XSY1262

题意

\(n\)排列，分解出的轮换个数的\(m\)次方的期望\(\times n!\)。\(n\le 10^5,m\le 30\)

做法

\(\begin{aligned} ans&=\sum_{i=1}^n\begin{bmatrix}n\\i\end{bmatrix}i^m\\ &=\sum_{i=1}^n\begin{bmatrix}n\\i\end{bmatrix}\sum_{j=1}^m\begin{Bmatrix}m\\j\end{Bmatrix}\binom{i}{j}j!\\ &=\sum_{i=1}^m\begin{Bmatrix}m\\i\end{Bmatrix}i!\sum_{j=1}^n\begin{bmatrix}n\\j\end{bmatrix}\binom{j}{i}\\ &=\sum_{i=1}^m\begin{Bmatrix}m\\i\end{Bmatrix}\begin{bmatrix}n+1\\i+1\end{bmatrix}i!\\ \end{aligned}\)