Luogu5591 小猪佩奇学数学 【单位根反演】

题目链接：洛谷

\[
Ans=\frac{1}{k}(\sum_{i=0}^n\binom{n}{i}p^ii-\sum_{i=0}^n\binom{n}{i}p^i(i \ \mathrm{mod} \ k))
\]

\[
\begin{aligned}
Ans&=\sum_{i=0}^n\binom{n}{i}p^i(i \ \mathrm{mod} \ k) \\
&=\sum_{d=0}^{k-1}\sum_{i=0}^n\binom{n}{i}p^id((i-d) \ \mathrm{mod} \ k=0) \\
&=\frac{1}{k}\sum_{d=0}^{k-1}\sum_{i=0}^n\binom{n}{i}p^id\sum_{j=0}^{k-1}w_k^{(i-d)j} \\
&=\frac{1}{k}\sum_{d=0}^{k-1}d\sum_{j=0}^{k-1}w_k^{-dj}\sum_{i=0}^n\binom{n}{i}(pw_k^{j})^i \\
&=\frac{1}{k}\sum_{d=0}^{k-1}d\sum_{j=0}^{k-1}w_k^{-dj}(pw_k^j+1)^n \\
&=\frac{1}{k}\sum_{i=0}^{k-1}(pw_k^i+1)^n\sum_{d=0}^{k-1}d(w_k^{-i})^d
\end{aligned}
\]

现在推推后面一部分。
\[
\begin{aligned}
S&=\sum_{i=0}^{k-1}ix^i \\
&=\sum_{i=0}^{k-1}(i+1)x^{i+1}-kx^k \\
&=x\sum_{i=0}^{k-1}ix^i+\sum_{i=0}^{k-1}x^i-kx^k \\
&=xS+\frac{1-x^k}{1-x}-kx^k \\
(1-x)S&=\frac{1-x^k}{1-x}-kx^k \\
\because x^k&=1\\
S&=\frac{k}{1-x} \\
Ans&=\frac{(p+1)^n(k-1)}{2}+\sum_{i=1}^{k-1}\frac{(pw_k^i+1)^n}{1-w_k^{-i}}
\end{aligned}
\]

还有一部分
\[
\begin{aligned}
Ans&=\sum_{i=0}^n\binom{n}{i}p^ii \\
&=np\sum_{i=0}^{n-1}\binom{n-1}{i}p^i \\
&=np(p+1)^{n-1}
\end{aligned}
\]