类欧几里得算法
类欧几里得算法
基础
求
\[f(a,b,c,n)=\sum_{i=0}^n \left\lfloor \dfrac{ai+b}{c}\right\rfloor
\]
\(a\ge c\)或\(b\ge c\)
\[\begin{aligned}
f(a,b,c,n)&=\sum_{i=0}^n \left\lfloor \dfrac{ai+b}{c}\right\rfloor\\
&=\sum_{i=0}^n \left\lfloor \dfrac{c\left\lfloor\frac{a}{c}\right\rfloor i+a\bmod c\times i+c\left\lfloor\frac{b}{c}\right\rfloor+b\bmod c}{c}\right\rfloor\\
&=\sum_{i=0}^n\left\lfloor\dfrac{a}{c}\right\rfloor i+\left\lfloor\frac{b}{c}\right\rfloor+\left\lfloor \dfrac{a\bmod c\times i+b\bmod c}{c}\right\rfloor\\
&=\left\lfloor\dfrac{a}{c}\right\rfloor\dfrac{n(n+1)}{2}+\left\lfloor\dfrac{b}{c}\right\rfloor(n+1)+f(a\bmod c,b\bmod c,c,n)
\end{aligned}
\]
\(a<c\)且\(b<c\)
把\(\left\lfloor \dfrac{ai+b}{c}\right\rfloor\)替换为\(\sum\limits_{j=0}\left[ j<\left\lfloor \dfrac{ai+b}{c}\right\rfloor\right]\)，其中\(j\)的上界为\(\left\lfloor \dfrac{an+b}{c}\right\rfloor-1\)。也就是求
\[f(a,b,c,n)=\sum_{i=0}^n\sum_{j=0}\left[j<\left\lfloor \dfrac{ai+b}{c}\right\rfloor\right]
\]
交换求和号
\[f(a,b,c,d)=\sum_{j=0}^{\lfloor\frac{an+b}{c}\rfloor-1}\sum_{i=0}^n\left[j<\left\lfloor\dfrac{ai+b}{c}\right\rfloor\right]\\
\]
乱搞一下后面那个东西，通过下取整与不等号的变换把\(i\)搞掉
\[j<\left\lfloor\dfrac{ai+b}{c}\right\rfloor \Leftrightarrow j+1\le \left\lfloor\dfrac{ai+b}{c}\right\rfloor\Leftrightarrow j+1\le \dfrac{ai+b}{c}\\
\Leftrightarrow \dfrac{cj+c-b}{a}\le i \Leftrightarrow i\ge \dfrac{cj+c-b}{a}\Leftrightarrow i>\left\lfloor\dfrac{cj+c-b-1}{a}\right\rfloor
\]
而\(i\)的上界是确定的，为\(n\)，也就是说\(j<\left\lfloor\dfrac{ai+b}{c}\right\rfloor\)的\(i\)的个数等于\(n-\left\lfloor\dfrac{cj+c-b-1}{a}\right\rfloor\)。
\[\begin{aligned}
f(a,b,c,d)&=\sum_{j=0}^{\lfloor \frac{an+b }{c} \rfloor-1}\sum_{i=0}^n\left[j<\left\lfloor\dfrac{ai+b}{c}\right\rfloor\right]\\
&=\sum_{j=0}^{\lfloor \frac{an+b }{c} \rfloor-1}n-\left\lfloor\dfrac{cj+c-b-1}{a}\right\rfloor\\
&= \left\lfloor\dfrac{an+b}{c}\right\rfloor n-\sum_{j=0}^{\lfloor\frac{an+b}{c}\rfloor-1}\left\lfloor\dfrac{cj+c-b-1}{a}\right\rfloor\\
&= \left\lfloor\dfrac{an+b}{c}\right\rfloor n-f(c,c-b-1,a,\lfloor\frac{an+b}{c}\rfloor -1)
\end{aligned}
\]
以后方便起见记\(\left\lfloor\dfrac{an+b}{c}\right\rfloor=m\)，记\(\left\lfloor\dfrac{cj+c-b-1}{a}\right\rfloor=t\)，即
\[f(a,b,c,n)=mn-f(c,c-b-1,a,m-1)
\]
递归求解即可，边界为\(a=0\)
复杂度\(O(\log n)\)，证明类似欧几里得算法，故算法得名
扩展
求
\[g(a,b,c,n)=\sum_{i=0}^n i\left\lfloor \dfrac{ai+b}{c}\right\rfloor\\
h(a,b,c,n)=\sum_{i=0}^n \left\lfloor \dfrac{ai+b}{c}\right\rfloor^2
\]
求\(g(a,b,c,n)\)
\(a\ge c\)或\(b\ge c\)
\[\begin{aligned}
g(a,b,c,n)&=\sum_{i=0}^n i\left\lfloor \dfrac{ai+b}{c}\right\rfloor\\
&=\sum_{i=0}^n i\left\lfloor \dfrac{c\left\lfloor\frac{a}{c}\right\rfloor i+a\bmod c\times i+c\left\lfloor\frac{b}{c}\right\rfloor+b\bmod c}{c}\right\rfloor\\
&=\sum_{i=0}^n\left\lfloor\dfrac{a}{c}\right\rfloor i^2+\left\lfloor\frac{b}{c}\right\rfloor i+\left\lfloor \dfrac{a\bmod c\times i+b\bmod c}{c}\right\rfloor i\\
&=\left\lfloor\dfrac{a}{c}\right\rfloor\dfrac{n(n+1)(2n+1)}{6}+\left\lfloor\dfrac{b}{c}\right\rfloor\dfrac{n(n+1)}{2}+g(a\% c,b\% c,c,n)
\end{aligned}
\]
\(a<c\)且\(b<c\)
\(m=\left\lfloor\dfrac{an+b}{c}\right\rfloor\)，\(t=\left\lfloor\dfrac{cj+c-b-1}{a}\right\rfloor\)
\[\begin{aligned}
g(a,b,c,n)&=\sum_{i=0}^ni\sum_{j=0}\left[j<\left\lfloor \dfrac{ai+b}{c}\right\rfloor\right]\\
&=\sum_{j=0}^{m-1}\sum_{i=0}^ni\left [j<\left\lfloor\dfrac{ai+b}{c}\right\rfloor\right]\\
&=\sum_{j=0}^{m-1}\sum_{i=0}^ni [i>t]\\
&=\sum_{j=0}^{m-1}\frac{n(n+1)}{2}-\frac{t(t+1)}{2}\\
&=\dfrac{1}{2}[mn(n+1)-\sum_{j=0}^{m-1}t^2-\sum_{j=0}^{m-1}t\ ]\\
&=\dfrac{1}{2}[mn(n+1)-h(c,c-b-1,a,m-1)-f(c,c-b-1,a,m-1)\ ]
\end{aligned}
\]
递归边界：\(a=0\)
求\(h(a,b,c,n)\)
\(a\ge c\)或\(b\ge c\)
\[\begin{aligned}
h(a,b,c,n)&=\sum_{i=0}^n \left\lfloor \dfrac{ai+b}{c}\right\rfloor^2\\
&=\sum_{i=0}^n\left(\left\lfloor\dfrac{a}{c}\right\rfloor i+\left\lfloor\frac{b}{c}\right\rfloor +\left\lfloor \dfrac{a\bmod c\times i+b\bmod c}{c}\right\rfloor\right)^2 \\
&=\sum_{i=0}^n\left\lfloor\dfrac{a}{c}\right\rfloor^2 i^2+\left\lfloor\frac{b}{c}\right\rfloor^2 +\left\lfloor \dfrac{a \% c\times i+b\% c}{c}\right\rfloor^2+2\left\lfloor\dfrac{a}{c}\right\rfloor\left\lfloor\frac{b}{c}\right\rfloor i\\&\ \ \ \ \ +2\left\lfloor\dfrac{a}{c}\right\rfloor\left\lfloor \dfrac{a\% c\times i+b\% c}{c}\right\rfloor i+2\left\lfloor\dfrac{b}{c}\right\rfloor\left\lfloor \dfrac{a\% c\times i+b\% c}{c}\right\rfloor\\
&=\frac{n(n+1)(2n+1)}{6}\left\lfloor\dfrac{a}{c}\right\rfloor^2+(n+1)\left\lfloor\frac{b}{c}\right\rfloor^2+n(n+1)\left\lfloor\dfrac{a}{c}\right\rfloor\left\lfloor\frac{b}{c}\right\rfloor+2\left\lfloor\dfrac{a}{c}\right\rfloor g(a\% c,b\%c,c,n)\\
&\ \ \ \ \ +2\left\lfloor\dfrac{b}{c}\right\rfloor f(a\%c,b\%c,c,n)+h(a\%c,b\%c,c,n)
\end{aligned}
\]
\(a<c\)且\(b<c\)
\(m=\left\lfloor\dfrac{an+b}{c}\right\rfloor\)，\(t=\left\lfloor\dfrac{cj+c-b-1}{a}\right\rfloor\)
\[h(a,b,c,n)=\sum_{i=0}^n \left\lfloor \dfrac{ai+b}{c}\right\rfloor^2
\]
将\(n^2\)写成\(\left(2\sum\limits_{i=1}^n i\right)-n\)
\[\begin{aligned}
h(a,b,c,n)&=\sum_{i=0}^n\left(2\sum_{j=1}^{\left\lfloor \frac{ai+b}{c}\right\rfloor}j\right)-\left\lfloor \dfrac{ai+b}{c}\right\rfloor\\
&=\left(2\sum_{i=0}^n\sum_{j=0}^{\left\lfloor \frac{ai+b}{c}\right\rfloor-1}j+1\right)-f(a,b,c,n)\\
&=\left(2\sum_{j=0}^{m-1}\sum_{i=0}^n (j+1)\left[j< \left\lfloor \frac{ai+b}{c}\right\rfloor\right]\right)-f(a,b,c,n)
\end{aligned}
\]
其中
\[\begin{aligned}
\sum_{j=0}^{m-1}\sum_{i=0}^n (j+1)\left[j< \left\lfloor \frac{ai+b}{c}\right\rfloor\right]&=\sum_{j=1}^{m-1}(j+1)\sum_{i=0}^n \left[j< \left\lfloor \frac{ai+b}{c}\right\rfloor\right]\\
&=\sum_{j=0}^{m-1}(j+1)\sum_{i=0}^n[i>t]\\
&=\sum_{j=0}^{m-1}(j+1)(n-t)\\
&=\dfrac{m(m+1)}{2}n-f(c,c-b-1,a,m-1) -g(c,c-b-1,a,m-1)
\end{aligned}
\]
所以
\[\begin{aligned}
h(a,b,c,n)&=m(m+1)n-2f(c,c-b-1,a,m-1)-2g(c,c-b-1,a,m-1)-f(a,b,c,n)
\end{aligned}
\]
递归边界：\(a=0\)
实现注意事项
因为\(g,h\)有相互调用，递归时把一组\((a,b,c,n)\)的\(f,g,h\)都求出来
