CF891E Lust

\[ans = \frac{1}{n^k} \prod_ia_i - \prod_i(a_i-b_i)\\ 考虑计算后面那个\\ 考虑一组 \sum_{b_i} = k 如何计算\\ \frac{k!}{\prod_{i = 1}^n b_i!} \prod_{i = 1}^n(a_i - b_i)\\ =\frac{k!}{n^k} \prod_{i = 1}^n\frac{(a_i - b_i)}{b_i!}\\ 考虑生成函数f_i(x) = \sum_{j = 0}^{\infty}\frac{a_i-j}{j!}x^j = (a_i - x)e^x\\ ans = [x^k]e^{nx}\prod_{i = 1}^{n}(a_i-x)\\ \]

将右边那个式子\(O(n^2)\)跑出来，然后与左边那个式子直接 卷积一下就好了