sum _{i=0}^{m+1}{f}_i\cdot n^i\cdot \prod _{j=1}^{\omega }\sum _{k=0}^{{\alpha }_j}\mu (p_j^k)\cdot p_j^{(m-i)\cdot k}\\ & =\sum _{i=0}^{m+1}{f}_i\cdot n^i\cdot \prod _{j=1}^{\omega }(1-p_j^{m-i})\\ & m\mathrm{的}\mathrm{范}\mathrm{围}\mathrm{比}\mathrm{较}\mathrm{小}\mathrm{因}\mathrm{此}\mathrm{可}\mathrm{以}\mathrm{直}\mathrm{接}\mathrm{用}\mathrm{拉}\mathrm{格}\mathrm{朗}\mathrm{日}\mathrm{插}\mathrm{值}\mathrm{求}\mathrm{出}{f}_i\end{array}$$

• $\mathrm{求}\prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^C{\left(\frac{\mathrm{lcm}(i,j)}{\gcd (i,k)}\right)}^{\gcd (i,j,k)}A,B,C\le {10}^5,T\mathrm{组}\mathrm{数}\mathrm{据},T\le 70$

  $$\begin{array}{rl}& \prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^C{\left(\frac{\mathrm{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{i\cdot j}{\gcd (i,k)\cdot \gcd (i,j)}\right)}^{\gcd (i,j,k)}\\ & \mathrm{可}\mathrm{以}\mathrm{拆}\mathrm{成}\mathrm{两}\mathrm{个}\mathrm{问}\mathrm{题}\prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^C{i}^{\gcd (i,j,k)}\mathrm{以}\mathrm{及}\prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^C\gcd (i,j{)}^{\gcd (i,j,k)}\\ & \prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^C{i}^{\gcd (i,j,k)}=\prod _{i=1}^A{i}^{\sum _{j=1}^B\sum _{k=1}^C\gcd (i,j,k)}\\ & =\prod _{i=1}^A{i}^{\sum _{d=1}^{min\{i,B,C\}}d\sum _{j=1}^{\lfloor \frac{B}{d}\rfloor }\sum _{k=1}^{\lfloor \frac{C}{d}\rfloor }[\gcd (\frac{i}{d},j,k)=1]}\\ & =\prod _{i=1}^A{i}^{\sum _{d=1}^{min\{i,B,C\}}d\sum _{j=1}^{\lfloor \frac{B}{d}\rfloor }\sum _{k=1}^{\lfloor \frac{C}{d}\rfloor }\sum _{t|\frac{i}{d},t|j,t|k}\mu (t)}\\ & =\prod _{i=1}^A{i}^{\sum _{s|i}\sum _{t|s}\mu (t)\cdot \lfloor \frac{B}{s}\rfloor \cdot \lfloor \frac{C}{s}\rfloor \cdot \frac{s}{t}}\\ & \mathrm{这}\mathrm{边}\mathrm{有}\mathrm{个}\mathrm{小}trick,\mu \ast id=\phi \\ & proof:\mathrm{\forall }d|n,\mathrm{满}\mathrm{足}1\le a\le n\mathrm{且}\gcd (a,n)=d\iff \gcd (\frac{a}{d},\frac{n}{d})\mathrm{的}a\mathrm{一}\mathrm{共}\mathrm{有}\phi \left(\frac{n}{d}\right)\mathrm{个}\\ & \mathrm{故}\sum _{d|n}\phi \left(\frac{n}{d}\right)=\sum _{d|n}\phi (d)=n\\ & \mathrm{原}\mathrm{式}=\prod _{i=1}^A{i}^{\sum _{s|i}\phi (s)\cdot \lfloor \frac{B}{s}\rfloor \cdot \lfloor \frac{C}{s}\rfloor }\\ & =\prod _{s=1}^A\prod _{i=1}^{\lfloor \frac{A}{s}\rfloor }{(i\cdot s)}^{\phi (s)\cdot \lfloor \frac{B}{s}\rfloor \cdot \lfloor \frac{C}{s}\rfloor }\\ & =\prod _{s=1}^A{[\left(\lfloor \frac{A}{s}\rfloor \right)!\cdot s^{\lfloor \frac{A}{s}\rfloor }]}^{\phi (s)\cdot \lfloor \frac{B}{s}\rfloor \cdot \lfloor \frac{C}{s}\rfloor }\\ & \mathrm{接}\mathrm{下}\mathrm{来}\mathrm{预}\mathrm{处}\mathrm{理}+\mathrm{数}\mathrm{论}\mathrm{分}\mathrm{块}\mathrm{就}\mathrm{可}\mathrm{以}\mathrm{了}\\ & \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}^{\lfloor \frac{A}{d}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{d}\rfloor }\prod _{k=1}^{\lfloor \frac{C}{d}\rfloor }[d\cdot \gcd (i,j){]}^{d\cdot [\gcd (i,j,k)=1]}\\ & =\prod _{d=1}^{min\{A,B,C\}}\prod _{i=1}^{\lfloor \frac{A}{d}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{d}\rfloor }[d\cdot \gcd (i,j){]}^{d\cdot \sum _{k=1}^{\lfloor \frac{C}{d}\rfloor }\sum _{t|i,t|j,t|k}\mu (t)}\\ & =\prod _{d=1}^{min\{A,B,C\}}\prod _{t=1}^{min\{\lfloor \frac{A}{d}\rfloor ,\lfloor \frac{B}{d}\rfloor ,\lfloor \frac{C}{d}\rfloor \}}\prod _{i=1}^{\lfloor \frac{A}{t\cdot d}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{t\cdot d}\rfloor }[d\cdot t\cdot \gcd (i,j){]}^{d\cdot \mu (t)\cdot \lfloor \frac{C}{t\cdot d}\rfloor }\\ & \mathrm{这}\mathrm{个}\mathrm{式}\mathrm{子}\mathrm{同}\mathrm{样}\mathrm{可}\mathrm{以}\mathrm{拆}\mathrm{成}\mathrm{两}\mathrm{部}\mathrm{分}\mathrm{来}\mathrm{计}\mathrm{算}，\prod _{d=1}^{min\{A,B,C\}}\prod _{t=1}^{min\{\lfloor \frac{A}{d}\rfloor ,\lfloor \frac{B}{d}\rfloor ,\lfloor \frac{C}{d}\rfloor \}}\prod _{i=1}^{\lfloor \frac{A}{t\cdot d}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{t\cdot d}\rfloor }[d\cdot t{]}^{d\cdot \mu (t)\cdot \lfloor \frac{C}{t\cdot d}\rfloor }\\ & \mathrm{以}\mathrm{及}\prod _{d=1}^{min\{A,B,C\}}\prod _{t=1}^{min\{\lfloor \frac{A}{d}\rfloor ,\lfloor \frac{B}{d}\rfloor ,\lfloor \frac{C}{d}\rfloor \}}\prod _{i=1}^{\lfloor \frac{A}{t\cdot d}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{t\cdot d}\rfloor }\gcd (i,j{)}^{d\cdot \mu (t)\cdot \lfloor \frac{C}{t\cdot d}\rfloor }\\ & \prod _{d=1}^{min\{A,B,C\}}\prod _{t=1}^{min\{\lfloor \frac{A}{d}\rfloor ,\lfloor \frac{B}{d}\rfloor ,\lfloor \frac{C}{d}\rfloor \}}\prod _{i=1}^{\lfloor \frac{A}{t\cdot d}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{t\cdot d}\rfloor }[d\cdot t{]}^{d\cdot \mu (t)\cdot \lfloor \frac{C}{t\cdot d}\rfloor }\\ & =\prod _{d=1}^{min\{A,B,C\}}\prod _{t=1}^{min\{\lfloor \frac{A}{d}\rfloor ,\lfloor \frac{B}{d}\rfloor ,\lfloor \frac{C}{d}\rfloor \}}[d\cdot t{]}^{d\cdot \mu (t)\cdot \lfloor \frac{A}{d\cdot t}\rfloor \cdot \lfloor \frac{B}{d\cdot t}\rfloor \cdot \lfloor \frac{C}{d\cdot t}\rfloor }\\ & =\prod _{s=1}^{min\{A,B,C\}}{s}^{\lfloor \frac{A}{s}\rfloor \cdot \lfloor \frac{B}{s}\rfloor \cdot \lfloor \frac{C}{s}\rfloor \cdot \sum _{t|s}\mu (t)\cdot \frac{s}{t}}\\ & =\prod _{s=1}^{min\{A,B,C\}}{s}^{\lfloor \frac{A}{s}\rfloor \cdot \lfloor \frac{B}{s}\rfloor \cdot \lfloor \frac{C}{s}\rfloor \cdot \phi (s)}\\ & \mathrm{第}\mathrm{一}\mathrm{个}\mathrm{式}\mathrm{子}\mathrm{这}\mathrm{样}\mathrm{已}\mathrm{经}\mathrm{可}\mathrm{以}\mathrm{数}\mathrm{论}\mathrm{分}\mathrm{块}\mathrm{做}\mathrm{了}。\\ & \prod _{d=1}^{min\{A,B,C\}}\prod _{t=1}^{min\{\lfloor \frac{A}{d}\rfloor ,\lfloor \frac{B}{d}\rfloor ,\lfloor \frac{C}{d}\rfloor \}}\prod _{i=1}^{\lfloor \frac{A}{t\cdot d}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{t\cdot d}\rfloor }\gcd (i,j{)}^{d\cdot \mu (t)\cdot \lfloor \frac{C}{t\cdot d}\rfloor }\\ & =\prod _{s=1}^{min\{A,B,C\}}\prod _{i=1}^{\lfloor \frac{A}{s}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{s}\rfloor }\gcd (i,j{)}^{\lfloor \frac{C}{s}\rfloor \cdot \phi (s)}\\ & =\prod _{s=1}^{min\{A,B,C\}}{\left[\prod _{d=1}^{min\{\lfloor \frac{A}{s}\rfloor ,\lfloor \frac{B}{s}\rfloor \}}{d}^{\sum _{i=1}^{\lfloor \frac{A}{s\cdot d}\rfloor }\sum _{j=1}^{\lfloor \frac{B}{s\cdot d}\rfloor }[\gcd (i,j)=1]}\right]}^{\phi (s)\cdot \lfloor \frac{C}{s}\rfloor }\\ & =\prod _{s=1}^{min\{A,B,C\}}{\left[\prod _{d=1}^{min\{\lfloor \frac{A}{s}\rfloor ,\lfloor \frac{B}{s}\rfloor \}}{d}^{\sum _{t=1}^{min\{\lfloor \frac{A}{s\cdot d}\rfloor ,\lfloor \frac{B}{s\cdot d}\rfloor \}}\mu (t)\cdot \lfloor \frac{A}{s\cdot d\cdot t}\rfloor \cdot \lfloor \frac{B}{s\cdot d\cdot t}\rfloor }\right]}^{\phi (s)\cdot \lfloor \frac{C}{s}\rfloor }\\ & =\prod _{s=1}^{min\{A,B,C\}}{\left[\prod _{m=1}^{min\lfloor \frac{A}{s}\rfloor ,\lfloor \frac{B}{s}\rfloor }{\left(\prod _{d|m}{d}^{\mu (\frac{m}{d})}\right)}^{\lfloor \frac{A}{s\cdot m}\rfloor \cdot \lfloor \frac{B}{s\cdot m}\rfloor }\right]}^{\phi (s)\cdot \lfloor \frac{C}{s}\rfloor }\\ & \mathrm{然}\mathrm{后}\mathrm{我}\mathrm{们}\mathrm{预}\mathrm{处}\mathrm{理}\mathrm{出}\mathrm{中}\mathrm{间}\mathrm{的}\prod _{d|m}{d}^{\mu (\frac{m}{d})}\mathrm{然}\mathrm{后}\mathrm{就}\mathrm{可}\mathrm{以}\mathrm{数}\mathrm{论}\mathrm{分}\mathrm{块}\mathrm{套}\mathrm{数}\mathrm{论}\mathrm{分}\mathrm{块}\mathrm{求}\mathrm{解}\mathrm{上}\mathrm{面}\mathrm{的}\mathrm{式}\mathrm{子}\mathrm{了}。\end{array}$$