CF235E Number Challenge

CF235E Number Challenge

题目简述

\[\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^cd(ijk) \]

\(a,b,c\leq 2\times 10^3\).

推导过程

首先需要证明

\[d(xyz)=\sum_{i|x}\sum_{j|y}\sum_{k|z}\bigl[\gcd(i,j)=1\bigr]\cdot\bigl[\gcd(j,k)=1\bigr]\cdot\bigl[\gcd(k,i)=1\bigr] \]

设

\[x=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_n^{\alpha_n} \]

\[y=p_1^{\beta_1}p_2^{\beta_2}\dots p_n^{\beta_n} \]

\[z=p_1^{\gamma_1}p_2^{\gamma_2}\dots p_n^{\gamma_n} \]

则

\[xyz=\prod_{i=1}^np_i^{\alpha_i+\beta_i+\gamma_i} \]

\[\begin{align*} &\sum_{i|x}\sum_{j|y}\sum_{k|z}\bigl[\gcd(i,j)=1\bigr]\cdot\bigl[\gcd(j,k)=1\bigr]\cdot\bigl[\gcd(k,i)=1\bigr] \\ =&\small\sum_{\substack{0\leq u_1\leq\alpha_1\\[2pt]0\leq u_2\leq\alpha_2\\[1pt]\dots\\[1pt]0\leq u_n\leq\alpha_n}}\sum_{\substack{0\leq v_1\leq\beta_1\\[2pt]0\leq v_2\leq\beta_2\\[1pt]\dots\\[1pt]0\leq v_n\leq\beta_n}}\sum_{\substack{0\leq w_1\leq\gamma_1\\[2pt]0\leq w_2\leq\gamma_2\\[1pt]\dots\\[1pt]0\leq w_n\leq\gamma_n}}\left[\gcd\!\left(\prod_{k=1}^np_k^{u_k},\,\prod_{k=1}^np_k^{v_k}\right)=1\right]\cdot\left[\gcd\!\left(\prod_{k=1}^np_k^{v_k},\;\prod_{k=1}^np_k^{w_k}\right)=1\right]\cdot\left[\gcd\!\left(\prod_{k=1}p_k^{w_k},\;\prod_{k=1}p_k^{u_k}\right)=1\right] \\ =&\sum_{\substack{0\leq u_1\leq\alpha_1\\[2pt]0\leq u_2\leq\alpha_2\\[1pt]\dots\\[1pt]0\leq u_n\leq\alpha_n}}\sum_{\substack{0\leq v_1\leq\beta_1\\[2pt]0\leq v_2\leq\beta_2\\[1pt]\dots\\[1pt]0\leq v_n\leq\beta_n}}\sum_{\substack{0\leq w_1\leq\gamma_1\\[2pt]0\leq w_2\leq\gamma_2\\[1pt]\dots\\[1pt]0\leq w_n\leq\gamma_n}}\left[\prod_{k=1}^np_k^{\min(u_k,v_k)}=\prod_{k=1}^np_k^{\,0}\right]\cdot\left[\prod_{k=1}^np_k^{\min(v_k,w_k)}=\prod_{k=1}^np_k^{\,0}\right]\cdot\left[\prod_{k=1}^np_k^{\min(w_k,u_k)}=\prod_{k=1}^np_k^{\,0}\right] \\ =&\sum_{\substack{0\leq u_1\leq\alpha_1\\[2pt]0\leq u_2\leq\alpha_2\\[1pt]\dots\\[1pt]0\leq u_n\leq\alpha_n}}\sum_{\substack{0\leq v_1\leq\beta_1\\[2pt]0\leq v_2\leq\beta_2\\[1pt]\dots\\[1pt]0\leq v_n\leq\beta_n}}\sum_{\substack{0\leq w_1\leq\gamma_1\\[2pt]0\leq w_2\leq\gamma_2\\[1pt]\dots\\[1pt]0\leq w_n\leq\gamma_n}}\prod_{k=1}^n\bigl[\min(u_k,v_k)=0\bigr]\cdot\bigl[\min(v_k,w_k)=0\bigr]\cdot\bigl[\min(w_k,u_k)=0\bigr] \\ =&\prod_{k=1}^n\sum_{u_k=0}^{\alpha_k}\sum_{v_k=0}^{\beta_k}\sum_{w_k=0}^{\gamma_k}\bigl[\min(u_k,v_k)=0\bigr]\cdot\bigl[\min(v_k,w_k)=0\bigr]\cdot\bigl[\min(w_k,u_k)=0\bigr] \\ =&\prod_{k=1}^n(\alpha_k+\beta_k+\gamma_k+1) \\ =&\space d\!\left(\prod_{k=1}^np_i^{\alpha_k+\beta_k+\gamma_k}\right) \\ =&\space d(xyz) \end{align*} \]

所以，原式

\[\begin{align*} \sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^cd(ijk)=&\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^c\sum_{x|i}\sum_{y|j}\sum_{z|k}\bigl[\gcd(x,y)=1\bigr]\cdot\bigl[\gcd(y,z)=1\bigr]\cdot\bigl[\gcd(z,x)=1\bigr] \\ =&\sum_{x=1}^a\sum_{y=1}^b\sum_{z=1}^c\bigl[\gcd(x,y)=1\bigr]\cdot\bigl[\gcd(y,z)=1\bigr]\cdot\bigl[\gcd(z,x)=1\bigr]\cdot\sum_{i=1}^{\lfloor a/x\rfloor}\sum_{j=1}^{\lfloor b/y\rfloor}\sum_{k=1}^{\lfloor c/z\rfloor}1 \\ =&\sum_{x=1}^a\sum_{y=1}^b\sum_{z=1}^c\bigl[\gcd(x,y)=1\bigr]\cdot\bigl[\gcd(y,z)=1\bigr]\cdot\bigl[\gcd(z,x)=1\bigr]\cdot\left\lfloor\frac{a}{x}\right\rfloor\cdot\left\lfloor\frac{b}{y}\right\rfloor\cdot\left\lfloor\frac{c}{z}\right\rfloor \\ =&\sum_{x=1}^a\left\lfloor\frac{a}{x}\right\rfloor\cdot\sum_{y=1}^b\bigl[\gcd(x,y)=1\bigr]\cdot\left\lfloor\frac{b}{y}\right\rfloor\cdot\sum_{z=1}^c\left\lfloor\frac{c}{z}\right\rfloor\cdot\bigl[\gcd(x,z)=1\bigr]\cdot\bigl[\gcd(y,z)=1\bigr] \\ =&\sum_{x=1}^a\left\lfloor\frac{a}{x}\right\rfloor\cdot\sum_{y=1}^b\bigl[\gcd(x,y)=1\bigr]\cdot\left\lfloor\frac{b}{y}\right\rfloor\cdot\sum_{z=1}^c\left\lfloor\frac{c}{z}\right\rfloor\cdot\bigl[\gcd(x,z)=1\bigr]\cdot\sum_{d|y,\,d|z}\mu(d) \\ =&\sum_{x=1}^a\left\lfloor\frac{a}{x}\right\rfloor\cdot\sum_{d=1}^{\min(b,c)}\mu(d)\cdot\sum_{y=1}^{\lfloor b/d\rfloor}\bigl[\gcd(x,dy)=1\bigr]\cdot\left\lfloor\frac{b}{dy}\right\rfloor\cdot\sum_{z=1}^{\lfloor c/d\rfloor}\bigl[\gcd(x,dz)=1\bigr]\cdot\left\lfloor\frac{c}{dz}\right\rfloor \end{align*} \]

因为

\[\bigl[\gcd(a,bc)=1\bigr]=\bigl[\gcd(a,b)=1\bigr]\cdot\bigl[\gcd(a,c)=1\bigr] \]

所以

\[\begin{align*} &\sum_{x=1}^a\left\lfloor\frac{a}{x}\right\rfloor\cdot\sum_{d=1}^{\min(b,c)}\mu(d)\cdot\sum_{y=1}^{\lfloor b/d\rfloor}\bigl[\gcd(x,dy)=1\bigr]\cdot\left\lfloor\frac{b}{dy}\right\rfloor\cdot\sum_{z=1}^{\lfloor c/d\rfloor}\bigl[\gcd(x,dz)=1\bigr]\cdot\left\lfloor\frac{c}{dz}\right\rfloor \\ =&\small\sum_{x=1}^a\left\lfloor\frac{a}{x}\right\rfloor\cdot\sum_{d=1}^{\min(b,c)}\mu(d)\cdot\sum_{y=1}^{\lfloor b/d\rfloor}\bigl[\gcd(x,d)=1\bigr]\cdot\bigl[\gcd(x,y)=1\bigr]\cdot\left\lfloor\frac{b}{dy}\right\rfloor\cdot\sum_{z=1}^{\lfloor c/d\rfloor}\bigl[\gcd(x,d)=1\bigr]\cdot\bigl[\gcd(x,z)=1\bigr]\cdot\left\lfloor\frac{c}{dz}\right\rfloor \\ =&\small\sum_{x=1}^a\left\lfloor\frac{a}{x}\right\rfloor\cdot\sum_{d=1}^{\min(b,c)}\mu(d)\cdot\Bigl(\bigl[\gcd(x,d)=1\bigr]\Bigr)^{\lfloor b/d\rfloor+\lfloor c/d\rfloor}\cdot\sum_{y=1}^{\lfloor b/d\rfloor}\bigl[\gcd(x,y)=1\bigr]\cdot\left\lfloor\frac{\lfloor b/d\rfloor}{y}\right\rfloor\cdot\sum_{z=1}^{\lfloor c/d\rfloor}\bigl[\gcd(x,z)=1\bigr]\cdot\left\lfloor\frac{\lfloor c/d\rfloor}{z}\right\rfloor \\ =&\sum_{x=1}^a\left\lfloor\frac{a}{x}\right\rfloor\cdot\sum_{d=1}^{\min(b,c)}\mu(d)\cdot\bigl[\gcd(x,d)=1\bigr]\cdot\sum_{y=1}^{\lfloor b/d\rfloor}\bigl[\gcd(x,y)=1\bigr]\cdot\left\lfloor\frac{\lfloor b/d\rfloor}{y}\right\rfloor\cdot\sum_{z=1}^{\lfloor c/d\rfloor}\bigl[\gcd(x,z)=1\bigr]\cdot\left\lfloor\frac{\lfloor c/d\rfloor}{z}\right\rfloor \end{align*} \]

令

\[\begin{align*} f(x,y)&=\sum_{i=1}^y\bigl[\gcd(x,i)=1\bigr]\cdot\left\lfloor\frac{y}{i}\right\rfloor \\ &=\sum_{i=1}^y\left\lfloor\frac{y}{i}\right\rfloor\cdot\sum_{t|x,\,t|i}\mu(t) \\ &=\sum_{t|x}\mu(t)\cdot\sum_{i=1}^{\lfloor y/t\rfloor}\left\lfloor\frac{y}{ti}\right\rfloor \\ &=\sum_{t|x}\mu(t)\cdot\sum_{i=1}^{\lfloor y/t\rfloor}\left\lfloor\frac{\lfloor y/t\rfloor}{i}\right\rfloor \end{align*} \]

令

\[S(n)=\sum_{i=1}^n\left\lfloor\frac{n}{i}\right\rfloor=\sum_{i=1}^n\tau(i) \]

则

\[f(x,y)=\sum_{t|x}\mu(t)\cdot S\!\left(\left\lfloor\frac{y}{t}\right\rfloor\right) \]

于是

\[\begin{align*} \sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^cd(ijk)&=\sum_{x=1}^a\left\lfloor\frac{a}{x}\right\rfloor\cdot\sum_{d=1}^{\min(b,c)}\mu(d)\cdot\sum_{y=1}^{\lfloor b/d\rfloor}\bigl[\gcd(x,dy)=1\bigr]\cdot\left\lfloor\frac{b}{dy}\right\rfloor\cdot\sum_{z=1}^{\lfloor c/d\rfloor}\bigl[\gcd(x,dz)=1\bigr]\cdot\left\lfloor\frac{c}{dz}\right\rfloor \\ &=\sum_{x=1}^a\left\lfloor\frac{a}{x}\right\rfloor\cdot\sum_{d=1}^{\min(b,c)}\mu(d)\cdot f\!\left(x,\left\lfloor\frac{b}{d}\right\rfloor\right)\cdot f\!\left(x,\left\lfloor\frac{c}{d}\right\rfloor\right) \end{align*} \]

计算这个的时间复杂度为 \(O(n^2\log n)\).