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LOJ138 类欧几里得算法

\(F_{k1,k2}(n,a,b,c)=\sum\limits_{i=0}^{n}i^{k_1}\lfloor\frac{ai+b}{c}\rfloor^{k_2}\)

\[\begin{aligned} \lfloor\frac{ax+b}{c}\rfloor^{k_2}&=(\lfloor\frac{(a\bmod c)x+b}{c}\rfloor+\lfloor\frac{a}{c}\rfloor x)^{k_2}\\ &=\sum_{i=0}^{k_2}\binom{k_2}{i}\lfloor\frac{a}{c}\rfloor^{i}\lfloor\frac{(a\bmod c)x+b}{c}\rfloor^{k_2-i}x^{i}\\ \end{aligned} \]

\[\begin{aligned} \lfloor\frac{ax+b}{c}\rfloor^{k_2}&=(\lfloor\frac{ax+(b\bmod c)}{c}\rfloor+\lfloor\frac{b}{c}\rfloor)^{k_2}\\ &=\sum_{i=0}^{k_2}\binom{k_2}{i}\lfloor\frac{b}{c}\rfloor^{i}\lfloor\frac{ax+(b\bmod c)}{c}\rfloor^{k_2-i}\\ \end{aligned} \]

不妨设 \(a<c,b<c\)

先求出所有 \(F_{k1,0}(n,a,b,c)\)。这个与 \(a,b,c\) 无关,是一个自然数幂和。

posted @ 2023-05-30 22:22  Semsue  阅读(11)  评论(0编辑  收藏  举报
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