莫比乌斯反演题目式子推导

约定：若有 $n,m$ 两参数，默认 $n\le m$

YY的GCD

$$\sum _{i=1}^n\sum _{j=1}^m[\gcd (i,j)=\text{prime}]$$

$$=\sum _{p\in \text{prime}}\sum _{i=1}^{\lfloor \frac{n}{p}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{p}\rfloor }[\gcd (i,j)=1]$$

$$=\sum _{p\in \text{prime}}\sum _{i=1}^{\lfloor \frac{n}{p}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{p}\rfloor }\sum _{d|\gcd (i,j)}\mu (d)$$

$$=\sum _{p\in \text{prime}}\sum _{i=1}^{min(\lfloor \frac{n}{p}\rfloor ,\lfloor \frac{n}{p}\rfloor )}\mu (i)\lfloor \frac{n}{pi}\rfloor \lfloor \frac{m}{pi}\rfloor$$

$$=\sum _{i=1}^{min(n,m)}\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{i}\rfloor \sum _{p\in \text{prime},p|i}\mu (\frac{i}{p})$$

预处理 ${\sum }_{p\in \text{prime},p|i}\mu (\frac{i}{p})$ 即可

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能量采集

$$\sum _{i=1}^n\sum _{j=1}^{m}2\gcd (i,j)-1$$

$$=\left(2\sum _{i=1}^n\sum _{j=1}^m\gcd (i,j)\right)-nm$$

$$=\left(2\sum _{i=1}^n\sum _{x=1}^{\lfloor \frac{n}{i}\rfloor }\sum _{y=1}^{\lfloor \frac{m}{i}\rfloor }i[\gcd (x,y)=1]\right)-nm$$

$$=\left(2\sum _{i=1}^n\sum _{x=1}^{\lfloor \frac{n}{i}\rfloor }\sum _{y=1}^{\lfloor \frac{m}{i}\rfloor }i\sum _{d|\gcd (x,y)}\mu (d)\right)-nm$$

$$=\left(2\sum _{i=1}^n\sum _{j=1}^{\lfloor \frac{n}{i}\rfloor }\mu (j)\lfloor \frac{n}{ij}\rfloor \lfloor \frac{m}{ij}\rfloor i\right)-nm$$

$$=\left(2\sum _{i=1}^n\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{i}\rfloor \sum _{d|i}d\mu (\frac{i}{d})\right)-nm$$

可以预处理 ${\sum }_{d|i}d\mu (\frac{i}{d})$ 的前缀和

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[SDOI2014]数表

$$\sum _{i=1}^n\sum _{j=1}^m\sum _{d|\gcd (i,j)}d[\sum _{d|\gcd (i,j)}d\le a]$$

$$=\sum _{i=1}^n{\sigma }_1(i)[{\sigma }_1(i)\le a]\sum _{x=1}^{\lfloor \frac{n}{i}\rfloor }\sum _{y=1}^{\lfloor \frac{m}{i}\rfloor }[gcd(x,y)=1]$$

$$=\sum _{i=1}^n{\sigma }_1(i)[{\sigma }_1(i)\le a]\sum _{x=1}^{\lfloor \frac{n}{i}\rfloor }\sum _{y=1}^{\lfloor \frac{m}{i}\rfloor }\sum _{d|gcd(x,y)}\mu (d)$$

$$=\sum _{i=1}^n{\sigma }_1(i)[{\sigma }_1(i)\le a]\sum _{j=1}^{\lfloor \frac{n}{i}\rfloor }\mu (j)\lfloor \frac{n}{ij}\rfloor \lfloor \frac{m}{ij}\rfloor$$

$$=\sum _{i=1}^n\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{i}\rfloor \sum _{d|i}\mu (\frac{i}{d}){\sigma }_1(d)[{\sigma }_1(d)\le a]$$

离线预处理 ${\sum }_{d|i}\mu (\frac{i}{d}){\sigma }_1(d)[{\sigma }_1(d)\le a]$ 即可

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[SDOI2017]数字表格

$$\prod _{i=1}^n\prod _{j=1}^{m}f[\gcd (i,j)]$$

$$=\prod _{i=1}^{n}f[i{]}^{{\sum }_{x=1}^{\lfloor \frac{n}{i}\rfloor }{\sum }_{y=1}^{\lfloor \frac{m}{i}\rfloor }[gcd(i,j)=1]}$$

$$=\prod _{i=1}^{n}f[i{]}^{{\sum }_{j=1}^{\lfloor \frac{n}{i}\rfloor }\mu (j)\lfloor \frac{n}{ij}\rfloor \lfloor \frac{m}{ij}\rfloor }$$

$$=\prod _{i=1}^n\prod _{j=1}^{\lfloor \frac{n}{i}\rfloor }f[i{]}^{\mu (j)\lfloor \frac{n}{ij}\rfloor \lfloor \frac{m}{ij}\rfloor }$$

$$=\prod _{i=1}^n(\prod _{d|i}f[d{]}^{\mu (\frac{i}{d})}{)}^{\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{i}\rfloor }$$

预处理 ${\prod }_{d|i}f[d{]}^{\mu (\frac{i}{d})}$ 的前缀积即可

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[POI2007]ZAP-Queries

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

$$=\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }[\gcd (i,j)=1]$$

$$=\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }\sum _{k|\gcd (i,j)}\mu (k)$$

$$=\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\mu (i)\lfloor \frac{n}{di}\rfloor \lfloor \frac{m}{di}\rfloor$$

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[HAOI2011]Problem b

$$\sum _{i=1}^b\sum _{j=1}^d[\gcd (i,j)=k]-\sum _{i=1}^a\sum _{j=1}^d[\gcd (i,j)=k]-\sum _{i=1}^b\sum _{j=1}^c[\gcd (i,j)=k]+\sum _{i=1}^a\sum _{j=1}^c[\gcd (i,j)=k]$$

同上即可

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[SDOI2015]约数个数和

$$\sum _{i=1}^n\sum _{j=1}^m{\sigma }_0(ij)$$

$$=\sum _{i=1}^n\sum _{j=1}^m\sum _{x|i}\sum _{y|j}[\gcd (x,y)=1]$$

$$=\sum _{x=1}^n\sum _{y=1}^m\sum _{i=1}^{\lfloor \frac{n}{x}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{y}\rfloor }[\gcd (x,y)=1]$$

$$=\sum _{x=1}^n\sum _{y=1}^m\lfloor \frac{n}{x}\rfloor \lfloor \frac{m}{y}\rfloor [\gcd (x,y)=1]$$

$$=\sum _{x=1}^n\sum _{y=1}^m\lfloor \frac{n}{x}\rfloor \lfloor \frac{m}{y}\rfloor \sum _{d|gcd(x,y)}\mu (d)$$

$$=\sum _{d=1}^n\mu (d)\sum _{x=1}^{\lfloor \frac{n}{d}\rfloor }\lfloor \frac{n}{xd}\rfloor \sum _{y=1}^{\lfloor \frac{m}{d}\rfloor }\lfloor \frac{m}{yd}\rfloor$$

$$=\sum _{d=1}^n\mu (d)divs(\lfloor \frac{n}{d}\rfloor )divs(\lfloor \frac{m}{d}\rfloor )$$

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[CQOI2015]选数

$$\sum _{x_1=l}^h\sum _{x_2=l}^h\cdots \sum _{x_n=l}^h[\gcd (x_1,x_2,\cdots ,x_n)=k]$$

$$=\sum _{x_1=\lceil \frac{l}{k}\rceil }^{\lfloor \frac{h}{k}\rfloor }\sum _{x_2=\lceil \frac{l}{k}\rceil }^{\lfloor \frac{h}{k}\rfloor }\cdots \sum _{x_n=\lceil \frac{l}{k}\rceil }^{\lfloor \frac{h}{k}\rfloor }\sum _{d|\gcd (x_1,x_2,\cdots ,x_n)}\mu (d)$$

$$=\sum _{d=\lceil \frac{l}{k}\rceil }^{\lfloor \frac{h}{k}\rfloor }\mu (d)(\lfloor \frac{h}{kd}\rfloor -\lceil \frac{\lceil \frac{l}{k}\rceil }{d}\rceil {)}^n$$

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常见积性函数与迪利克雷卷积（用于杜教筛）

$$ϵ(i)=[i=1]$$

$$\text{I}(i)=1$$

$$\text{id}(i)=i$$

$$\mu \ast \text{I}=ϵ$$

$$\mu \ast \text{id}=\phi$$

$$\phi \ast \text{I}=\text{id}$$

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简单的数学题

$$\sum _{i=1}^n\sum _{j=1}^{n}ij\gcd (i,j)$$

$$=\sum _{i=1}^n\sum _{x=1}^{\lfloor \frac{n}{i}\rfloor }\sum _{y=1}^{\lfloor \frac{n}{i}\rfloor }xy{i}^3[\gcd (x,y)==1]$$

$$=\sum _{i=1}^n\sum _{x=1}^{\lfloor \frac{n}{i}\rfloor }\sum _{y=1}^{\lfloor \frac{n}{i}\rfloor }xy{i}^3\sum _{d|\gcd (x,y)}\mu (d)$$

$$=\sum _{i=1}^n{i}^3\sum _{j=1}^{\lfloor \frac{n}{i}\rfloor }{j}^2\mu (j)\sum _{x=1}^{\lfloor \frac{n}{ij}\rfloor }x\sum _{y=1}^{\lfloor \frac{n}{ij}\rfloor }y$$

$$=\sum _{i=1}^n{i}^3\sum _{j=1}^{\lfloor \frac{n}{i}\rfloor }{j}^2\mu (j)\frac{(\lfloor \frac{n}{ij}\rfloor {)}^2(\lfloor \frac{n}{ij}\rfloor +1{)}^2}{4}$$

$$=\sum _{i=1}^n{i}^2\frac{(\lfloor \frac{n}{i}\rfloor {)}^2(\lfloor \frac{n}{i}\rfloor +1{)}^2}{4}\sum _{d|i}d\mu (\frac{i}{d})$$

$$=\sum _{i=1}^n{i}^2\phi (i)\frac{(\lfloor \frac{n}{i}\rfloor {)}^2(\lfloor \frac{n}{i}\rfloor +1{)}^2}{4}$$

杜教筛 $i^2\phi (i)$ 前缀和，整除分块

如何杜教筛 $i^2\phi (i)$：

$$f(i)=i^2\phi (i)$$

$$g(i)=\text{id}(i^2)$$

$$(f\ast g)(i)=\sum _{d|i}(d^2\phi (d))(\frac{i}{d}{)}^2=i^2\sum _{d|i}\phi (d)=i^3$$