[egin{aligned}
F(a, b, c, n)&=sum_{i=0}^n lfloor frac{ai+b}c
floor\
&=frac{n(n+1)}2lfloorfrac ac
floor + (n+1)lfloorfrac bc
floor+F(a',b',c,n)\
F(a, b, c, n)&=sum_{i=0}^n lfloor frac{ai+b}c
floor\
&=sum_{i=0}^nlfloorfrac{ai+b}c
floor\
&=sum_{i=0}^nsum_{j=1}^m[cjleq ai+b']\
&=sum_{j=1}^m[ai>cj-b-1]\
&=sum_{j=0}^{m-1}n-lfloorfrac{cj+c-b-1}a
floor\
&=nm-F(c, c-b-1,a, m-1)
end{aligned}
]
[egin{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]^2frac{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}
]
[egin{aligned}
G(a,b,c,n)&=sum_{i=0}^n[frac{ai+b}c]^2\
&=sum_{i=0}^n2(sum_{j=1}^m[cjleq 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}
]
[egin{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}
]
[egin{aligned}
H(a,b,c,n)&=sum_{i=0}^ni[frac{ai+b}c]\
&=sum_{i=0}^nsum_{j=1}^m [cjleq 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）：
[egin{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}^nsum_{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}
]
[egin{aligned}
G_{x,y}(a,b,c,n)&=sum_{i=0}^n[frac{ai+b}c]^overline xi^overline y\
&=sum_{i=0}^nxsum_{j=1}^m[cjleq 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+yleq k)即可。
究极加倍版：
[egin{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。