類歐幾里得算法

類歐幾里得算法

• By Winniechen

一種用來快速求某些帶有特殊性質的式子的和，通常複雜度爲$O(\log n)$，有些特殊時刻，須要用到$\gcd$，所以爲$O(\log^2 n)$

第一類

$f(a,b,c,n)=\sum\limits_{i=0}^n\lfloor\frac{a\times i+b}{c}\rfloor$

顯然，若$a\ge c$，那麼：​
$$
f(a,b,c,n)=\sum\limits_{i=0}^n\lfloor\frac{(c\times k+t)\times i+b}{c}\rfloor \ =\sum\limits_{i=0}^nk\times i+\lfloor\frac{t\times i+b}{c}\rfloor \ =\frac{i\times (i+1)\times k}{2}+\sum\limits_{i=0}^n\lfloor\frac{t\times i+b}{c}\rfloor \ =\frac{i\times (i+1)\times k}{2}+f(t,b,c,n)
$$

一樣，若$b\ge c$，那麼：$f(a,b,c,n)=\sum\limits_{i=0}^n{\lfloor\frac{a\times i+c\times k+t}{c}\rfloor}=f(a,t,c,n)+k\times n$

所以，咱們能夠將上面兩種狀況轉化爲下面的這種：

$f(a,b,c,n),a< c,b<c$

$f(a,b,c,n)=\sum\limits_{i=0}^n\lfloor\frac{a\times i+b}{c}\rfloor=\sum\limits_{i=0}^n\sum\limits_{j=1}^m[\lfloor\frac{a\times i+b}{c}\rfloor\ge j],(m=\frac{a\times n+b}{c})$
$$
f(a,b,c,n) \ =\sum\limits_{i=0}^n\sum\limits_{j=0}^{m-1}{[a\times i > j\times c+c-b-1]} \ =\sum\limits_{i=0}^n\sum\limits_{j=0}^{m-1}[i> \frac{j\times c+c-b-1}{a}] \ = \sum\limits_{j=0}^{m-1}n-\lfloor\frac{c\times j-b+c-1}{a}\rfloor \ = n\times m-f(c,c-b-1,a,m-1)​
$$

顯然，對於項：$a,c$，通過了相似$\gcd$的$a=c,c=a%c$

所以，複雜度得以保證。

第二類

$g(a,b,c,n)=\sum\limits_{i=0}^ni\times \lfloor\frac{a\times i+b}{c}\rfloor$

相似前面的，推出$a,b\ge c$的狀況。

$g(a,b,c,n)=\frac{n\times (n + 1)\times (2\times n+1)}{6} k_a+\frac{n\times (n+1)}{2}k_b+g(t_a,t_b,c,n)$

$g(a,b,c,n)=\sum\limits_{i=0}^ni\times \lfloor\frac{a\times i+b}{c}\rfloor=\sum\limits_{i=0}^ni\sum\limits_{j=1}^m[\lfloor\frac{a\times i+n}{c}\rfloor\ge j]$，$m$同上

發現後面的同上，全部我就跳過幾步.jpg
$$
      g(a,b,c,n)\ =\sum\limits_{i=0}^ni\sum\limits_{j=0}^{m-1}[i>\lfloor\frac{c\times j+c-b-1}{a}\rfloor ] \ =\sum\limits_{j=0}^{m-1} \frac{n\times (n+1)}{2}-\frac{x\times (x+1)}{2}(x=\lfloor\frac{c\times j+c-b-1}{a}\rfloor) \ =\frac{n\times (n+1)\times m+\sum\limits_{j=0}^{m-1}x+x^2}{2}\ = \frac{n\times (n+1)\times m-f(c,c-b-1,a,m-1)-h(c,c-b-1,a,m-1)}{2}
$$

$h(a,b,c,n)=\sum\limits_{i=0}^n \lfloor\frac{a\times i+b}{c}\rfloor ^2$，這個立刻介紹...

可是顯然，這個$g(a,b,c,d)$就只差$h(a,b,c,d)$了，其餘都沒有問題，均可以在$\log n$內解決

第三類

$h(a,b,c,n)=\sum\limits_{i=0}^n \lfloor\frac{a\times i+b}{c}\rfloor ^2$

一樣相似上面的那個...
$$
h(a,b,c,n)=\ (a/c)^2n(n+1)(2n+1)/6+\ (b/c)^2(n+1)+(a/c)(b/c)n(n+1)+\ h(a%c,b%c,c,n)+2(a/c)g(a%c,b%c,c,n)+\ 2(b/c)f(a%c,b%c,c,n)$​
$$
其實也沒啥，就是麻煩了點...

就是須要稍微構造一下，由於正常推的話，顯然是不能推的...

$n^2=2\times\frac{n\times (n+1)}{2}-n=2(\sum\limits_{i=0}^n i) -n$

$h(a,b,c,n)=\sum\limits_{i=0}^n2\sum\limits_{j=1}^xj-f(a,b,c,n)$

$h(a,b,c,n)=\sum\limits_{j=0}^{m-1}2\times (j+1)\sum\limits_{i=0}^n[\lfloor\frac{a\times i+b}{c}\rfloor \ge j+1]-f(a,b,c,n)$

$h(a,b,c,n)=2\sum\limits_{j=0}^{m-1}(j+1)\times(n- \lfloor\frac{c\times j+c-b-1}{a}])-f(a,b,c,n)$
$$
h(a,b,c,n)=\ n\times m\times (m+1) \ - 2\times g(c,c-b-1,a,m-1) \ - 2\times f(c,c-b-1,a,m-1) \ - f(a,b,c,n)
$$
完事了！

第一種，難寫難調跑得慢...（而且複雜度多了一個$\log$

#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\n",(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 
// 很是慢.jpg

第二種，好寫好調跑得快...

#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\n",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();}