[P5170] 类欧几里得算法

“类欧几里得算法”第二题 P5170

【题意】已知\(n,a,b,c\),求

\[\begin{aligned}
f_{1}(a,b,c,n)&=\sum_{i=0}^n\lfloor\dfrac{ai+b}{c}\rfloor\\
f_{2}(a,b,c,n)&=\sum_{i=0}^n\lfloor\dfrac{ai+b}{c}\rfloor^2\\
f_{3}(a,b,c,n)&=\sum_{i=0}^n\lfloor\dfrac{ai+b}{c}\rfloor*i\\
\end{aligned}
\]

【预备】

设\(m=\lfloor\dfrac{a\times n+b}{c}\rfloor,\ t_{1}=\lfloor\dfrac{a}{c}\rfloor,\ t_{2}=\lfloor\dfrac{b}{c}\rfloor\)。

定义\([\text{expression}]\)为真值表达式。

简单的引理，当\(a,b,c\in Z​\)时

• \(a\le\lfloor\dfrac{b}{c}\rfloor \Rightarrow ac\le b ​\)。
• \(a< bc \Rightarrow t_{1}<b\)。

【限界】a=0时直接计算。

【式一】对原式变形

\[f_{1}(a,b,c,n)
=\sum_{i=0}^nt_{1}\times i+t_{2}+\lfloor\dfrac{(a\bmod c)\times i+(b\bmod c)}{c}\rfloor\\
=t_{1}\times\dfrac{n(n+1)}{2}+t_{2}\times(n+1)+ f_{1}(a\bmod c,b\bmod c,c,n)
\]

当\(t1=t2=0\)即\(a<c\)且\(b<c\)时，

\[f_{1}(a,b,c,n)=
\sum_{i=0}^n\sum_{j=1}^m [j\le\dfrac{ai+b}{c}]
=\sum_{i=0}^n\sum_{j=1}^m [j\le\lfloor\dfrac{ai+b}{c}\rfloor]\\
=\sum_{j=1}^m\sum_{i=0}^n[j\le\lfloor\dfrac{ai+b}{c}\rfloor]
=\sum_{j=1}^m\sum_{i=0}^n[cj-b\le ai]
=\sum_{j=1}^m\sum_{i=0}^n[cj-b-1< ai]\\
=\sum_{j=1}^m\sum_{i=0}^n[\lfloor\dfrac{cj-b-1}{a}\rfloor< i]
=\sum_{j=1}^m(n-\lfloor\dfrac{cj-b-1}{a}\rfloor)\\
=mn-\sum_{i=1}^m\lfloor\dfrac{ci-b-1}{a}\rfloor
=mn-\sum_{i=0}^{m-1}\lfloor\dfrac{ci+c-b-1}{a}\rfloor\\
=mn-f_{1}(c,c-b-1,a,m-1)
\]

【式二】对原式变形

\[f_{2}(a,b,c,n)
=\sum_{i=0}^n(t_{1}\times i+t_{2}+\lfloor\dfrac{(a\bmod c)\times i+(b\bmod c)}{c}\rfloor)^2\\
=\sum_{i=0}^n \begin{cases}
(t_{1}\times i)^2+t_{2}^2+\lfloor\dfrac{(a\bmod c)\times i+(b\bmod c)}{c}\rfloor^2\\
+2t_{1}t_{2}*i\\
+2t_{1}i\lfloor\dfrac{(a\bmod c)\times i+(b\bmod c)}{c}\rfloor\\
+2t_{2}\lfloor\dfrac{(a\bmod c)\times i+(b\bmod c)}{c}\rfloor
\end{cases}\\
=\begin{cases}
t_{1}^2\sum_{i=0}^ni^2+t_{2}^2*(n+1)+f_{2}(a\bmod c,b\bmod c,c,n)\\
+2t_{1}t_{2}\sum_{i=0}^n i\\
+2t_{1}f_{3}(a\bmod c,b \bmod c,c,n)\\
+2t_{2}f_{1}(a\bmod c,b \bmod c,c,n)
\end{cases}\\
\]

当\(t1=t2=0\)即\(a<c\)且\(b<c​\)时，

\[f_{2}(a,b,c,d)=
\sum_{i=0}^n\lfloor\dfrac{ai+b}{c}\rfloor
=\sum_{i=0}^n\sum_{j=1}^m\sum_{k=1}^m[\lfloor\dfrac{cj-b-1}{a}\rfloor< i\text{ and }\lfloor\dfrac{ck-b-1}{a}\rfloor< i]\\
=\sum_{i=0}^n\sum_{j=1}^m\sum_{k=1}^m [\max(\lfloor\dfrac{cj-b-1}{a}\rfloor,\lfloor\dfrac{ck-b-1}{a}\rfloor)< i]\\
=\sum_{j=1}^m\sum_{k=1}^m \sum_{i=0}^n[\max(\lfloor\dfrac{cj-b-1}{a}\rfloor,\lfloor\dfrac{ck-b-1}{a}\rfloor)< i]\\
=\sum_{j=1}^m\sum_{k=1}^m n-\max(\lfloor\dfrac{cj-b-1}{a}\rfloor,\lfloor\dfrac{ck-b-1}{a}\rfloor)\\
=nm^2-\sum_{j=1}^m\sum_{k=1}^m\max(\lfloor\dfrac{cj-b-1}{a}\rfloor,\lfloor\dfrac{ck-b-1}{a}\rfloor)\\
=nm^2-2*\sum_{j=1}^m\lfloor\dfrac{cj-b-1}{a}\rfloor*(j-1)-\sum_{j=1}^m \lfloor\dfrac{cj-b-1}{a}\rfloor\\
=nm^2-\sum_{j=0}^{m-1} \lfloor\dfrac{cj+c-b-1}{a}\rfloor*j-\sum_{j=0}^{m-1} \lfloor\dfrac{cj+c-b-1}{a}\rfloor\\
=nm^2-f_{1}(c,c-b-1,a,m-1)-2*f_{3}(c,c-b-1,a,m-1)
\]

【式三】对原式变形

\[f_{3}(a,b,c,n)=\sum_{i=0}^n\lfloor\dfrac{ai+b}{c}\rfloor*i
=\sum_{i=0}^n (t_{1}\times i+t_{2}+\lfloor\dfrac{(a\bmod c)\times i+(b\bmod c)}{c}\rfloor)*i\\
=\sum_{i=0}^n t_{1}\times i^2+t_{2}\times i+\lfloor\dfrac{(a\bmod c)\times i+(b\bmod c)}{c}\rfloor\times i\\
=t_{1}\sum_{i=0}^ni^2+t_{2}\sum_{i=0}^ni+f_{3}(a\bmod c,b\bmod c,c,n)
\]

当\(t1=t2=0\)即\(a<c\)且\(b<c\)时，定义\(p(j)=\lfloor\dfrac{cj-b-1}{a}\rfloor\)。

\[f3(a,b,c,d)
=\sum_{i=0}^n\sum_{j=1}^m [j\le\lfloor\dfrac{ai+b}{c}\rfloor]*i
=\sum_{j=1}^m\sum_{i=0}^n[\lfloor\dfrac{cj-b-1}{a}\rfloor< i]*i\\
=\sum_{j=1}^m\sum_{i=p(j)+1}^ni
=\sum_{j=1}^m \dfrac{1}{2}(p(j)+1+n)(n-p(j))\\
=\sum_{j=1}^m \dfrac{1}{2}(n\times p(j)-p^2(j)+n-p(j)+n^2-n\times p(j))\\
=\sum_{j=1}^m \dfrac{1}{2}(-p^2(j)+n-p(j)+n^2)\\
=\dfrac{-f_{2}(c,c-b-1,a,m-1)-f_{1}(c,c-b-1,a,m-1)+nm+n^2m}{2}
\]

【时间复杂度】如果每个都单独搜索的话，大概因该会炸吧。。考虑到三个函数的递归模式都很**，干脆用一个结构体存下三个值。再参考第一题的分析，状态数目是\(\log\)级别的。

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const LL mod=998244353;
const LL I2=499122177;
const LL I6=166374059;

inline LL s1(LL n) {return I2*n%mod*(n+1)%mod;}
inline LL s2(LL n) {return I6*n%mod*(n+1)%mod*(n+n+1)%mod;}

struct node {
	LL f1,f2,f3;
	node(LL f1=0,LL f2=0,LL f3=0):f1(f1),f2(f2),f3(f3){
//		assert(0<=f1 && 0<=f2 && 0<=f3);
//		assert(f1<mod && f2<mod && f3<mod);
	}
};
node dfs(LL a,LL b,LL c,LL n) {
	if(!a||!n) return node(
		(b/c)*(n+1)%mod,
		(b/c)*(b/c)%mod*(n+1)%mod,
		(b/c)*s1(n)%mod
	);
	if(a>=c || b>=c) {
		LL t1=a/c, t2=b/c;
		node tmp=dfs(a%c,b%c,c,n);
		return node(
			(t1*s1(n)%mod+t2*(n+1)%mod+tmp.f1)%mod,
			(t1*t1%mod*s2(n)%mod
				+t2*t2%mod*(n+1)%mod
				+tmp.f2
				+2*t1%mod*t2%mod*s1(n)%mod
				+2*t1%mod*tmp.f3%mod
				+2*t2%mod*tmp.f1%mod
			)%mod,
			(t1*s2(n)%mod+t2*s1(n)%mod+tmp.f3)%mod
		);
	} else {
		LL m=(a*n+b)/c;
		node tmp=dfs(c,c-b-1,a,m-1);
		return node(
			(n*m%mod-tmp.f1+mod)%mod,
			(n*m%mod*m%mod-tmp.f1-2*tmp.f3%mod+mod+mod)%mod,
			(n*m%mod+n*n%mod*m%mod-tmp.f1-tmp.f2+mod+mod)%mod*I2%mod
		);
	}
} 

int main() {
	int T,a,b,c,n;
	scanf("%d",&T);
	while(T--) {
		scanf("%d%d%d%d",&n,&a,&b,&c);
		node tmp=dfs(a,b,c,n);
		printf("%lld %lld %lld\n",tmp.f1,tmp.f2,tmp.f3) ;
	}
	return 0;
}