莫比乌斯反演常见的三个模型

莫比乌斯反演常见模型

模型1:\(\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=t]\)

\[\begin{aligned} \sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=t]&=\sum_{i=1}^{\lfloor\frac{n}{t}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{t}\rfloor}[gcd(i,j)=1]\\ &=\sum_{i=1}^{\lfloor\frac{n}{t}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{t}\rfloor}\sum_{d|gcd(i,j)}\mu(d)\\ &=\sum_{d=1}^{min(n,m)}\sum_{d|i}^{\lfloor\frac{n}{t}\rfloor}\sum_{d|j}^{\lfloor\frac{m}{t}\rfloor}\mu(d)\\ &=\sum_{d=1}^{min(n,m)}\lfloor\frac{n}{td}\rfloor\lfloor\frac{m}{td}\rfloor\mu(d) \end{aligned} \]

模型2:\(\sum_{i=1}^n\sum_{j=1}^mij[gcd(i,j)=t]\)

\[\begin{aligned} \sum_{i=1}^n\sum_{j=1}^mij[gcd(i,j)=t]&= \sum_{d=1}^{min(n,m)}\mu(d)\sum_{d|i}^{\lfloor\frac{n}{t}\rfloor}\sum_{d|j}^{\lfloor\frac{m}{t}\rfloor}ij\\ &=\frac{\sum_{d=1}^{min(n,m)}\lfloor\frac{n}{td}\rfloor\lfloor\frac{m}{td}\rfloor(\lfloor\frac{n}{td}\rfloor+1)(\lfloor\frac{m}{td}\rfloor+1)\mu(d)d^2}{4} \end{aligned} \]

至于推导的话就是等差数列求和而已

模型3:\(\sum_{i=1}^n\sum_{j=1}^mf[gcd(i,j)=t]\)

这个的推导就总结向上面一样照葫芦画瓢推就行了

答案如下:

\[\sum_{u=1}^{min(n,m)}\lfloor\frac{n}{u}\rfloor\lfloor\frac{m}{u}\rfloor\sum_{d|u}f(d)\mu(\frac{u}{d}) \]

注意所有的求和都要用到整除分块

posted @ 2023-05-11 16:38  6penny  阅读(39)  评论(0编辑  收藏  举报