类欧几里得算法
- By Winniechen
一种用来快速求某些带有特殊性质的式子的和，一般复杂度为$O(log n)$，有些特殊时刻，需要用到$gcd$，因此为$O(log^2 n)$
第一类
$f(a,b,c,n)=sumlimits_{i=0}^nlfloorfrac{a imes i+b}{c} floor$
显然，若$age c$，那么：
$$
f(a,b,c,n)=sumlimits_{i=0}^nlfloorfrac{(c imes k+t) imes i+b}{c}
floor =sumlimits_{i=0}^nk imes i+lfloorfrac{t imes i+b}{c}
floor =frac{i imes (i+1) imes k}{2}+sumlimits_{i=0}^nlfloorfrac{t imes i+b}{c}
floor =frac{i imes (i+1) imes k}{2}+f(t,b,c,n)
$$
同样，若$bge c$，那么：$f(a,b,c,n)=sumlimits_{i=0}^n{lfloorfrac{a imes i+c imes k+t}{c} floor}=f(a,t,c,n)+k imes n$
因此，我们可以将上面两种情况转化为下面的这种：
$f(a,b,c,n),a< c,b<c$
$f(a,b,c,n)=sumlimits_{i=0}^nlfloorfrac{a imes i+b}{c}
floor=sumlimits_{i=0}nsumlimits_{j=1}m[lfloorfrac{a imes i+b}{c}
floorge j],(m=frac{a imes n+b}{c})$
$$
f(a,b,c,n) \ =sumlimits_{i=0}nsumlimits_{j=0}{m-1}{[a imes i > j imes c+c-b-1]} \ =sumlimits_{i=0}nsumlimits_{j=0}{m-1}[i> frac{j imes c+c-b-1}{a}] \ = sumlimits_{j=0}^{m-1}n-lfloorfrac{c imes j-b+c-1}{a}
floor \ = n imes m-f(c,c-b-1,a,m-1)
$$
显然，对于项：$a,c$，经过了类似$gcd$的$a=c,c=a%c$
因此，复杂度得以保证。
第二类
$g(a,b,c,n)=sumlimits_{i=0}^ni imes lfloorfrac{a imes i+b}{c} floor$
类似前面的，推出$a,bge c$的情况。
$g(a,b,c,n)=frac{n imes (n + 1) imes (2 imes n+1)}{6} k_a+frac{n imes (n+1)}{2}k_b+g(t_a,t_b,c,n)$
$g(a,b,c,n)=sumlimits_{i=0}^ni imes lfloorfrac{a imes i+b}{c} floor=sumlimits_{i=0}nisumlimits_{j=1}m[lfloorfrac{a imes i+n}{c} floorge j]$，$m$同上
发现后面的同上，所有我就跳过几步.jpg
$$
g(a,b,c,n)\ =sumlimits_{i=0}nisumlimits_{j=0}{m-1}[i>lfloorfrac{c imes j+c-b-1}{a}
floor ] \ =sumlimits_{j=0}^{m-1} frac{n imes (n+1)}{2}-frac{x imes (x+1)}{2}(x=lfloorfrac{c imes j+c-b-1}{a}
floor) \ =frac{n imes (n+1) imes m+sumlimits_{j=0}{m-1}x+x2}{2}\ = frac{n imes (n+1) imes m-f(c,c-b-1,a,m-1)-h(c,c-b-1,a,m-1)}{2}
$$
$h(a,b,c,n)=sumlimits_{i=0}^n lfloorfrac{a imes i+b}{c} floor ^2$，这个马上介绍...
但是显然，这个$g(a,b,c,d)$就只差$h(a,b,c,d)$了，其他都没有问题，都可以在$log n$内解决
第三类
$h(a,b,c,n)=sumlimits_{i=0}^n lfloorfrac{a imes i+b}{c} floor ^2$
同样类似上面的那个...
$$
h(a,b,c,n)= (a/c)^2n(n+1)(2n+1)/6+ (b/c)^2(n+1)+(a/c)(b/c)n(n+1)+ h(a%c,b%c,c,n)+2(a/c)g(a%c,b%c,c,n)+ 2(b/c)f(a%c,b%c,c,n)$
$$
其实也没啥，就是麻烦了点...
就是需要稍微构造一下，因为正常推的话，显然是不能推的...
$n^2=2 imesfrac{n imes (n+1)}{2}-n=2(sumlimits_{i=0}^n i) -n$
$h(a,b,c,n)=sumlimits_{i=0}n2sumlimits_{j=1}xj-f(a,b,c,n)$
$h(a,b,c,n)=sumlimits_{j=0}^{m-1}2 imes (j+1)sumlimits_{i=0}^n[lfloorfrac{a imes i+b}{c} floor ge j+1]-f(a,b,c,n)$
$h(a,b,c,n)=2sumlimits_{j=0}^{m-1}(j+1) imes(n- lfloorfrac{c imes j+c-b-1}{a}])-f(a,b,c,n)$
$$
h(a,b,c,n)= n imes m imes (m+1) - 2 imes g(c,c-b-1,a,m-1) - 2 imes f(c,c-b-1,a,m-1) - f(a,b,c,n)
$$
完事了！
第一种，难写难调跑得慢...（并且复杂度多了一个$log$
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <iostream>
#include <bitset>
#include <map>
using namespace std;
#define ll long long
#define mod 998244353
#define inv2 499122177
#define inv6 166374059
struct node
{
ll a,b,c,d;
node(){}
node(ll x,ll y,ll z,ll w){a=x,b=y,c=z,d=w;}
bool operator < (const node &t) const {return a==t.a?(t.b==b?(t.c==c?d<t.d:c<t.c):b<t.b):a<t.a;}
};
map<node ,ll >F,G,H;
ll f(ll n,ll a,ll b,ll c)
{
if(F.find(node(n,a,b,c))!=F.end())return F[node(n,a,b,c)];
if(b>=c)return F[node(n,a,b,c)]=(f(n,a,b%c,c)+(b/c)*(n+1)%mod)%mod;
if(a>=c)return F[node(n,a,b,c)]=(f(n,a%c,b,c)+(a/c)*(n*(n+1)%mod*inv2%mod)%mod)%mod;
if(!a)return 0;
ll m=(a*n+b)/c;return F[node(n,a,b,c)]=(n*m%mod-f(m-1,c,c-b-1,a))%mod;
}
ll g(ll n,ll a,ll b,ll c);
ll h(ll n,ll a,ll b,ll c);
ll g(ll n,ll a,ll b,ll c)
{
if(G.find(node(n,a,b,c))!=G.end())return G[node(n,a,b,c)];
if(a>=c)return G[node(n,a,b,c)]=(g(n,a%c,b,c)+(a/c)*n%mod*(n+1)%mod*(n*2+1)%mod*inv6)%mod;
if(b>=c)return G[node(n,a,b,c)]=(g(n,a,b%c,c)+(b/c)*n%mod*(n+1)%mod*inv2)%mod;
if(!a)return 0;
ll m=(a*n+b)/c;return G[node(n,a,b,c)]=(n*(n+1)%mod*m%mod-f(m-1,c,c-b-1,a)+mod-h(m-1,c,c-b-1,a)+mod)*inv2%mod;
}
ll h(ll n,ll a,ll b,ll c)
{
if(H.find(node(n,a,b,c))!=H.end())return H[node(n,a,b,c)];
if(a>=c||b>=c)
return H[node(n,a,b,c)]=((a/c)*(a/c)%mod*n%mod*(n+1)%mod*(n*2+1)%mod*inv6%mod+
(b/c)*(b/c)%mod*(n+1)%mod+(a/c)*(b/c)%mod*n%mod*(n+1)%mod+
h(n,a%c,b%c,c)%mod+2*(a/c)%mod*g(n,a%c,b%c,c)%mod+
2*(b/c)*f(n,a%c,b%c,c)%mod)%mod;
if(!a)return 0;
ll m=(a*n+b)/c;
return H[node(n,a,b,c)]=((n*m%mod*(m+1)%mod-2*g(m-1,c,c-b-1,a)-2*f(m-1,c,c-b-1,a))%mod-f(n,a,b,c)+mod+mod)%mod;
}
int main()
{
int T;scanf("%d",&T);
while(T--)
{
ll a,b,c,n;scanf("%lld%lld%lld%lld",&n,&a,&b,&c);
printf("%lld %lld %lld
",(f(n,a,b,c)+mod)%mod,(h(n,a,b,c)+mod)%mod,(g(n,a,b,c)+mod)%mod);
F.clear();G.clear();H.clear();
}
}
// TLE
// 非常慢.jpg
第二种，好写好调跑得快...
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <iostream>
#include <bitset>
using namespace std;
#define ll long long
#define mod 998244353
#define inv2 499122177
#define inv6 166374059
struct node
{
ll f,g,h;
node(){f=0,g=0,h=0;}
node(ll a,ll b,ll c){f=a,g=b,h=c;}
void print(){printf("%lld %lld %lld
",f,h,g);}
};
node solve(ll n,ll a,ll b,ll c)
{
if(!a)return node((b/c)*(n+1)%mod,(n*(n+1)%mod*inv2)%mod*(b/c)%mod,(b/c)*(b/c)%mod*(n+1)%mod);
node ret=node();
if(a>=c||b>=c)
{
ll t1=a/c,t2=b/c,s1=n*(n+1)%mod*inv2%mod,s2=n*(n+1)%mod*(2*n+1)%mod*inv6%mod;
node tmp=solve(n,a%c,b%c,c);
ret.f=(tmp.f+(n+1)*t2%mod+s1*t1%mod)%mod;
ret.g=(tmp.g+s1*t2%mod+s2*t1%mod)%mod;
ret.h=(t1*t1%mod*s2%mod+(n+1)*t2%mod*t2%mod+2*t1*t2%mod*s1%mod+2*t1*tmp.g%mod+2*t2*tmp.f%mod+tmp.h)%mod;
return ret;
}
ll m=(a*n+b)/c;node tmp=solve(m-1,c,c-b-1,a);
ret.f=(n*m%mod-tmp.f+mod)%mod;
ret.g=((n*(n+1)%mod*m%mod-tmp.f-tmp.h)%mod+mod)*inv2%mod;
ret.h=((n*m%mod*m%mod-tmp.g*2%mod-tmp.f)%mod+mod)%mod;
return ret;
}
ll a,b,c,n;
int main(){int T;scanf("%d",&T);while(T--)scanf("%lld%lld%lld%lld",&n,&a,&b,&c),solve(n,a,b,c).print();}