类欧几里得算法及其拓展
\[\begin{aligned}
F(a, b, c, n)&=\sum_{i=0}^n \lfloor \frac{ai+b}c\rfloor\\
&=\frac{n(n+1)}2\lfloor\frac ac\rfloor + (n+1)\lfloor\frac bc\rfloor+F(a',b',c,n)\\
F(a, b, c, n)&=\sum_{i=0}^n \lfloor \frac{ai+b}c\rfloor\\
&=\sum_{i=0}^n\lfloor\frac{ai+b}c\rfloor\\
&=\sum_{i=0}^n\sum_{j=1}^m[cj\leq ai+b']\\
&=\sum_{j=1}^m[ai>cj-b-1]\\
&=\sum_{j=0}^{m-1}n-\lfloor\frac{cj+c-b-1}a\rfloor\\
&=nm-F(c, c-b-1,a, m-1)
\end{aligned}
\]
\[\begin{aligned}
G(a,b,c,n)&=\sum_{i=0}^n[\frac{ai+b}c]^2\\
&=\sum_{i=0}^n([\frac ac]i+\frac bc+[\frac{a'i+b'}c])^2\\
&=\sum_{i=0}^n[\frac ac]^2i^2+2[\frac ac][\frac bc]i+2[\frac ac]i[\frac{a'i+b'}c]+[\frac bc]^2+2[\frac bc][\frac{a'i+b'}c]+[\frac{a'i+b'}c]^2\\
&=[\frac ac]^2\frac{n(n+1)(2n+1)}6+[\frac ac][\frac bc]n(n+1)+2[\frac ac]H(a',b',c,n)+(n+1)[\frac bc]+2[\frac bc]F(a',b',c,n)+G(a',b',c,n)
\end{aligned}
\]
\[\begin{aligned}
G(a,b,c,n)&=\sum_{i=0}^n[\frac{ai+b}c]^2\\
&=\sum_{i=0}^n2(\sum_{j=1}^m[cj\leq ai+b]j)-[\frac{ai+b}c]\\
&=\sum_{i=0}^n2(\sum_{j=0}^{m-1}[ai>cj+c-b-1](j+1))-[\frac{ai+b}c]\\
&=-F(a, b, c, n)+\sum_{j=0}^{m-1}2(n-[\frac {cj+c-b-1}a])(j+1)\\
&=-F(a, b, c, n)+m(m+1)n-2H(c, c-b-1, a, m-1)-2F(c, c-b-1, a, m-1)
\end{aligned}
\]
\[\begin{aligned}
H(a,b,c,n)&=\sum_{i=0}^ni[\frac{ai+b}c]\\
&=\sum_{i=0}^ni^2[\frac ac]+i[\frac bc]+i[\frac{a'i+b'}c]\\
&=[\frac ac]\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2[\frac bc]+H(a',b',c,n)
\end{aligned}
\]
\[\begin{aligned}
H(a,b,c,n)&=\sum_{i=0}^ni[\frac{ai+b}c]\\
&=\sum_{i=0}^n\sum_{j=1}^m [cj\leq ai+b]i\\
&=\sum_{j=0}^{m-1}\sum_{i=0}^n[ai>cj+c-b-1]i\\
&=\sum_{j=0}^{m-1}\frac{n(n+1)}2-\frac12[\frac{cj+c-b-1}a]^2-\frac12[\frac{cj+c-b-1}a]\\
&=\frac12mn(n+1)-\frac12G(c,c-b-1,a,m-1)-\frac12F(c,c-b-1,a,m-1)
\end{aligned}
\]
一些拓展（我自己想到的）：
\[\begin{aligned}
G_{x,y}(a,b,c,n)&=\sum_{i=0}^n[\frac{ai+b}c]^\overline xi^\overline y\\
&=\sum_{i=0}^n([\frac ac]i+\frac bc+[\frac{a'i+b'}c])^\overline xi^\overline y\\
&=\sum_{i=0}^n\sum_{d+e+f=x}^kcoef_{d,e,f}[\frac ac]^di^\overline {d+y}[\frac bc]^e[\frac{a'i+b'}c]^\overline f\\
\end{aligned}
\]
\[\begin{aligned}
G_{x,y}(a,b,c,n)&=\sum_{i=0}^n[\frac{ai+b}c]^\overline xi^\overline y\\
&=\sum_{i=0}^nx\sum_{j=1}^m[cj\leq ai+b]j^\overline{x-1}i^\overline y\\
&=\sum_{j=0}^{m-1}\frac x{y+1}(n^\overline{y+1}-[\frac{cj+c-b-1}a]^\overline {y+1}j^\overline {x-1})\\
&=\frac x{y+1}(mn^\overline{y+1}-G_{y+1,x-1}(c,c-b-1,a,n))
\end{aligned}
\]
只需要维护所有的\(G_{x,y}\)其中\(x+y\leq k\)即可。
究极加倍版：
\[\begin{aligned}
G_{x,y}(A,B,C,N,X,Y)&=\sum_{v_k=0}^{n_k}\prod_k[\frac{a_kv_k+b_k}{c_k}]^\overline {x_k}v_k^\overline {y_k}\\
&=\sum_{v_k=0}^{n_k}\prod_k([\frac {a_k}{c_k}]v_k+\frac {b_k}{c_k}+[\frac{{a_k}'i_k+{b_k}'}{c_k}])^\overline {x_k}v_k^\overline {y_k}\\
&=\sum_{v_k}^{n_k}\sum_{D,E,F}coef_{D,E,F}[\frac {a_k}{c_k}]^{d_k}i^\overline {d_k+y_k}[\frac {b_k}{c_k}]^{e_k}[\frac{a_k'v_k+b_k'}{c_k}]^\overline {f_k}\\
\end{aligned}
\]
等心情好了考虑出成题。
