CF235E

莫比乌斯反演题

约定:

\((x,y) = [\gcd(x,y)=1]\)

\[\sum_{i=1}^A\sum_{j=1}^B\sum_{k=1}^C d(ijk) \\ = \sum_{i=1}^A \sum_{j=1}^B \sum_{k=1}^C \sum_{x|i}\sum_{y|j}\sum_{z|k}(x,y)(y,z)(z,x) \\ = \sum_{x=1}^A\sum_{y=1}^B\sum_{z=1}^C(x,y)(y,z)(z,x)\sum_{x|i}\sum_{y|j}\sum_{z|k}1 \\ = \sum_{x=1}^A\sum_{y=z}^B\sum_{z=1}^C(x,y)(y,z)(z,x)\left \lfloor \frac{A}{x} \right \rfloor \left \lfloor \frac{B}{y} \right \rfloor \left \lfloor \frac{C}{y} \right \rfloor \]

注意到\(A,B,C \le 2000\), 那我们枚举x，然后将后面的式子化简得:

\[\sum_{y=1}^B \sum_{z=1}^C(x,y)(x,z)(y,z)\left \lfloor \frac{A}{x} \right \rfloor \left \lfloor \frac{B}{y} \right \rfloor \left \lfloor \frac{C}{y} \right \rfloor \]

因为可以将$\left \lfloor \frac{A}{x} \right \rfloor \(提到x后，而\)(y,z)$是需要计算的，那么我们将他反演

\[\sum_{y=1}^B\left \lfloor \frac{B}{y} \right \rfloor \sum_{z=1}^C \left \lfloor \frac{C}{y} \right \rfloor (x,y)(x,z)\sum_{d|\gcd(y,z)} \mu(d)\\ =\sum_{d=1}^{\mathrm{min} (B,C)} \mu(d) \sum_{y=1}^{\left \lfloor B/d \right \rfloor}\left \lfloor \frac{B}{dy} \right \rfloor \sum_{z=1}^{\left \lfloor C/d \right \rfloor}\left \lfloor \frac{C}{dz} \right \rfloor(x,y)(x,z)(x,d)^2\\ = \sum_{d=1}^{\mathrm{min} (b,c)} \mu(d) \sum_{y=1}^{\left \lfloor B/d \right \rfloor}\left \lfloor \frac{B}{dy} \right \rfloor \sum_{z=1}^{\left \lfloor C/d \right \rfloor}\left \lfloor \frac{C}{dz} \right \rfloor(x,y)(x,z)(x,d) \]

将\((x,d)\)提到前面d的求和号后，注意到\(y,z\)的式子形式几乎相同，那么设:

\[g(x,y) = \sum_{i=1}^y(x,i)\left \lfloor \frac{y}{i} \right \rfloor \]

那么:

\[原式=\sum_{x=1}^A\left \lfloor \frac{x}{i} \right \rfloor \sum_{d=1}^\mathrm{min(B,C)}\mu(d)g(x,\left \lfloor \frac{B}{d} \right \rfloor) g(x,\left \lfloor \frac{C}{d} \right \rfloor) \]

注意到\(g(x,y)\)的差分数组只需\(O(n^2\log n)\)的时间获得，那么枚举\(x,d\)即可

时间复杂度的瓶颈在于求\(g(x,y)\)和枚举\(x,d\),时间为\(O(n^2\log n)\),可以通过本题

EOF