类欧几里得算法&洛谷P5170题解
好久没更博客了说，不能这样颓废了(flag
Problem
给定(n,a,b,c),求(f(a,b,c,n)=sumlimits_{i=0}^nlfloorfrac{ai+b}c floor,g(a,b,c,n)=sumlimits_{i=0}^nlfloorfrac{ai+b}c floor^2,h(a,b,c,n)=sumlimits_{i=0}^nilfloorfrac{ai+b}c floor)
Algorithm
先考虑看上去最可做的第一个柿子。
首先可以通过一些trivial的讨论把(a,b)对(c)取模。
[egin{align}
f(a,b,c,n)&=sumlimits_{i=0}^nlfloorfrac{ai+b}c
floor\
&=sumlimits_{i=0}^nlfloorfrac{(a\%c)i+b\%c}c
floor+frac{(n+1)*n}2lfloorfrac ac
floor+(n+1)lfloorfrac bc
floor\
&=f(a\%c,b\%c,c,n)+frac{(n+1)*n}2lfloorfrac ac
floor+(n+1)lfloorfrac bc
floor
end{align}
]
我们接下来考虑(a,b<c)的情况,考虑此时的(lfloorfrac{ai+b}c floor)取值不会超过(n),可以枚举取值。
[egin{align}
f(a,b,c,n)
&=sumlimits_{i=0}^nlfloorfrac{ai+b}c
floor\
&=sumlimits_{i=0}^nsumlimits_{d=1}^{lfloorfrac{an+b}{c}
floor}[lfloorfrac{ai+b}c
floorge d]\
两边大力乘上a^{-1}c
&=sumlimits_{i=0}^nsumlimits_{d=1}^{lfloorfrac{an+b}{c}
floor}[a^{-1}clfloorfrac{ai+b}c
floorge a^{-1}cdgt a^{-1}(cd-1)]\
&=sumlimits_{i=0}^nsumlimits_{d=1}^{lfloorfrac{an+b}{c}
floor}[igt frac{(cd-b-1)}a]\
&=sumlimits_{i=0}^nsumlimits_{d=0}^{lfloorfrac{an+b}{c}
floor-1}[igt frac{(cd+c-b-1)}a]\
&=sumlimits_{d=0}^{lfloorfrac{an+b}{c}
floor-1}(n-lfloorfrac{(cd+c-b-1)}a
floor)\
&=nlfloorfrac{an+b}{c}
floor-sumlimits_{d=0}^{lfloorfrac{an+b}{c}
floor-1}lfloorfrac{(cd+c-b-1)}a
floor\
&=nlfloorfrac{an+b}{c}
floor-f(c,c-b-1,a,lfloorfrac{an+b}{c}
floor-1)\
end{align}
]
于是我们就可以在(log(n))的渐进时间复杂度下求出(f)了
好的，相信(g,h)不是问题。
于是我们看看(g)
[g(a,b,c,n)=sumlimits_{i=0}^nlfloorfrac{ai+b}c
floor^2
]
套路一波。
[egin{aligned}g(a,b,c,n)&=sumlimits_{i=0}^n(lfloorfrac{(a\%c)i+b\%c}c
floor+ilfloorfrac ac
floor+lfloorfrac bc
floor)^2\大力展开,上式&=sumlimits_{i=0}^n(lfloorfrac{(a\%c)i+b\%c}c
floor^2+i^2lfloorfrac ac
floor^2+lfloorfrac bc
floor^2+2ilfloorfrac ac
floorlfloorfrac{(a\%c)i+b\%c}c
floor+2ilfloorfrac bc
floorlfloorfrac{(a\%c)i+b\%c}c
floor+2ilfloorfrac ac
floorlfloorfrac bc
floor)^2\设S1(n)&=sumlimits_{i=0}^ni=frac{i*(i+1)}{2},S2(n)=sumlimits_{i=0}^ni^2=frac {i*(i+1)*(2i+1)}6\整理,原式&=g(a\%c,b\%c,c,n)+S2(n)lfloorfrac ac
floor^2+(n+1)lfloorfrac bc
floor^2+2lfloorfrac ac
floor h(a\%c,b\%c,c,n)+2lfloorfrac bc
floor f(a\%c,b\%c,c,n)+2S1(n)lfloorfrac ac
floorlfloorfrac bc
floorend{aligned}
]
你发现事情变得辣手了起来，我们似乎并不知道这个(h)怎么搞。
但是你先不管它,假设你知道了(h)怎么算。
[egin{aligned}g(a,b,c,n)&=sumlimits_{i=0}^nlfloorfrac{ai+b}c
floor^2\&=sumlimits_{i=0}^n(2sumlimits_{d=1}^{lfloorfrac{ai+b}c
floor}d-lfloorfrac{ai+b}c
floor)\&=sumlimits_{i=0}^n(2sumlimits_{d=0}^{lfloorfrac{ai+b}c
floor-1}(d+1)-lfloorfrac{ai+b}c
floor)\&=2sum_{d=0}^{lfloorfrac {an+b}c
floor-1}(d+1)sumlimits_{i=0}^n[lfloorfrac{ai+b}c
floorge d+1]-f(a,b,c,n)\&=2sum_{d=0}^{lfloorfrac {an+b}c
floor-1}(d+1)(n-lfloorfrac{cd+c-b-1}a
floor)-f(a,b,c,n)\&=n*lfloorfrac {an+b}c
floor*(lfloorfrac {an+b}c
floor+1)-2sum_{d=0}^{lfloorfrac {an+b}c
floor-1}(d+1)lfloorfrac{cd+c-b-1}a
floor-f(a,b,c,n)\&=n*lfloorfrac {an+b}c
floor*(lfloorfrac {an+b}c
floor+1)-2h(c,c-b-1,a,lfloorfrac{an+b}c
floor-1)-2f(c,c-b-1,a,lfloorfrac{an+b}c
floor-1)-f(a,b,c,n)\end{aligned}
]
我们最后考虑一下(h),首先大力展开
[egin{aligned}h(a,b,c,n)=S2(n)lfloorfrac ac
floor+S1(n)lfloorfrac bc
floor+h(a\%c,b\%c,c,n)end{aligned}
]
好的,继续
[egin{aligned}
h(a,b,c,n)&=sumlimits_{i=0}^nilfloorfrac {ai+b}c
floor\
&=sumlimits_{i=0}^nisumlimits_{d=1}^{lfloorfrac{an+b}{c}
floor}[a^{-1}clfloorfrac{ai+b}c
floorge a^{-1}cdgt a^{-1}(cd-1)]\
&=sumlimits_{i=0}^nisumlimits_{d=0}^{lfloorfrac{an+b}{c}
floor-1}[igtfrac{cd+c-b-1}{a}]\
&=S1(n)lfloorfrac {an+b}c
floor-sumlimits_{d=0}^{lfloorfrac{an+b}{c}
floor-1}sumlimits_{i=0}^ni[frac{cd+c-b-1}age i]\
&=S1(n)lfloorfrac {an+b}c
floor-sumlimits_{d=0}^{lfloorfrac{an+b}{c}
floor-1}frac{(frac{cd+c-b-1}a)*(frac{cd+c-b-1}a+1)}2\
&=S1(n)lfloorfrac {an+b}c
floor-frac12f(c,c-b-1,a,lfloorfrac{an+b}{c}
floor-1)-frac12\g(c,c-b-1,a,lfloorfrac{an+b}{c}
floor-1)
end{aligned}
]
如此递归，我们就可以在(O(logn))的渐进时间复杂度内把(f,g,h)都求出来了。
代码
/*
@Date : 2020-02-13 08:16:26
@Author : Adscn (adscn@qq.com)
@Link : https://www.cnblogs.com/LLCSBlog
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define int ll
#define IL inline
#define RG register
#define gi geti<int>()
#define gl geti<ll>()
#define gc getchar()
#define File(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
template<typename T>IL bool chkmax(T &x,const T &y){return x<y?x=y,1:0;}
template<typename T>IL bool chkmin(T &x,const T &y){return x>y?x=y,1:0;}
template<typename T>
IL T geti()
{
RG T xi=0;
RG char ch=gc;
bool f=0;
while(!isdigit(ch))ch=='-'?f=1:f,ch=gc;
while(isdigit(ch))xi=xi*10+ch-48,ch=gc;
return f?-xi:xi;
}
template<typename T>
IL void pi(T k,char ch=0)
{
if(k<0)k=-k,putchar('-');
if(k>=10)pi(k/10);
putchar(k%10+'0');
if(ch)putchar(ch);
}
const int P=998244353,i2=(P+1)>>1;
inline void exgcd(int a,int b,int &x,int &y){
if(!b)return x=1,y=0,void();
exgcd(b,a%b,y,x);
y=(y-a/b*x)%P;
}
inline int inv(int a){
int x,y;
exgcd(a,P,x,y);
return (x+P)%P;
}
const int i6=inv(6);
struct data{ll a,b,c;};
inline data solve(int a,int b,int c,int n)
{
ll n1=(n+1)%P,S1=n1*n%P*i2%P,S2=n*n1%P*(n+n1)%P*i6%P;
ll m=(1ll*a*n+b)/c,ac=a/c,bc=b/c,acq=ac*ac%P,bcq=bc*bc%P,nm=n*m%P;
if(!a)return (data){n1*bc%P,n1*bcq%P,bc*S1%P};
if(a>=c||b>=c){
data ret=solve(a%c,b%c,c,n);
return (data){
(ret.a+S1*ac%P +n1*bc%P)%P,
(ret.b
+S2*acq%P
+n1*bcq%P
+2*ac*ret.c%P
+2*bc*ret.a%P
+2*S1*ac%P*bc%P)%P,
(ret.c+S2*ac%P +S1*bc%P)%P};
}
auto lst=solve(c,c-b-1,a,m-1);
data ret;
ret.a=(nm-lst.a)%P;
ret.b=(nm*(m+1)%P-2ll*(lst.a+lst.c)%P-ret.a)%P;
ret.c=(S1*m%P-1ll*i2*(lst.a+lst.b)%P)%P;
return ret;
}
signed main(void)
{
#ifndef ONLINE_JUDGE
// File("");
#endif
int T=gi;
while(T--)
{
int n=gi,a=gi,b=gi,c=gi;
auto ret=solve(a,b,c,n);
printf("%lld %lld %lld
",(ret.a+P)%P,(ret.b+P)%P,(ret.c+P)%P);
}
return 0;
}