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

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