莫比乌斯反演专题题解

「SDOI2015」约数个数和

这个题关键就在于知不知道这个 $d_{ij}={\sum }_{x|i}{\sum }_{y|j}(gcd(x,y)==1)$ 结论 以及 $gcd==1$ 怎么用莫比乌斯函数表示的 然后推式子就很愉快了
$$\begin{gathered} \begin{array}{rl}ans& =\sum _{i=1}^n\sum _{j=1}^m{d}_{ij}\\ & =\sum _{i=1}^n\sum _{j=1}^m\sum _{x|i}\sum _{y|j}(gcd(x,y)==1)\\ & =\sum _{i=1}^n\sum _{j=1}^m\sum _{x|i}\sum _{y|j}\sum _{d|gcd(x,y)}{u}_d\\ & =\sum _{d=1}^{min(n,m)}{u}_d\sum _{d|x}\sum _{d|y}\sum _{i|x}\sum _{j|y}\\ & =\sum _{d=1}^{min(n,m)}{u}_d\sum _{d|x}\sum _{d|y}\lfloor \frac{n}{x}\rfloor \lfloor \frac{m}{y}\rfloor \end{array} \\ g(k)=\sum _{i=1}^k\lfloor \frac{k}{i}\rfloor \\ ans=\sum _{d=1}^{min(n,m)}{u}_d\times g(\lfloor \frac{n}{d}\rfloor )\times g(\lfloor \frac{m}{d}\rfloor ) \end{gathered}$$
$g(x)$ 函数用整出分块预处理 然后就是套路了

「SDOI2017」数字表格

$$\begin{array}{rl}ans& =\prod _{i=1}^n\prod _{j=1}^{m}fib[gcd(i,j)]\\ & =\prod _{t=1}^{n}fib[t{]}^{{\sum }_{i=1}^n{\sum }_{j=1}^m[gcd(i,j)==1]}\\ & =\prod _{t=1}^{n}fib[t{]}^{{\sum }_{t|d}^{n}u(\frac{d}{t})\lfloor \frac{n}{d}\rfloor \lfloor \frac{m}{d}\rfloor }\\ & =\prod _{t=1}^n\prod _{t|d}fib[t{]}^{u(\frac{d}{t})\lfloor \frac{n}{d}\rfloor \lfloor \frac{m}{d}\rfloor }\\ & =\prod _{d=1}^n\prod _{t|d}fib[t{]}^{u(\frac{d}{t})\lfloor \frac{n}{d}\rfloor \lfloor \frac{m}{d}\rfloor }\\ & =\prod _{d=1}^n{[\prod _{t|d}fib[t{]}^{u(\frac{d}{t})}]}^{\lfloor \frac{n}{d}\rfloor \lfloor \frac{m}{d}\rfloor }\end{array}$$

这道题的关键是是第三步到第四步 推式子的过程中交换枚举顺序很重要

然后反演里面整除分块的式子常用但不限于四则运算

时间 $O(T\sqrt{n}+n\log n)$

嗷嗷对了 还有对函数 $g(d)={\prod }_{t|d}fib[t{]}^{u(\frac{d}{t})}$ 的预处理可以用埃式筛来做

「BZOJ2693」jzptab

$$\begin{array}{rl}ans& =\sum _{i=1}^n\sum _{j=1}^{m}lcm(i,j)\\ & =\sum _{i=1}^n\sum _{j=1}^m\frac{i\times j}{gcd(i,j)}\\ & =\sum _{t=1}^{n}t\sum _{i=1}^{it\le n}\sum _{j=1}^{jt\le m}i\times j\times (gcd(i,j)==1)\\ & =\sum _{t=1}^{n}t\sum _{i=1}^{it\le n}\sum _{j=1}^{jt\le m}i\times j\times \sum _{d|id|j}{u}_d\\ & =\sum _{d=1}^n{u}_d\times d^2\sum _{td\le n}t\sum _{idt\le n}\sum _{jdt\le m}i\times j\end{array}$$

令 $x=\lfloor \frac{n}{d}\rfloor ,y=\lfloor \frac{m}{d}\rfloor ,sum(a,b)={\sum }_{i=1}^a{\sum }_{j=1}^{b}i\times j$
$$\begin{array}{rl}ans& =\sum _{d=1}^n{u}_d\times d^2\sum _{t=1}^{x}t\sum _{it\le x}\sum _{jt\le y}i\times j\\ & =\sum _{d=1}^n{u}_d\times d^2\sum _{t=1}^{x}t\times sum(\lfloor \frac{x}{t}\rfloor ,\lfloor \frac{y}{t}\rfloor )\\ & =\sum _{d=1}^n\sum _{t=1}^{min(x,y)}{u}_d\times d^2\times t\times sum(\lfloor \frac{n}{dt}\rfloor ,\lfloor \frac{m}{dt}\rfloor )①\\ & =\sum _{t=1}^n\sum _{dt\le n}{u}_d\times d^2\times t\times sum(\lfloor \frac{n}{dt}\rfloor ,\lfloor \frac{m}{dt}\rfloor )\\ & =\sum _{dt\le n}sum(\lfloor \frac{n}{dt}\rfloor ,\lfloor \frac{m}{dt}\rfloor )\sum _{t|dt}{u}_d\times d^2\times t②\end{array}$$
最后化成可以整除分块的式子的时候 如果有函数不能直接求或者预处理 要把该函数提到最前面 如这道题 ① -> ②这个过程

数字表格那道题同样也用了这个技巧

「SDOI2014」数表

$$\begin{array}{rl}ans& =\sum _{i=1}^n\sum _{j=1}^{m}g(gcd(i,j))\\ & =\sum _{t=1}^{n}g(t)\sum _{i=1}^{it\le n}\sum _{j=1}^{jt\le m}(gcd(i,j)==1)\\ & =\sum _{t=1}^{n}g(t)\sum _{i=1}^{\lfloor \frac{n}{t}\rfloor }{u}_i\lfloor \frac{n}{it}\rfloor \lfloor \frac{m}{it}\rfloor \\ & =\sum _{t=1}^n\sum _{i=1}^{\lfloor \frac{n}{t}\rfloor }g(t)u_i\lfloor \frac{n}{it}\rfloor \lfloor \frac{m}{it}\rfloor \\ & =\sum _{it=1}^n\sum _{t|it}g(t)u_i\lfloor \frac{n}{it}\rfloor \lfloor \frac{m}{it}\rfloor \\ & =\sum _{it=1}^n\lfloor \frac{n}{it}\rfloor \lfloor \frac{m}{it}\rfloor \sum _{t|it}g(t)u_{\frac{it}{t}}\end{array}$$

离线有点麻烦 常数有点卡（指卡我的常数