类欧几里得算法
类欧几里得算法
定义
\[\displaystyle\begin{aligned}
f(a,b,c,n) &= \sum\limits_{i = 0}^{n}\left\lfloor\dfrac{ai + b}{c}\right\rfloor \\
g(a,b,c,n) &= \sum\limits_{i = 0}^{n}{\left\lfloor\dfrac{ai + b}{c}\right\rfloor}^2 \\
h(a,b,c,n) &= \sum\limits_{i = 0}^{n}i\left\lfloor\dfrac{ai + b}{c}\right\rfloor
\end{aligned}\]
给定 \(n,a,b,c\)，求 \(f(a,b,c,n),g(a,b,c,n),h(a,b,c,n)\)，对 \(998244353\) 取模（\(1 \le T \le 10^5,0 \le n,a,b,c \le 10^9, c \neq 0\)）。
考虑函数 \(f\)。首先考虑 \(a > c \lor b > c\) 的情况
\[\displaystyle \begin{aligned}
f(a,b,c,n) &= \sum\limits_{i = 0}^{n}\left\lfloor\dfrac{ai + b}{c}\right\rfloor \\
&= \sum\limits_{i = 0}^{n}\left\lfloor\dfrac{{(\left\lfloor\dfrac{a}{c}\right\rfloor c + a \bmod c)i + \left\lfloor\dfrac{b}{c}\right\rfloor c + b \bmod c}}{c}\right\rfloor \\
&= \sum\limits_{i = 0}^{n}(\left\lfloor\dfrac{a}{c}\right\rfloor i + \left\lfloor\dfrac{b}{c}\right\rfloor + \left\lfloor\dfrac{(a \bmod c) i + (b \bmod c)}{c}\right\rfloor) \\
&= \dfrac{n(n + 1)}{2}\left\lfloor\dfrac{a}{c}\right\rfloor + (n + 1)\left\lfloor\dfrac{b}{c}\right\rfloor + \sum\limits_{i = 0}^{n}\left\lfloor\dfrac{{a \bmod ci + b \bmod c}}{c}\right\rfloor \\
&= \dfrac{n(n + 1)}{2}\left\lfloor\dfrac{a}{c}\right\rfloor + (n + 1)\left\lfloor\dfrac{b}{c}\right\rfloor + f(a \bmod c,b \bmod c, c, n) \\
\end{aligned}\]
接下来考虑 \(a \le c \land b \le c\) 的情况
\[\displaystyle \begin{aligned}
f(a,b,c,n) &= \sum\limits_{i = 0}^{n}\left\lfloor\dfrac{ai + b}{c}\right\rfloor \\
&= \sum\limits_{i = 0}^{n}\sum\limits_{j = 0}^{\left\lfloor\frac{ai + b}{c}\right\rfloor - 1}1 \\
&= \sum\limits_{j}\sum\limits_{i = 0}^{n}{\left[j < \left\lfloor\frac{ai + b}{c}\right\rfloor \right]} \\
&= \sum\limits_{j = 0}^{\left\lfloor\frac{an + b}{c}\right\rfloor - 1}\sum\limits_{i = 0}^{n}{\left[j < \left\lfloor\frac{ai + b}{c}\right\rfloor \right]} \\
\end{aligned}\]
考虑限制条件 \(j < \left\lfloor\frac{ai + b}{c}\right\rfloor\)
\[\displaystyle \begin{aligned}
j < \left\lfloor\frac{ai + b}{c}\right\rfloor
& \Leftrightarrow j < \frac{ai + b}{c} \\
& \Leftrightarrow j + 1 \le \frac{ai + b}{c} \\
& \Leftrightarrow jc + c \le ai + b \\
& \Leftrightarrow jc + c - b \le ai \\
& \Leftrightarrow jc + c - b - 1 < ai \\
& \Leftrightarrow \dfrac{jc + c - b - 1}{a} < i \\
& \Leftrightarrow \left\lfloor\dfrac{jc + c - b - 1}{a}\right\rfloor < i
\end{aligned}\]
记 \(\displaystyle m = \left\lfloor\frac{an + b}{c}\right\rfloor,t = \left\lfloor\dfrac{jc + c - b - 1}{a}\right\rfloor\)，则
\[\displaystyle \begin{aligned}
f(a,b,c,n) &= \sum\limits_{j = 0}^{\left\lfloor\frac{an + b}{c}\right\rfloor - 1}\sum\limits_{i = 0}^{n}{\left[j < \left\lfloor\frac{ai + b}{c}\right\rfloor \right]} \\
&= \sum\limits_{j = 0}^{m - 1}\sum\limits_{i = 0}^{n}{\left[i > t\right]} \\
&= \sum\limits_{j = 0}^{m - 1}(n - t) \\
&= nm - \sum\limits_{j = 0}^{m - 1}\left\lfloor\dfrac{jc + c - b - 1}{a}\right\rfloor \\
&= nm - f(c,c - b - 1, a, m - 1)
\end{aligned}\]
观察到当 \(a > c\) 时，\(a \rightarrow a \bmod c\)；当 \(a \le c\) 时，\(a,c\) 互换位置，故复杂度为 \(\mathcal{O}(\log\max(a,c))\)。
考虑函数 \(g\)。首先考虑 \(a > c \lor b > c\) 的情况
\[\displaystyle \begin{aligned}
g(a,b,c,n) &= \sum\limits_{i = 0}^{n}{\left\lfloor\dfrac{ai + b}{c}\right\rfloor}^2 \\
&= \sum\limits_{i = 0}^{n}{\left\lfloor\dfrac{{(\left\lfloor\dfrac{a}{c}\right\rfloor c + a \bmod c)i + \left\lfloor\dfrac{b}{c}\right\rfloor c + b \bmod c}}{c}\right\rfloor}^2 \\
&= \sum\limits_{i = 0}^{n}(\left\lfloor\dfrac{a}{c}\right\rfloor i + \left\lfloor\dfrac{b}{c}\right\rfloor + \left\lfloor\dfrac{(a \bmod c) i + (b \bmod c)}{c}\right\rfloor)^2 \\
&= \dfrac{n(n + 1)(2n + 1)}{6}\left\lfloor\dfrac{a}{c}\right\rfloor^2 + \dfrac{n(n + 1)}{2}\left\lfloor\dfrac{b}{c}\right\rfloor^2 + g(a \bmod c,b \bmod c, c, n) \\
&+ \dfrac{n(n + 1)}{2}\left\lfloor\dfrac{a}{c}\right\rfloor\left\lfloor\dfrac{b}{c}\right\rfloor+ 2 \left\lfloor\dfrac{a}{c}\right\rfloor h(a \bmod c,b \bmod c, c, n) + 2 \left\lfloor\dfrac{b}{c}\right\rfloor f(a \bmod c,b \bmod c, c, n)
\end{aligned}\]
接下来考虑 \(a \le c \land b \le c\) 的情况，观察到
\[n^2 = 2\dfrac{n(n + 1)}{2} - n = 2\left(\sum_{i = 1}^{n}i\right) - n
\]
那么
\[\displaystyle \begin{aligned}
g(a,b,c,n) &= \sum\limits_{i = 0}^{n}{\left\lfloor\dfrac{ai + b}{c}\right\rfloor}^2 \\
&= \sum\limits_{i = 0}^{n}\left[2\left(\sum\limits_{j=1}^{\left\lfloor\frac{ai + b}{c}\right\rfloor}j\right) - {\left\lfloor\frac{ai + b}{c}\right\rfloor} \right] \\
\end{aligned}\]
考虑处理 \(\displaystyle \sum\limits_{i = 0}^{n}\sum\limits_{j=1}^{\left\lfloor\frac{ai + b}{c}\right\rfloor}j\)，记 \(\displaystyle m = \left\lfloor\frac{an + b}{c}\right\rfloor,t = \left\lfloor\dfrac{jc + c - b - 1}{a}\right\rfloor\)
\[\displaystyle \begin{aligned}
& \sum\limits_{i = 0}^{n}\sum\limits_{j=1}^{\left\lfloor\frac{ai + b}{c}\right\rfloor}j \\
=& \sum\limits_{i = 0}^{n}\sum\limits_{j=0}^{\left\lfloor\frac{ai + b}{c}\right\rfloor - 1}\left(j + 1\right) \\
=& \sum\limits_{j = 0}^{m - 1}\left(j + 1\right)\sum\limits_{i = 0}^{n}\left[i > t\right] \\
=& \sum\limits_{j = 0}^{m - 1}\left(j + 1\right)\left(n - t\right) \\
=& \dfrac{1}{2}nm(m + 1) + h(c,c - b - 1,a,m - 1) - f(c,c - b - 1,a,m - 1)
\end{aligned}\]
那么
\[\displaystyle \begin{aligned}
g(a,b,c,n) &= \sum\limits_{i = 0}^{n}\left[2\left(\sum\limits_{j=1}^{\left\lfloor\frac{ai + b}{c}\right\rfloor}j\right) - {\left\lfloor\frac{ai + b}{c}\right\rfloor} \right] \\
&= nm(m + 1) + 2h(c,c - b - 1,a,m - 1) - 2f(c,c - b - 1,a,m - 1) + f(a,b,c,n)
\end{aligned}\]
考虑函数 \(h\)。首先考虑 \(a > c \lor b > c\) 的情况
\[\displaystyle \begin{aligned}
h(a,b,c,n) &= \sum\limits_{i = 0}^{n}i\left\lfloor\dfrac{ai + b}{c}\right\rfloor \\
&= \sum\limits_{i = 0}^{n}i\left\lfloor\dfrac{{(\left\lfloor\dfrac{a}{c}\right\rfloor c + a \bmod c)i + \left\lfloor\dfrac{b}{c}\right\rfloor c + b \bmod c}}{c}\right\rfloor \\
&= \sum\limits_{i = 0}^{n}i(\left\lfloor\dfrac{a}{c}\right\rfloor i + \left\lfloor\dfrac{b}{c}\right\rfloor + \left\lfloor\dfrac{(a \bmod c) i + (b \bmod c)}{c}\right\rfloor) \\
&= \dfrac{n(n + 1)(2n + 1)}{6}\left\lfloor\dfrac{a}{c}\right\rfloor + \dfrac{n(n + 1)}{2}\left\lfloor\dfrac{b}{c}\right\rfloor + \sum\limits_{i = 0}^{n}i\left\lfloor\dfrac{{a \bmod ci + b \bmod c}}{c}\right\rfloor \\
&= \dfrac{n(n + 1)(2n + 1)}{6}\left\lfloor\dfrac{a}{c}\right\rfloor + \dfrac{n(n + 1)}{2}\left\lfloor\dfrac{b}{c}\right\rfloor + h(a \bmod c,b \bmod c, c, n) \\
\end{aligned}\]
接下来考虑 \(a \le c \land b \le c\) 的情况，记 \(\displaystyle m = \left\lfloor\frac{an + b}{c}\right\rfloor,t = \left\lfloor\dfrac{jc + c - b - 1}{a}\right\rfloor\)
\[\displaystyle \begin{aligned}
h(a,b,c,n) &= \sum\limits_{i = 0}^{n}i\left\lfloor\dfrac{ai + b}{c}\right\rfloor \\
&= \sum\limits_{j = 0}^{m - 1}\sum_{i = 0}^{n}\left[i > t\right]i \\
&= \sum\limits_{j = 0}^{m - 1}\dfrac{(n + t + 1)(n - t)}{2} \\
&= \dfrac{1}{2}\left[mn(n + 1) - \sum\limits_{i = 0}^{n}t^2 - \sum\limits_{i = 0}^{n}t\right] \\
&= \dfrac{1}{2}\left[mn(n + 1) - g(c,c - b - 1,a,m - 1) - f(c,c - b - 1,a,m - 1)\right] \\
\end{aligned}\]
