注意答案为：

$$ \begin{aligned} &\sum_{d\geq 1}\sum_{l_B}\sum_{l_Bd\leq l_R\leq R-d(B-l_B)}{l_B+l_R\choose l_B}{B-l_B+R-l_R\choose B-l_B}\\ =&\sum_{l_B}\sum_{d\geq 1}\sum_{0\leq t\leq R-dB}{(d+1)l_B+t\choose l_B}{B+R-(d+1)l_B-t\choose B-l_B} \end{aligned} $$

用 $\mathcal{B}_t(z)$ 提取系数（$\mathcal{[z^n]\frac{B_t(z)^r}{1-t+tB_t^{-1}(z)}}={tn+r\choose n}$）：

$$ \begin{aligned} &\sum_{l_B}\sum_{d\geq 1}\sum_{0\leq t\leq R-dB}{(d+1)l_B+t\choose l_B}{B+R-(d+1)l_B-t\choose B-l_B}\\ =&\sum_{l_B}\sum_{d\geq 1}\sum_{0\leq t\leq R-dB}\left([z^{l_B}]\frac{\mathcal{B}*{d+1}(z)^t}{(d+1)\mathcal{B}*{d+1}^{-1}(z)-d}\right)\left([z^{B-l_B}]\frac{\mathcal{B}*{d+1}(z)^{R-dB-t}}{(d+1)\mathcal{B}*{d+1}^{-1}(z)-d}\right)\\ =&\sum_{d\geq 1}(R-dB+1)\left([z^{B}]\frac{\mathcal{B}*{d+1}(z)^{R-dB}}{((d+1)\mathcal{B}*{d+1}^{-1}(z)-d)^2}\right) \end{aligned} $$

令 $F(z)+1=\mathcal{B}*{d+1}(z)$，那么 $\mathcal{\frac{B*{d+1}(z)^{R-dB}}{((d+1)B_{d+1}^{-1}(z)-d)^2}}$ 等价于 $\frac{(1+F(z))^{R-dB+2}}{(1-dF(z))^2}$。用另类拉格朗日反演提取系数：

$$ \begin{aligned} [z^B]\frac{(1+F(z))^{R-dB+2}}{(1-dF(z))^2} &=[z^B]\frac{(1+z)^{R-dB+2}}{(1-dz)^2}\frac{(1-dz)}{(z+1)^{d+2}}\left(z+1\right)^{(d+1)(B+1)}\\ &=[z^B]\frac{(1+z)^{R+B+1}}{1-dz}=\sum_{0\leq i\leq B}{R+B+1\choose i}d^{B-i} \end{aligned} $$

代回去：

$$ \sum_{0\leq i\leq B}{R+B+1\choose i}\left((R+1)\sum_{1\leq d\leq \lfloor{\frac{R}{B}\rfloor}}d^{B-i}-B\sum_{1\leq d\leq \lfloor{\frac{R}{B}\rfloor}}d^{B-i+1}\right) $$

是自然数幂和。即 $\sum\limits_{i=1}^{n}i^k=[\frac{x^k}{k!}]\frac{e^{x}-e^{(n+1)x}}{1-e^x}$ 即可。

时间复杂度 $O(B\log B)$。 提交记录：Submission #97802 - QOJ.ac。