Card

\(Ans=\frac{\sum\limits_{i=0}^ni^k(m-1)^{n-i}\binom ni}{m^k}\)
\(F(x)=\sum\limits_{t\ge0}\frac{x^t}{t!}\sum\limits_{i=0}^ni^t\binom ni(m-1)^{n-i}\)
\(=\sum\limits_{i=0}^n\binom ni(m-1)^{n-i}e^{ix}\)
\(=(e^x+m-1)^n\)
\(现在求[x^k]F(x),我们试用EI介绍的方法进行处理:\)
\(看成复合形态G(e^x),G(x)=(x+m-1)^n\)
\(注意G(e^k)=G((e^k-1)+1)\)
\(考虑H(x)=G(x+1)=(x+m)^n\)
\(nH(x)=(x+m)H'(x)\)
\(nH_i=iH_i+m(i+1)H_{i+1}\)
\(m(i+1)H_{i+1}+(i-n)H_i=0\)
\(而Q(x)=H(x)\%x^{n+1}\)
\(那么m(i+1)Q_{i+1}+(i-n)Q_i=[i==k](k-n)Q_k\)
\(Q'(x)(x+m)-Q(x)=(k-n)m^{n-k}\binom nkx^k\)
\(令P(x)=Q(x-1)\)
\(P'(x)(x+m-1)-P(x)=(k-n)m^{n-k}\binom nk(x-1)^k\)
\(解出P之后就容易做了\)

标签：nH,ni,frac,limits,sum,binom,Card
来源： https://www.cnblogs.com/ldxcaicai/p/15306551.html