一个数论题的更优做法

定义一个正整数序列 \(a_1,a_2,...,a_n\) 是好的，当且仅当 \(\gcd(a_1,a_2,...,a_n) \leq q\) 且 \(\operatorname{lcm}(a_1,a_2,...,a_n) \geq p\)。

对于一个正整数序列 \(a_1,a_2,...,a_n\)，定义其价值为 \(\prod_{i=1}^{n}a_i\)。

给定 \(n,m,p,q\)，对于所有长度为 \(n\) 的正整数序列 \(a_1,a_2,...,a_n\)，满足对于任意 \(1 \leq i \leq n\)，均有
\(1 \leq a_i \leq m\)，统计其中所有好的序列的价值和。答案有可能很大，对 \(998244353\) 取模。

\[\begin{aligned} &\sum\limits_{x=1}^{p}\sum\limits_{y=q}\sum\limits_{a\in[m]^n}[\gcd(a_i)=x][\text{lcm}(a_i)=y]\prod a_i\\ =&\sum\limits_{x=1}^{p}\sum\limits_{y=q}\sum\limits_{a\in\left[\left\lfloor\frac{m}{x}\right\rfloor\right]^n}[\gcd(a_i)=1][\text{lcm}(a_i)x=y]\left(\prod a_i\right)x^n\\ =&\sum\limits_{x=1}^{p}\sum\limits_{y=q}\sum\limits_{a\in\left[\left\lfloor\frac{m}{x}\right\rfloor\right]^n}\sum\limits_{d|a_1,d|a_2,\cdots,d|a_n}\mu(d)[\text{lcm}(a_i)x=y]\left(\prod a_i\right)x^n\\ =&\sum\limits_{T=1}^{m}\sum\limits_{d|T}\left[\dfrac{T}{d}\le x\right]\mu(d)T^n \sum\limits_{a\in\left[\left\lfloor\frac{m}{T}\right\rfloor\right]^n}[\text{lcm}(a_i)T\ge y]\prod a_i\\ =&\sum\limits_{T=1}^{m}\sum\limits_{d|T}\left[\dfrac{T}{d}\le x\right]\mu(d)T^n \left(S_1^n\left(\left\lfloor\frac{m}{T}\right\rfloor\right)-\sum\limits_{i=1}^{\left\lfloor\frac{y-1}{T}\right\rfloor}\sum\limits_{a\in\left[\left\lfloor\frac{m}{T}\right\rfloor\right]^n}[\text{lcm}(a)=i]\prod a_i\right)\\ =&\sum\limits_{T=1}^{m}\sum\limits_{d|T}\left[\dfrac{T}{d}\le x\right]\mu(d)T^n \left(S_1^n\left(\left\lfloor\frac{m}{T}\right\rfloor\right)-\sum\limits_{i=1}^{\left\lfloor\frac{y-1}{T}\right\rfloor}\sum\limits_{\forall t,a_t|i}\left[\gcd\limits_{k=1}^{n}\left(\dfrac{i}{a_k}\right)=1\right]\prod a_i\right)\\ =&\sum\limits_{T=1}^{m}\sum\limits_{d|T}\left[\dfrac{T}{d}\le x\right]\mu(d)T^n \left(S_1^n\left(\left\lfloor\frac{m}{T}\right\rfloor\right)-\sum\limits_{i=1}^{\left\lfloor\frac{y-1}{T}\right\rfloor}\sum\limits_{\forall t,a_t|i}\sum\limits_{D|a_1,D|a_2,\cdots,D|a_k}\mu(D)\dfrac{i^n}{\prod a_i}\right)\\ =&\sum\limits_{T=1}^{m}\sum\limits_{d|T}\left[\dfrac{T}{d}\le x\right]\mu(d)T^nS_1^n\left(\left\lfloor\frac{m}{T}\right\rfloor\right)-\sum\limits_{T'=1}^{m}\sum\limits_{T|T'}\sum\limits_{d|T}\left[\dfrac{T}{d}\le x\right]\mu(d)T^n\mu\left(\dfrac{T'}{T}\right)\sum\limits_{\frac{T'}{T}|i}^{\left\lfloor\frac{y-1}{T}\right\rfloor}\sum\limits_{\forall t,\frac{T'}{T}|a_t|i}\dfrac{i^n}{\prod a_i}\\ =&\sum\limits_{T=1}^{m}\sum\limits_{d|T}\left[\dfrac{T}{d}\le x\right]\mu(d)T^nS_1^n\left(\left\lfloor\frac{m}{T}\right\rfloor\right)-\sum\limits_{T'=1}^{m}\sum\limits_{T|T'}\sum\limits_{d|T}\left[\dfrac{T}{d}\le x\right]\mu(d)T^n\mu\left(\dfrac{T'}{T}\right)\sum\limits_{i=1}^{\left\lfloor\frac{y-1}{T'}\right\rfloor}\sum\limits_{\forall t,a_t|i}\dfrac{i^n}{\prod a_i}\\ =&\sum\limits_{T=1}^{m}\sum\limits_{d|T}\left[\dfrac{T}{d}\le x\right]\mu(d)T^nS_1^n\left(\left\lfloor\frac{m}{T}\right\rfloor\right)-\sum\limits_{T'=1}^{m}\sum\limits_{T|T'}\sum\limits_{d|T}\left[\dfrac{T}{d}\le x\right]\mu(d)T^n\mu\left(\dfrac{T'}{T}\right)\sum\limits_{i=1}^{\left\lfloor\frac{y-1}{T'}\right\rfloor}i^nF(i)^n\\ &(\text{where }F_i=\sum\limits_{x|i}\dfrac{1}{x})\\ =&\sum\limits_{T=1}^{m}\sum\limits_{d|T}\left[\dfrac{T}{d}\le x\right]\mu(d)T^nS_1^n\left(\left\lfloor\frac{m}{T}\right\rfloor\right)-\sum\limits_{T'=1}^{m}\sum\limits_{T|T'}\sum\limits_{d|T}\left[\dfrac{T}{d}\le x\right]\mu(d)T^n\mu\left(\dfrac{T'}{T}\right)S\left(\left\lfloor\frac{y-1}{T'}\right\rfloor\right)\\ &(\text{where }S_i=\sum\limits_{x=1}^{i}(xF_x)^n)=\sum\limits_{x=1}^{i}d_1(x)^n\\ &\text{Let }f_T=\left(\sum\limits_{d|T\land\frac{T}{d}\le x}\mu(d)\right)T^n\text{, so}\\ =&\sum\limits_{T=1}^{m}f(T)\left(S_1^n\left(\left\lfloor\dfrac{m}{T}\right\rfloor\right)-\sum\limits_{T|T'}\mu\left(\dfrac{T'}{T}\right)S\left(\left\lfloor\dfrac{y-1}{T'}\right\rfloor\right)\right)\\ =&\sum\limits_{T=1}^{m}f(T)\left(S_1^n\left(\left\lfloor\dfrac{m}{T}\right\rfloor\right)-\sum\limits_{T|T'}\mu\left(\dfrac{T'}{T}\right)S\left(\left\lfloor\dfrac{\left\lfloor\frac{y-1}{T}\right\rfloor}{T'/T}\right\rfloor\right)\right) \end{aligned} \]

\(f_x,S_x\) 都可轻易 \(\mathcal{O}(m)\) 预处理，对后半部分使用整除分块即可，时间复杂度为

\[\mathcal{O}(m)+\int_{1}^{m}\mathcal{O}\left(\sqrt{\dfrac{m}{x}}\right)\text{ d}x=\mathcal{O}(m) \]