类欧几里得算法

类欧几里得算法

定义

$${\displaystyle \begin{array}{rl}f(a,b,c,n)& =\sum _{i=0}^n\lfloor {\displaystyle \frac{ai+b}{c}}\rfloor \\ g(a,b,c,n)& =\sum _{i=0}^n{\lfloor {\displaystyle \frac{ai+b}{c}}\rfloor }^2\\ h(a,b,c,n)& =\sum _{i=0}^{n}i\lfloor {\displaystyle \frac{ai+b}{c}}\rfloor \end{array}}$$

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（$1\le T\le {10}^5,0\le n,a,b,c\le {10}^9,c\ne 0$）。

考虑函数 $f$。首先考虑 $a>c\vee b>c$ 的情况

$${\displaystyle \begin{array}{rl}f(a,b,c,n)& =\sum _{i=0}^n\lfloor {\displaystyle \frac{ai+b}{c}}\rfloor \\ & =\sum _{i=0}^n\lfloor {\displaystyle \frac{(\lfloor {\displaystyle \frac{a}{c}}\rfloor c+amodc)i+\lfloor {\displaystyle \frac{b}{c}}\rfloor c+bmodc}{c}}\rfloor \\ & =\sum _{i=0}^n(\lfloor {\displaystyle \frac{a}{c}}\rfloor i+\lfloor {\displaystyle \frac{b}{c}}\rfloor +\lfloor {\displaystyle \frac{(amodc)i+(bmodc)}{c}}\rfloor )\\ & ={\displaystyle \frac{n(n+1)}{2}}\lfloor {\displaystyle \frac{a}{c}}\rfloor +(n+1)\lfloor {\displaystyle \frac{b}{c}}\rfloor +\sum _{i=0}^n\lfloor {\displaystyle \frac{amodci+bmodc}{c}}\rfloor \\ & ={\displaystyle \frac{n(n+1)}{2}}\lfloor {\displaystyle \frac{a}{c}}\rfloor +(n+1)\lfloor {\displaystyle \frac{b}{c}}\rfloor +f(amodc,bmodc,c,n)\end{array}}$$

接下来考虑 $a\le c\wedge b\le c$ 的情况

$${\displaystyle \begin{array}{rl}f(a,b,c,n)& =\sum _{i=0}^n\lfloor {\displaystyle \frac{ai+b}{c}}\rfloor \\ & =\sum _{i=0}^n\sum _{j=0}^{\lfloor \frac{ai+b}{c}\rfloor -1}1\\ & =\sum _j\sum _{i=0}^n[j<\lfloor \frac{ai+b}{c}\rfloor ]\\ & =\sum _{j=0}^{\lfloor \frac{an+b}{c}\rfloor -1}\sum _{i=0}^n[j<\lfloor \frac{ai+b}{c}\rfloor ]\end{array}}$$

考虑限制条件 $j<\lfloor \frac{ai+b}{c}\rfloor$

$${\displaystyle \begin{array}{rl}j<\lfloor \frac{ai+b}{c}\rfloor & \iff j<\frac{ai+b}{c}\\ & \iff j+1\le \frac{ai+b}{c}\\ & \iff jc+c\le ai+b\\ & \iff jc+c-b\le ai\\ & \iff jc+c-b-1<ai\\ & \iff {\displaystyle \frac{jc+c-b-1}{a}}<i\\ & \iff \lfloor {\displaystyle \frac{jc+c-b-1}{a}}\rfloor <i\end{array}}$$

记 ${\displaystyle m=\lfloor \frac{an+b}{c}\rfloor ,t=\lfloor {\displaystyle \frac{jc+c-b-1}{a}}\rfloor }$，则

$${\displaystyle \begin{array}{rl}f(a,b,c,n)& =\sum _{j=0}^{\lfloor \frac{an+b}{c}\rfloor -1}\sum _{i=0}^n[j<\lfloor \frac{ai+b}{c}\rfloor ]\\ & =\sum _{j=0}^{m-1}\sum _{i=0}^n[i>t]\\ & =\sum _{j=0}^{m-1}(n-t)\\ & =nm-\sum _{j=0}^{m-1}\lfloor {\displaystyle \frac{jc+c-b-1}{a}}\rfloor \\ & =nm-f(c,c-b-1,a,m-1)\end{array}}$$

观察到当 $a>c$ 时，$a\to amodc$；当 $a\le c$ 时，$a,c$ 互换位置，故复杂度为 $\mathcal{O}(\log max(a,c))$。

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考虑函数 $g$。首先考虑 $a>c\vee b>c$ 的情况

$${\displaystyle \begin{array}{rl}g(a,b,c,n)& =\sum _{i=0}^n{\lfloor {\displaystyle \frac{ai+b}{c}}\rfloor }^2\\ & =\sum _{i=0}^n{\lfloor {\displaystyle \frac{(\lfloor {\displaystyle \frac{a}{c}}\rfloor c+amodc)i+\lfloor {\displaystyle \frac{b}{c}}\rfloor c+bmodc}{c}}\rfloor }^2\\ & =\sum _{i=0}^n(\lfloor {\displaystyle \frac{a}{c}}\rfloor i+\lfloor {\displaystyle \frac{b}{c}}\rfloor +\lfloor {\displaystyle \frac{(amodc)i+(bmodc)}{c}}\rfloor {)}^2\\ & ={\displaystyle \frac{n(n+1)(2n+1)}{6}}{\lfloor {\displaystyle \frac{a}{c}}\rfloor }^2+{\displaystyle \frac{n(n+1)}{2}}{\lfloor {\displaystyle \frac{b}{c}}\rfloor }^2+g(amodc,bmodc,c,n)\\ & +{\displaystyle \frac{n(n+1)}{2}}\lfloor {\displaystyle \frac{a}{c}}\rfloor \lfloor {\displaystyle \frac{b}{c}}\rfloor +2\lfloor {\displaystyle \frac{a}{c}}\rfloor h(amodc,bmodc,c,n)+2\lfloor {\displaystyle \frac{b}{c}}\rfloor f(amodc,bmodc,c,n)\end{array}}$$

接下来考虑 $a\le c\wedge b\le c$ 的情况，观察到

$$n^2=2{\displaystyle \frac{n(n+1)}{2}}-n=2\left(\sum _{i=1}^{n}i\right)-n$$

那么

$${\displaystyle \begin{array}{rl}g(a,b,c,n)& =\sum _{i=0}^n{\lfloor {\displaystyle \frac{ai+b}{c}}\rfloor }^2\\ & =\sum _{i=0}^n[2\left(\sum _{j=1}^{\lfloor \frac{ai+b}{c}\rfloor }j\right)-\lfloor \frac{ai+b}{c}\rfloor ]\end{array}}$$

考虑处理 ${\displaystyle \sum _{i=0}^n\sum _{j=1}^{\lfloor \frac{ai+b}{c}\rfloor }j}$，记 ${\displaystyle m=\lfloor \frac{an+b}{c}\rfloor ,t=\lfloor {\displaystyle \frac{jc+c-b-1}{a}}\rfloor }$

$${\displaystyle \begin{array}{rl}& \sum _{i=0}^n\sum _{j=1}^{\lfloor \frac{ai+b}{c}\rfloor }j\\ =& \sum _{i=0}^n\sum _{j=0}^{\lfloor \frac{ai+b}{c}\rfloor -1}(j+1)\\ =& \sum _{j=0}^{m-1}(j+1)\sum _{i=0}^n[i>t]\\ =& \sum _{j=0}^{m-1}(j+1)(n-t)\\ =& {\displaystyle \frac{1}{2}}nm(m+1)+h(c,c-b-1,a,m-1)-f(c,c-b-1,a,m-1)\end{array}}$$

那么

$${\displaystyle \begin{array}{rl}g(a,b,c,n)& =\sum _{i=0}^n[2\left(\sum _{j=1}^{\lfloor \frac{ai+b}{c}\rfloor }j\right)-\lfloor \frac{ai+b}{c}\rfloor ]\\ & =nm(m+1)+2h(c,c-b-1,a,m-1)-2f(c,c-b-1,a,m-1)+f(a,b,c,n)\end{array}}$$

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考虑函数 $h$。首先考虑 $a>c\vee b>c$ 的情况

$${\displaystyle \begin{array}{rl}h(a,b,c,n)& =\sum _{i=0}^{n}i\lfloor {\displaystyle \frac{ai+b}{c}}\rfloor \\ & =\sum _{i=0}^{n}i\lfloor {\displaystyle \frac{(\lfloor {\displaystyle \frac{a}{c}}\rfloor c+amodc)i+\lfloor {\displaystyle \frac{b}{c}}\rfloor c+bmodc}{c}}\rfloor \\ & =\sum _{i=0}^{n}i(\lfloor {\displaystyle \frac{a}{c}}\rfloor i+\lfloor {\displaystyle \frac{b}{c}}\rfloor +\lfloor {\displaystyle \frac{(amodc)i+(bmodc)}{c}}\rfloor )\\ & ={\displaystyle \frac{n(n+1)(2n+1)}{6}}\lfloor {\displaystyle \frac{a}{c}}\rfloor +{\displaystyle \frac{n(n+1)}{2}}\lfloor {\displaystyle \frac{b}{c}}\rfloor +\sum _{i=0}^{n}i\lfloor {\displaystyle \frac{amodci+bmodc}{c}}\rfloor \\ & ={\displaystyle \frac{n(n+1)(2n+1)}{6}}\lfloor {\displaystyle \frac{a}{c}}\rfloor +{\displaystyle \frac{n(n+1)}{2}}\lfloor {\displaystyle \frac{b}{c}}\rfloor +h(amodc,bmodc,c,n)\end{array}}$$

接下来考虑 $a\le c\wedge b\le c$ 的情况，记 ${\displaystyle m=\lfloor \frac{an+b}{c}\rfloor ,t=\lfloor {\displaystyle \frac{jc+c-b-1}{a}}\rfloor }$

$${\displaystyle \begin{array}{rl}h(a,b,c,n)& =\sum _{i=0}^{n}i\lfloor {\displaystyle \frac{ai+b}{c}}\rfloor \\ & =\sum _{j=0}^{m-1}\sum _{i=0}^n[i>t]i\\ & =\sum _{j=0}^{m-1}{\displaystyle \frac{(n+t+1)(n-t)}{2}}\\ & ={\displaystyle \frac{1}{2}}[mn(n+1)-\sum _{i=0}^n{t}^2-\sum _{i=0}^{n}t]\\ & ={\displaystyle \frac{1}{2}}[mn(n+1)-g(c,c-b-1,a,m-1)-f(c,c-b-1,a,m-1)]\end{array}}$$