没啥好说的，暴解

$$\begin{eqnarray}
&& \prod_{i=l}^{r}\prod_{j=l}^{r} \frac{ij}{gcd(i,j)}^{\frac{ij}{gcd(i,j)}} \
\end{eqnarray}$$
设 $f(x,y)=\prod_{i=1}^{x}\prod_{j=1}^{y} \frac{ij}{gcd(i,j)}^{\frac{ij}{gcd(i,j)}}$

那么答案则为

$$\frac{f(r,r)f(l-1,l-1)}{f(l-1,r)f(r,l-1)}$$

$$\begin{eqnarray}
f(x,y)&=&\prod_{i=1}^{x}\prod_{j=1}^{y} \frac{ij}{gcd(i,j)}^{\frac{ij}{gcd(i,j)}}\
&=&\prod_{d=1}^{y}\prod_{i=1}^{x}\prod_{j=1}^{y} \frac{ij}{d}^{\frac{ij}{d}[gcd(i,j)=d]}\
&=&\prod_{d=1}^{y}\prod_{i=1}^{\left\lfloor\frac{x}{d}\right\rfloor}\prod_{j=1}^{\left\lfloor\frac{y}{d}\right\rfloor} dij^{dij[gcd(i,j)=1]}\
&=&\prod_{d=1}^{y}\prod_{i=1}^{\left\lfloor\frac{x}{d}\right\rfloor}\prod_{j=1}^{\left\lfloor\frac{y}{d}\right\rfloor} dij^{dij\sum\limits_{p \mid gcd(i,j)}\mu(p)}\
&=&\prod_{d=1}^{y}\prod_{i=1}^{\left\lfloor\frac{x}{d}\right\rfloor}\prod_{j=1}^{\left\lfloor\frac{y}{d}\right\rfloor} dij^{dij\sum\limits_{p=1}^{\left\lfloor \frac{y}{d} \right\rfloor}\mu(p)[p \mid i][\mid j]}\
&=&\prod_{d=1}^{y}\prod\limits_{p=1}^{\left\lfloor \frac{y}{d} \right\rfloor}\prod_{i=1}^{\left\lfloor\frac{x}{dp}\right\rfloor}\prod_{j=1}^{\left\lfloor\frac{y}{dp}\right\rfloor} (ijdp^2)^{ijdp^2\mu(p)}\
&=&\prod_{T=1}^{y}\prod\limits_{p\mid T}\prod_{i=1}^{\left\lfloor\frac{x}{T}\right\rfloor}\prod_{j=1}^{\left\lfloor\frac{y}{T}\right\rfloor} (ijTp)^{ijTp\mu(p)}\
&=&\prod_{T=1}^{y}\prod\limits_{p\mid T}\left(\prod_{i=1}^{\left\lfloor\frac{x}{T}\right\rfloor}\prod_{j=1}^{\left\lfloor\frac{y}{T}\right\rfloor} (ij)^{ij}\times(Tp)^{S(\left\lfloor\frac{x}{T}\right\rfloor)S(\left\lfloor\frac{y}{T}\right\rfloor)}\right)^{Tp\mu(p)}\
\end{eqnarray}$$

不妨设 $g(x,y)=\prod_{i=1}^{x}\prod_{j=1}^{y} (ij)^{ij},D(x)=\prod_{i=1}^{x}i^i$ ，那么有

$$g(x,y)=D(x)^{S_{y}} \times D(y)^{S_{x}}$$
将其替换

$$\begin{eqnarray}
&=&\prod_{T=1}^{y}\prod\limits_{p\mid T}\left(g\left(\left\lfloor \frac{x}{T} \right\rfloor,\left\lfloor \frac{y}{T} \right\rfloor\right) \times(Tp)^{S(\left\lfloor\frac{x}{T}\right\rfloor)S(\left\lfloor\frac{y}{T}\right\rfloor)}\right)^{Tp\mu(p)}\

\end{eqnarray}$$
不妨设 $h(x)=\sum_{d \mid x}d\mu(d),w(x)=\prod_{d \mid x}d^{d\mu(d)}$，那么有

$$\begin{eqnarray}
&=&\prod_{T=1}^{y}\left(g\left(\left\lfloor \frac{x}{T} \right\rfloor,\left\lfloor \frac{y}{T} \right\rfloor\right)^{Th(T)}(T^{h(T)}\prod\limits_{p\mid T}p^{p\mu(p)})^{TS(\left\lfloor\frac{x}{T}\right\rfloor)S(\left\lfloor\frac{y}{T}\right\rfloor)}\right)\
&=&\prod_{T=1}^{y}g\left(\left\lfloor \frac{x}{T} \right\rfloor,\left\lfloor \frac{y}{T} \right\rfloor\right)^{Th(T)}\left(T^{h(T)}w(T)\right)^{TS(\left\lfloor\frac{x}{T}\right\rfloor)S(\left\lfloor\frac{y}{T}\right\rfloor)}\

\end{eqnarray}$$
最后再设 $E(x)=\left(T^{h(T)}w(T)\right)^T,H(x)=Th(T)$ ，就变成了

$$\begin{eqnarray}

&=&\prod_{T=1}^{y}g\left(\left\lfloor \frac{x}{T} \right\rfloor,\left\lfloor \frac{y}{T} \right\rfloor\right)^{Th(T)}E(T)^{S(\left\lfloor\frac{x}{T}\right\rfloor)S(\left\lfloor\frac{y}{T}\right\rfloor)}\

\end{eqnarray}$$

于是我们需要预处理以下东西

$h(i),E(i)$ 的前缀和

$D(i),w(i)$