标签:tip 复杂 多点 play inf spl display text 顺序
\(x\)的\(n\)阶下降幂\(x^{\underline n}=\prod_0^{n-1}(x-i) = \frac{x!}{(x-n)!}\)
一个下降幂多项式\(F(x)=\sum a_ix^{\underline i}\)
点值的\(\text{EGF}\)为\(\begin{aligned} EGF(F(x))=\sum_0^{\infty}F(i)x^i \end{aligned}\)
\(\begin{aligned}EGF(F(x))=\sum_{i=0}^{\infty}\frac{x^i}{i!}\sum_{j=0}^{n} \frac{i!}{(i-j)!}\cdot F_j\end{aligned}\)
\(\begin{aligned}EGF(F(x))=\sum_{i=0}^{\infty}x^i \sum_{j=0}^{n} \frac{1}{(i-j)!}\cdot F_j\end{aligned}\)
换一下顺序
\(\begin{aligned}EGF(F(x))=\sum_{i=0}^{n} F_i \sum_{j=i}^{\infty}\frac{1}{(j-i)!} x^j\end{aligned}\)
\(\begin{aligned}EGF(F(x))=\sum_{i=0}^{n} F_i \cdot x^i \sum_{j=0}^{\infty}\frac{1}{j!} x^j\end{aligned}\)
\(\begin{aligned}EGF(F(x))=\sum_{i=0}^{n} F_i \cdot x^i e^x\end{aligned}\)
那么直接和\(e^x\)卷积就可以得到\(F(x)\)的\(\text{EGF}\)
Tips: \(e^x\)直接带入展开式\(\begin{aligned} e^{ax}=\sum_0^{\infty}\frac{(ax)^i}{i!} \end {aligned}\)
如果要从\(\text{EGF}\)得到\(F(x)\)
\(\begin{aligned}EGF(F(x))=\sum_{i=0}^{n} F_i \cdot x^ie^x\end{aligned}\)
\(F_i=\frac{EGF(F(x))}{x^ie^x}\)
那么就直接卷上\(e^{-x}\)就可以了
求出\(\text{EGF}\),然后点值对应相乘(注意乘完之后要补上一个\(i!\)),最后再反求\(F(x)\)
带入\(x^n\)的第二类斯特林数展开式,\(\begin{aligned} x^n=\sum C(x,i)i!S(n,i)=\sum x^{\underline i}S(n,i) \end{aligned}\)
\(\begin{aligned} G(x)=\sum [x^i]F(x)\sum x^{\underline j}S(i,j) \end{aligned}\)
再带入\(S(i,j)\)的通项公式
\(\begin{aligned} G(x)=\sum [x^i]F(x)\sum x^{\underline j}\sum \frac{(-1)^{j-k} C(j,k) k^i}{j!}\end{aligned}\)
\(\begin{aligned} G(x)=\sum [x^i]F(x)\sum x^{\underline j}\sum \frac{(-1)^{j-k} k^i}{k!(j-k)!}\end{aligned}\)
是一个三元的卷积优化,比较复杂
\(\begin{aligned} i\rightarrow j,[x^{\underline j}]G(x)=\frac{(-1)^{j-k}[x^i]F(x) k^i}{(j-k)!k!}\end{aligned}\)
考虑分步优化
先计算出\(H(k)=\sum [x^i]F(x) k^i\)
发现\([x^i]H(x)=F(i)\),多项式多点求值得到
对于已经得到的\(H(x)\)
\(\begin{aligned} [x^{\underline j}]G(x)=\sum \frac{(-1)^{j-k}[x^k]H(x)}{k!(j-k)!}\end{aligned}\)
可以直接卷积得到
求出\(F(x)\)的\(EGF\),然后带入前\(n\)项的值,快速插值回来即可
标签:tip 复杂 多点 play inf spl display text 顺序
原文地址:https://www.cnblogs.com/chasedeath/p/13073206.html