杂题

题意

\(x=0\)，\(n\)次操作，有\(\frac{Q}{Q+1}\)将\(x\)加\(1\)。求\(E(\sum\limits_{i=1}^x i^k)\times (Q+1)^n\)

做法一

令\(f_{n,i}\)为\(n\)次操作后\(E(x^i)\)
显然有：\(f_{n,i}=\frac{1}{1+Q}f_{n-1,i}+\frac{Q}{1+Q}\sum\limits_{k=0}^i {i\choose k}f_{n-1,k}\)
令\(F_n(x)=\sum\limits_{i}f_{n,i}\frac{x^i}{i!},G(x)=\frac{1}{1+Q}+\sum\limits_{i}\frac{x^i}{i!}\)，有\(F_n(x)=F_{n-1}(x)G(x)\)
然后就可以快速幂了
\(\sum\limits_{i=1}^x i^k\)，可以快速插值出\(k+1\)次系数，因为期望是线性的，所以带入就好了

做法二

这个没什么意思
\(\begin{aligned} Ans&=\sum\limits_{i=0}^n {n\choose i}Q^i\sum\limits_{j=1}^i j^k\\ &=\sum\limits_{i=0}^n {n\choose i}Q^i\sum\limits_{j=1}^i \sum\limits_{t=0}^k S_{k,t}{j\choose t}t!\\ &=\sum\limits_{t=0}^k t!S_{k,t}\sum\limits_{i=t}^n {n\choose i}Q^i\sum\limits_{j=0}^i {j\choose t}\\ &=\sum\limits_{t=0}^k t!S_{k,t}\sum\limits_{i=t}^n {n\choose i}Q^i{i+1\choose t+1}\\ &=\sum\limits_{t=0}^k \frac{S_{k,t}}{t+1}Q^t n^{\underline t}\sum\limits_{i=0}^{n-t}{n-t\choose i}(i+t+1)Q^i\\ &=\sum\limits_{t=0}^k \frac{S_{k,t}}{t+1}Q^t n^{\underline t}((t+1)(Q+1)^{n-t}+(n-t)Q(Q+1)^{n-t-1})\\ \end{aligned}\)