类欧几里得算法

模板题

设

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

• 当 $a=0$

$$\begin{gathered} f(a,b,c,n)=(n+1)\lfloor \frac{b}{c}\rfloor \\ g(a,b,c,n)=(n+1){\lfloor \frac{b}{c}\rfloor }^2 \\ h(a,b,c,n)=\frac{n(n+1)}{2}\lfloor \frac{b}{c}\rfloor \end{gathered}$$

• 当 $a\ge c$ 或 $b\ge c$

令 $a^{\prime }=a\mathrm{m}\mathrm{o}\mathrm{d}c,b^{\prime }=b\mathrm{m}\mathrm{o}\mathrm{d}c$

$$\begin{gathered} f(a,b,c,n)=\sum _{i=0}^n\lfloor \frac{ai+b}{c}\rfloor \\ =\sum _{i=0}^n\left(\lfloor \frac{a}{c}\rfloor i+\lfloor \frac{b}{c}\rfloor +\lfloor \frac{a^{\prime }i+b^{\prime }}{c}\rfloor \right) \\ =\frac{n(n+1)}{2}\lfloor \frac{a}{c}\rfloor +(n+1)\lfloor \frac{b}{c}\rfloor +f(a^{\prime },b^{\prime },c,n) \end{gathered}$$
$$\begin{gathered} g(a,b,c,n)=\sum _{i=0}^n{\lfloor \frac{ai+b}{c}\rfloor }^2 \\ =\sum _{i=0}^n{\left(\lfloor \frac{a}{c}\rfloor i+\lfloor \frac{b}{c}\rfloor +\lfloor \frac{a^{\prime }i+b^{\prime }}{c}\rfloor \right)}^2 \\ =\sum _{i=0}^n\left({\lfloor \frac{a}{c}\rfloor }^2{i}^2+{\lfloor \frac{b}{c}\rfloor }^2+{\lfloor \frac{a^{\prime }i+b^{\prime }}{c}\rfloor }^2+2i\lfloor \frac{a}{c}\rfloor \lfloor \frac{b}{c}\rfloor +2i\lfloor \frac{a}{c}\rfloor \lfloor \frac{a^{\prime }i+b^{\prime }}{c}\rfloor +2\lfloor \frac{b}{c}\rfloor \lfloor \frac{a^{\prime }i+b^{\prime }}{c}\rfloor \right) \\ =\frac{n(n+1)(2n+1)}{6}{\lfloor \frac{a}{c}\rfloor }^2+(n+1){\lfloor \frac{b}{c}\rfloor }^2+g(a^{\prime },b^{\prime },c,n)+n(n+1)\lfloor \frac{a}{c}\rfloor \lfloor \frac{b}{c}\rfloor +2\lfloor \frac{a}{c}\rfloor h(a^{\prime },b^{\prime },c,n)+2\lfloor \frac{b}{c}\rfloor f(a^{\prime },b^{\prime },c,n) \end{gathered}$$
$$\begin{gathered} h(a,b,c,n)=\sum _{i=0}^{n}i\lfloor \frac{ai+b}{c}\rfloor \\ =\sum _{i=0}^{n}i\left(\lfloor \frac{a}{c}\rfloor i+\lfloor \frac{b}{c}\rfloor +\lfloor \frac{a^{\prime }i+b^{\prime }}{c}\rfloor \right) \\ =\frac{n(n+1)(2n+1)}{6}\lfloor \frac{a}{c}\rfloor +\frac{n(n+1)}{2}\lfloor \frac{b}{c}\rfloor +h(a^{\prime },b^{\prime },c,n) \end{gathered}$$

• 当 $a<c$ 且 $b<c$

令 $m=\lfloor \frac{an+b}{c}\rfloor$

$$\begin{gathered} f(a,b,c,n)=\sum _{i=0}^n\lfloor \frac{ai+b}{c}\rfloor \\ =\sum _{i=0}^n\sum _{j=1}^m\left[j\le \lfloor \frac{ai+b}{c}\rfloor \right] \\ =\sum _{i=0}^n\sum _{j=1}^m\left[j\le \frac{ai+b}{c}\right] \\ =\sum _{i=0}^n\sum _{j=1}^m\left[cj\le ai+b\right] \\ =\sum _{i=0}^n\sum _{j=0}^{m-1}\left[cj+c\le ai+b\right] \\ =\sum _{i=0}^n\sum _{j=0}^{m-1}\left[cj+c-1<ai+b\right] \\ =\sum _{j=0}^{m-1}\sum _{i=0}^n\left[i>\frac{cj+c-b-1}{a}\right] \\ =\sum _{j=0}^{m-1}\left(n+1-\sum _{i=0}^n\left[i\le \frac{cj+c-b-1}{a}\right]\right) \\ =\sum _{j=0}^{m-1}\left(n-\sum _{i=1}^n\left[i\le \frac{cj+c-b-1}{a}\right]\right) \\ =\sum _{j=0}^{m-1}\left(n-\lfloor \frac{cj+c-b-1}{a}\rfloor \right) \\ =nm-f(c,c-b-1,a,m-1) \end{gathered}$$

剩下两个主体类似，故略去了一些中间步骤。

首先有： $n^2=2{\sum }_{i=1}^{n}i-n$

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

$$\begin{gathered} h(a,b,c,n)=\sum _{i=0}^{n}i\lfloor \frac{ai+b}{c}\rfloor \\ =\sum _{i=0}^{n}i\sum _{j=1}^m\left[j\le \lfloor \frac{ai+b}{c}\rfloor \right] \\ =\sum _{i=0}^{n}i\sum _{j=0}^{m-1}\left[cj+c-1<ai+b\right] \\ =\sum _{j=0}^{m-1}\sum _{i=0}^{n}i\left[cj+c-1<ai+b\right] \\ =\sum _{j=0}^{m-1}\left(\frac{n(n+1)}{2}-\frac{1}{2}\lfloor \frac{cj+c-b-1}{a}\rfloor \left(\lfloor \frac{cj+c-b-1}{a}\rfloor +1\right)\right) \\ =\frac{1}{2}\left(mn(n+1)-g(c,c-b-1,a,m-1)-f(c,c-b-1,a,m-1)\right) \end{gathered}$$

递归算即可。$a,c$ 两维每 $\mathrm{O}(1)$ 次至少减半，所以复杂度是 $\mathrm{O}(\log n)$ 级别的。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cctype>
#define MAXN 10000005
using namespace std;
inline int read()
{
	int ans=0;
	char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
	return ans;
}
const int MOD=998244353,INV2=(MOD+1)/2,INV6=(MOD+1)/6;
typedef long long ll;
struct node{int f,g,h;inline node(const int& f=0,const int& g=0,const int& h=0):f(f),g(g),h(h){}};
node query(int n,int a,int b,int c)
{
	if (!a) return node((n+1ll)*(b/c)%MOD,(n+1ll)*(b/c)%MOD*(b/c)%MOD,n*(n+1ll)/2%MOD*(b/c)%MOD);
	if (a>=c||b>=c)
	{
		node t=query(n,a%c,b%c,c);
		return node(
		(n*(n+1ll)/2%MOD*(a/c)%MOD+(n+1ll)*(b/c)%MOD+t.f)%MOD,
		(n*(n+1ll)%MOD*(2ll*n+1)%MOD*INV6%MOD*(a/c)%MOD*(a/c)%MOD+(n+1ll)*(b/c)%MOD*(b/c)%MOD+t.g+n*(n+1ll)%MOD*(a/c)%MOD*(b/c)%MOD+2ll*(a/c)%MOD*t.h%MOD+2ll*(b/c)%MOD*t.f%MOD)%MOD,
		(n*(n+1ll)%MOD*(2ll*n+1)%MOD*INV6%MOD*(a/c)%MOD+n*(n+1ll)/2%MOD*(b/c)%MOD+t.h)%MOD
		);
	}	
	int m=((ll)a*n+b)/c;
	node t=query(m-1,c,c-b-1,a),ans;
	ans.f=((ll)n*m%MOD+MOD-t.f)%MOD;
	ans.g=(m*(m+1ll)%MOD*n%MOD+MOD-(2ll*t.h%MOD+2ll*t.f%MOD+ans.f)%MOD)%MOD;
	ans.h=(n*(n+1ll)%MOD*m%MOD+MOD-(t.f+t.g)%MOD)*INV2%MOD;
	return ans;
}
int main()
{
	for (int T=read();T;T--)
	{
		int n,a,b,c;
		n=read(),a=read(),b=read(),c=read();
		node ans=query(n,a,b,c);
		printf("%d %d %d\n",ans.f,ans.g,ans.h);
	}
	return 0;
}