求
\[
\sum_{i=0}^n f(i){n\choose i} x^i (1-x)^{n-i}
\]
模\(998244353\)
\(n\leq 10^9,m\leq 2*10^4\)
式子后面长得很像二项式定理，我们要想办法把\(f(i)\)分离出来。
一种高妙的做法是把\(f\)转成下降幂形式。可能是我做题太少没见过吧
令
\[
f(x)=\sum_{i=0}^m g_i x^{\underline{i}}
\]
则
\[
\begin{aligned}
ans&=\sum_{i=0}^n \sum_{j=0}^m g_j i^{\underline{j}} {n\choose i} x^i (1-x)^{n-i}\&=\sum_{i=0}^n \sum_{j=0}^m g_j \frac{i!}{(i-j)!} \frac{n!}{i!(n-i)!} x^i (1-x)^{n-i}\&=\sum_{i=0}^n \sum_{j=0}^m g_j \frac{n!}{(n-j)!} \frac{(n-j)!}{(i-j)!(n-i)!} x^i (1-x)^{n-i}\&=\sum_{i=0}^n \sum_{j=0}^m g_j n^{\underline{j}} {n-j \choose n-i} x^i (1-x)^{n-i}\\end{aligned}
\]
原文：https://www.cnblogs.com/zzqtxdy/p/12040532.html