初探莫比乌斯反演

莫比乌斯反演的题目大多可以使用莫比乌斯函数性质求出

莫比乌斯函数性质就是

\([n=1]=\displaystyle \sum _{i|n}\mu(i)\)

比如说你要求

\(\displaystyle \sum _{i=1}^{n}\displaystyle \sum _{j=1}^{m}gcd(i,j)\)
其中\(n<m\)

你可以先枚举gcd

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

考虑将\(i=i*d,j=j*d\)这里说的不详细可以看下面

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

由莫比乌斯函数性质，原式可化为

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

更换枚举项顺序

\(\displaystyle \sum _{d=1}^{n}d\displaystyle \sum _{t=1}^{\lfloor \frac{n}{d}\rfloor}\mu(t) \displaystyle \sum _{i=1}^{\lfloor \frac{n}{d}\rfloor}\displaystyle \sum _{j=1}^{\lfloor \frac{m}{d}\rfloor}[t|i]*[t|j]\)

看后面那个鬼

我们设\(i=s*t,j=r*t\)，此时因为\(i\leq n\) ，\(s*t\leq n\)，\(s\leq \frac {n}{t}\)

此处因为\(t\nmid i\) 贡献为零，我们只考虑整除情况

\(\displaystyle \sum _{d=1}^{n}d\displaystyle \sum _{t=1}^{\lfloor \frac{n}{d}\rfloor}\mu(t)\displaystyle \sum _{s=1}^{\lfloor \frac{n}{td}\rfloor}\displaystyle \sum _{r=1}^{\lfloor \frac{m}{td}\rfloor}1\)

后面\(\displaystyle \sum _{s=1}^{\lfloor \frac{n}{td}\rfloor}\displaystyle \sum _{r=1}^{\lfloor \frac{m}{td}\rfloor}1=\displaystyle \sum _{s=1}^{\lfloor \frac{n}{td}\rfloor}1\displaystyle \sum _{r=1}^{\lfloor \frac{m}{td}\rfloor}1=\displaystyle \sum _{s=1}^{\lfloor \frac{n}{td}\rfloor}1*\displaystyle \sum _{r=1}^{\lfloor \frac{m}{td}\rfloor}1=\lfloor\frac{n}{td}\rfloor\lfloor \frac{m}{td}\rfloor\)

原式等于

\(\displaystyle \sum _{d=1}^{n}d\displaystyle \sum _{t=1}^{\lfloor \frac{n}{d}\rfloor}\mu(t)\lfloor\frac{n}{td}\rfloor\) \(\lfloor \frac{m}{td}\rfloor\)

看起来不够简洁，莫反题有一种套路

就是当\(\displaystyle \sum _{d=1}^{n}f(d)\displaystyle \sum _{t=1}^{\lfloor \frac{n}{d}\rfloor}h(t)\)时\(f,h\)为两个关于\(d,t\)的函数

我们可以设 \(g=d\times t\)

原式为\(\displaystyle \sum _{g=1}^{n}\sum _{d\mid g} f(d)*h(\frac{g}{d})\)\(=\)\(\displaystyle \sum _{g=1}^{n}\sum _{t\mid g} h(t)*f(\frac{g}{t})\)

\(\displaystyle \sum _{g=1}^{n} \lfloor\frac{n}{g}\rfloor\lfloor \frac{m}{g}\rfloor\sum _{d\mid g} d*\mu(\frac{g}{d})\)

后面那个相当于\(id*\mu\) 其中\(*\)是迪利克雷卷积

迪利克雷卷积常见结论\(id*\mu=\varphi\)

原式为\(\displaystyle \sum _{g=1}^{n} \lfloor\frac{n}{g}\rfloor\lfloor \frac{m}{g}\rfloor\varphi(g)\)

处理完\(\varphi\)前缀和后，剩下的就是整除分块了