类欧几里得算法学习笔记

类欧几里得算法

类欧几里得算法可以在\(O(log_2\max\{a,b\})\)的时间内求解形如\(\sum_{i=0}^n\lfloor \frac{a\cdot i+b}{c} \rfloor\)的式子，而他的求解过程酷似欧几里得算法，故而得名。

基础款

令\(f\left( a,b,c,n \right) = \sum_{i=0}^n \left\lfloor \frac{a\cdot i+b}{c} \right\rfloor\)

• case 1：\(a=0\)

  \[f\left( a,b,c,n \right) = \left\lfloor \frac{b}{c} \right\rfloor\cdot \left( n+1 \right) \]

• case 2：\(a\ge c {\kern 5pt} || {\kern 5pt} b\ge c\)

  \[\begin{aligned} f\left( a,b,c,n \right)&=\sum_{i=0}^n\left\lfloor \frac{a\cdot i+b}{c} \right\rfloor \\ &=\sum_{i=0}^n\left\lfloor \frac{(\left\lfloor \frac{a}{c} \right\rfloor \cdot c+a\%c)\cdot i+\left\lfloor \frac{b}{c} \right\rfloor\cdot c+b\%c}{c} \right\rfloor \\ &=\sum_{i=0}^n\left\lfloor \frac{a}{c} \right\rfloor\cdot i+\sum_{i=0}^n\left\lfloor \frac{b}{c} \right\rfloor+\sum_{i=0}^n\left\lfloor \frac{(a\%c)\cdot i+b\%c}{c} \right\rfloor \\ &=\left\lfloor \frac{a}{c} \right\rfloor\cdot \frac{n\cdot (n+1)}{2}+\left\lfloor \frac{b}{c} \right\rfloor\cdot (n+1)+f\left( a\%c,b\%c,c,n \right) \end{aligned} \]

• case 3：\(a < c {\kern 5pt} \&\& {\kern 5pt} b < c\)
  先考虑一个问题：已知\(\left\lfloor \frac{a\cdot i+b}{c} \right\rfloor=k\)，求\(i\)的范围，以上均为正整数。

  \[\begin{aligned} &{\kern 16pt} k\le \frac{a\cdot i+b}{c}<k+1 \\ &\Leftrightarrow k\cdot c-b\le a\cdot i<c\cdot (k+1)-b \\ &\Leftrightarrow\left\lfloor \frac{k\cdot c-b+a-1}{a} \right\rfloor\le i \le \left\lfloor \frac{c\cdot (k+1)-b+a-1}{a} \right\rfloor-1 \end{aligned} \]

  我们变换求和的方法，枚举\(k\)

  \[\begin{aligned} f\left( a,b,c,n \right) &=\sum_{i=1}^n\left\lfloor \frac{a\cdot i+b}{c} \right\rfloor \\ &=\sum_{k=0}^{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor-1}\left( \left\lfloor \frac{c\cdot (k+1)-b+a-1}{a} \right\rfloor-1-\left\lfloor \frac{k\cdot c-b+a-1}{a} \right\rfloor+1\right)\cdot k+ \\ &{\kern 15pt}\left( n-\left\lfloor \frac{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor\cdot c-b+a-1}{a} \right\rfloor+1 \right)\cdot \left\lfloor \frac{a\cdot n+b}{c} \right\rfloor \\ &=\sum_{k=1}^{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor}\left\lfloor \frac{c\cdot k-b-1}{a} \right\rfloor \cdot k-\sum_{k=0}^{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor-1} \left\lfloor \frac{c\cdot k-b-1}{a} \right\rfloor \cdot k-\sum_{k=0}^{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor-1}\left\lfloor \frac{c\cdot (k+1)-b-1}{a} \right\rfloor+\\ &{\kern 15pt}\left( n-\left\lfloor \frac{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor\cdot c-b-1}{a} \right\rfloor \right)\cdot \left\lfloor \frac{a\cdot n+b}{c} \right\rfloor \\ &=\left\lfloor\frac{a\cdot n+b}{c}\right\rfloor\cdot n-\sum_{k=0}^{\left\lfloor\frac{a\cdot n+b}{c}\right\rfloor-1}\left\lfloor \frac{c\cdot (k+1)-b-1}{a} \right\rfloor \\ &=\left\lfloor\frac{a\cdot n+b}{c}\right\rfloor\cdot n-f\left( c,c-b-1,a, \left\lfloor\frac{a\cdot n+b}{c}\right\rfloor-1\right) \end{aligned} \]

我们发现上面的计算过程与欧几里得算法类似，复杂度是\(O(log_2\max\{a,b\})\)

进阶款

$g\left( a,b,c,n \right)=\sum_{i=0}^n\left\lfloor \frac{a\cdot i+b}{c} \right\rfloor^2 $

$h\left( a,b,c,n \right)=\sum_{i=0}^n\left\lfloor \frac{a\cdot i+b}{c} \right\rfloor \cdot i $

等会我们会看到这两个式子具有关联性。

• case 1：\(a=0\)
  \( \begin{aligned} &g\left( a,b,c,n \right) =\left\lfloor \frac{b}{c} \right\rfloor^2\cdot(n+1)\\ &h\left( a,b,c,n \right) =\left\lfloor \frac{b}{c} \right\rfloor\cdot \frac{n\cdot (n+1)}{2} \end{aligned} \)

• case 2：\(a\ge c {\kern 5pt} || {\kern 5pt} b\ge c\)

  \[\begin{aligned} g\left( a,b,c,n \right) &=\sum_{i=0}^n\left\lfloor \frac{a\cdot i+b}{c} \right\rfloor^2 \\ &=\sum_{i=0}^n\left\lfloor \frac{(\left\lfloor \frac{a}{c} \right\rfloor \cdot c+a\%c)\cdot i+\left\lfloor \frac{b}{c} \right\rfloor\cdot c+b\%c}{c} \right\rfloor^2 \\ &=\sum_{i=0}^n\left( \left\lfloor \frac{a}{c} \right\rfloor\cdot i+\left\lfloor \frac{b}{c} \right\rfloor+\left\lfloor \frac{(a\%c)\cdot i+b\%c}{c} \right\rfloor \right)^2 \\ &=\left\lfloor \frac{a}{c} \right\rfloor^2\cdot \frac{n\cdot (n+1)\cdot (2n+1)}{6}+\left\lfloor \frac{a}{c} \right\rfloor\cdot \left\lfloor \frac{b}{c} \right\rfloor\cdot n\cdot \left( n+1 \right) + \left\lfloor \frac{b}{c} \right\rfloor^2\cdot (n+1) +\\ &{\kern 15pt}g\left( a\%c,b\%c,c,n \right)+2\cdot \left\lfloor \frac{a}{c} \right\rfloor\cdot h\left( a\%c,b\%c,c,n \right) +2\cdot \left\lfloor \frac{b}{c} \right\rfloor \cdot f\left( a\%c,b\%c,c,n \right) \\ h\left( a,b,c,n \right) &=\sum_{i=0}^n\left\lfloor \frac{a\cdot i+b}{c} \right\rfloor\cdot i \\ &=\sum_{i=0}^n\left\lfloor \frac{(\left\lfloor \frac{a}{c} \right\rfloor \cdot c+a\%c)\cdot i+\left\lfloor \frac{b}{c} \right\rfloor\cdot c+b\%c}{c} \right\rfloor\cdot i \\ &=\left\lfloor \frac{a}{c} \right\rfloor\cdot \frac{n\cdot (n+1)\cdot (2n+1)}{6}+\left\lfloor \frac{b}{c} \right\rfloor\cdot \frac{n\cdot (n+1)}{2}+h\left( a\%c,b\%c,c,n \right) \end{aligned} \]

• case 3：\(a < c {\kern 5pt} \&\& {\kern 5pt} b < c\)

  \[\begin{aligned} g\left( a,b,c,n \right) &=\sum_{i=1}^n\left\lfloor \frac{a\cdot i+b}{c} \right\rfloor^2 \\ &=\sum_{k=0}^{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor-1}\left( \left\lfloor \frac{c\cdot (k+1)-b+a-1}{a} \right\rfloor-1-\left\lfloor \frac{k\cdot c-b+a-1}{a} \right\rfloor+1\right)\cdot k^2+ \\ &{\kern 15pt}\left( n-\left\lfloor \frac{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor\cdot c-b+a-1}{a} \right\rfloor+1 \right)\cdot \left\lfloor \frac{a\cdot n+b}{c} \right\rfloor^2 \\ &=\sum_{k=1}^{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor}\left\lfloor \frac{c\cdot k-b-1}{a} \right\rfloor \cdot k^2-\sum_{k=0}^{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor-1} \left\lfloor \frac{c\cdot k-b-1}{a} \right\rfloor \cdot k^2-\sum_{k=0}^{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor-1}\left\lfloor \frac{c\cdot (k+1)-b-1}{a} \right\rfloor-\\ &{\kern 15pt}\sum_{k=0}^{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor-1}2k\cdot \left\lfloor \frac{c\cdot (k+1)-b-1}{a} \right\rfloor+\left( n-\left\lfloor \frac{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor\cdot c-b-1}{a} \right\rfloor \right)\cdot \left\lfloor \frac{a\cdot n+b}{c} \right\rfloor^2 \\ &=\left\lfloor\frac{a\cdot n+b}{c}\right\rfloor^2\cdot n-f\left( c,c-b-1,a, \left\lfloor\frac{a\cdot n+b}{c}\right\rfloor-1\right)-2\cdot h\left( c,c-b-1,a, \left\lfloor\frac{a\cdot n+b}{c}\right\rfloor-1\right) \\ h\left( a,b,c,n \right) &=\sum_{i=0}^n\left\lfloor \frac{a\cdot i+b}{c} \right\rfloor \cdot i \\ &=\sum_{k=0}^{\left\lfloor \frac{a\cdot n+b}{c} \right\rfloor-1}\frac{k}{2}\cdot \left( \left\lfloor \frac{c\cdot (k+1)-b-1}{a} \right\rfloor -\left\lfloor \frac{c\cdot k-b-1}{a} \right\rfloor\right)\cdot \left( \left\lfloor \frac{c\cdot (k+1)-b-1}{a} \right\rfloor+\left\lfloor \frac{c\cdot k-b-1}{a} \right\rfloor+1 \right) + \\ &{\kern 15pt}\frac{1}{2}\cdot \left\lfloor \frac{n\cdot a+b}{c} \right\rfloor\cdot \left( n-\left\lfloor \frac{\left\lfloor \frac{n\cdot a+b}{c} \right\rfloor\cdot c-b-1}{a} \right\rfloor\right) \cdot \left( n+\left\lfloor \frac{\left\lfloor \frac{n\cdot a+b}{c} \right\rfloor\cdot c-b-1}{a} \right\rfloor+1 \right) \\ &=\frac{1}{2}\cdot \sum_{i=0}^{\left\lfloor \frac{n\cdot a+b}{c} \right\rfloor-1}i\cdot \left( \left\lfloor \frac{c\cdot (i+1)-b-1}{a} \right\rfloor^2-\left\lfloor \frac{c\cdot i-b-1}{a} \right\rfloor^2 \right) + \frac{1}{2}\cdot \sum_{i=0}^{\left\lfloor \frac{n\cdot a+b}{c} \right\rfloor-1}i\cdot \left( \left\lfloor \frac{c\cdot (i+1)-b-1}{a} \right\rfloor-\left\lfloor \frac{c\cdot i-b-1}{a} \right\rfloor \right) \\ &{\kern 15pt} \frac{1}{2}\cdot \left\lfloor \frac{n\cdot a+b}{c} \right\rfloor\cdot \left( n-\left\lfloor \frac{\left\lfloor \frac{n\cdot a+b}{c} \right\rfloor\cdot c-b-1}{a} \right\rfloor\right) \cdot \left( n+\left\lfloor \frac{\left\lfloor \frac{n\cdot a+b}{c} \right\rfloor\cdot c-b-1}{a} \right\rfloor+1 \right) \\ &=\left\lfloor \frac{n\cdot a+b}{c} \right\rfloor\cdot \frac{n\cdot (n+1)}{2}-\frac{1}{2}\cdot g\left( c,c-b-1,a, \left\lfloor\frac{a\cdot n+b}{c}\right\rfloor-1\right) - \frac{1}{2}\cdot f\left( c,c-b-1,a, \left\lfloor\frac{a\cdot n+b}{c}\right\rfloor-1\right) \end{aligned} \]

综上，

\[\begin{aligned} f\left( a,b,c,n \right)&=\begin{cases} \left\lfloor \frac{b}{c} \right\rfloor\cdot \left( n+1 \right), & a=0 \\ \left\lfloor \frac{a}{c} \right\rfloor\cdot \frac{n\cdot (n+1)}{2}+\left\lfloor \frac{b}{c} \right\rfloor\cdot (n+1)+f\left( a\%c,b\%c,c,n \right), & a\ge c {\kern 5pt} || {\kern 5pt} b\ge c \\ \left\lfloor\frac{a\cdot n+b}{c}\right\rfloor\cdot n-f\left( c,c-b-1,a, \left\lfloor\frac{a\cdot n+b}{c}\right\rfloor-1\right), & a < c {\kern 5pt} \&\& {\kern 5pt} b < c \end{cases} \\ g\left( a,b,c,n \right)&=\begin{cases} \left\lfloor \frac{b}{c} \right\rfloor^2\cdot (n+1), & a=0 \\ \left\lfloor \frac{a}{c} \right\rfloor^2\cdot \frac{n\cdot (n+1)\cdot (2n+1)}{6}+\left\lfloor \frac{a}{c} \right\rfloor\cdot \left\lfloor \frac{b}{c} \right\rfloor\cdot n\cdot \left( n+1 \right) + \left\lfloor \frac{b}{c} \right\rfloor^2\cdot (n+1) + \\g\left( a\%c,b\%c,c,n \right)+2\cdot \left\lfloor \frac{a}{c} \right\rfloor\cdot h\left( a\%c,b\%c,c,n \right) +2\cdot \left\lfloor \frac{b}{c} \right\rfloor \cdot f\left( a\%c,b\%c,c,n \right), & a\ge c {\kern 5pt} || {\kern 5pt} b\ge c \\ \left\lfloor\frac{a\cdot n+b}{c}\right\rfloor^2\cdot n-f\left( c,c-b-1,a, \left\lfloor\frac{a\cdot n+b}{c}\right\rfloor-1\right)-2\cdot h\left( c,c-b-1,a, \left\lfloor\frac{a\cdot n+b}{c}\right\rfloor-1\right), & a < c {\kern 5pt} \&\& {\kern 5pt} b < c \end{cases} \\ h\left( a,b,c,n \right) &=\begin{cases} \left\lfloor \frac{b}{c} \right\rfloor\cdot \frac{n\cdot (n+1)}{2}, & a=0 \\ \left\lfloor \frac{a}{c} \right\rfloor\cdot \frac{n\cdot (n+1)\cdot (2n+1)}{6}+\left\lfloor \frac{b}{c} \right\rfloor\cdot \frac{n\cdot (n+1)}{2}+h\left( a\%c,b\%c,c,n \right), & a\ge c {\kern 5pt} || {\kern 5pt} b\ge c \\ \left\lfloor \frac{n\cdot a+b}{c} \right\rfloor\cdot \frac{n\cdot (n+1)}{2}-\frac{1}{2}\cdot g\left( c,c-b-1,a, \left\lfloor\frac{a\cdot n+b}{c}\right\rfloor-1\right) - \frac{1}{2}\cdot f\left( c,c-b-1,a, \left\lfloor\frac{a\cdot n+b}{c}\right\rfloor-1\right), & a < c {\kern 5pt} \&\& {\kern 5pt} b < c \end{cases} \end{aligned} \]

code：P5170 【模板】类欧几里得算法

#include <cstdio>
#include <iostream>

using namespace std;

const int mo = 998244353;

struct Mint {
    int a;
    Mint() {}
    Mint(int o) { a = o; }
    Mint operator+(const Mint& o) const { return (a + o.a) % mo; }
    Mint operator+(const int& o) const { return (o + a) % mo; }
    Mint operator-(const Mint& o) const {
        return Mint(((a - o.a) % mo) + mo) % mo;
    }
    Mint operator*(const Mint& o) const { return Mint(1ll * a * o.a % mo); }
    Mint operator/(const Mint& o) const { return Mint(a / o.a); }
    operator int() { return a; }
};

struct func {
    Mint f, g, h;
    func() {}
    func(Mint o, Mint p, Mint q) {
        f = o;
        g = p;
        h = q;
    }
};

const Mint inv2 = Mint((mo + 1) >> 1);
const Mint inv3 = Mint((mo + 1) / 3);
const Mint inv6 = Mint(inv2 * inv3);

func calc(int A, int B, int C, int N) {
    Mint a = A;
    Mint b = B;
    Mint c = C;
    Mint n = N;
    if (!A) {
        return func((b / c) * (n + 1), (b / c) * (b / c) * (n + 1),
                    (b / c) * n * (n + 1) * inv2);
    }
    func ans;
    if (A >= C || B >= C) {
        ans = calc(A % C, B % C, C, N);
        return func((a / c) * n * (n + 1) * inv2 + (b / c) * (n + 1) + ans.f,
                    n * (n + 1) * (n + n + 1) * inv6 * (a / c) * (a / c) +
                        (a / c) * (b / c) * n * (n + 1) +
                        (b / c) * (b / c) * (n + 1) + ans.g +
                        (a / c) * ans.h * Mint(2) + (b / c) * ans.f * Mint(2),
                    (a / c) * n * (n + 1) * (n + n + 1) * inv6 +
                        (b / c) * n * (n + 1) * inv2 + ans.h);
    }
    int m = (1ll * A * N + B) / C;
    Mint M = Mint(m);
    ans = calc(C, C - B - 1, A, m - 1);
    return func(M * n - ans.f, n * M * M - ans.h * Mint(2) - ans.f,
                M * n * n * inv2 + M * n * inv2 - ans.g * inv2 - ans.f * inv2);
}

int main() {
    int T;
    cin >> T;
    while (T--) {
        int a, b, c, n;
        scanf("%d%d%d%d", &n, &a, &b, &c);
        func ans = calc(a, b, c, n);
        printf("%d %d %d\n", ans.f, ans.g, ans.h);
    }
    return 0;
}