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$O(lo{g}_{2}max\{a,b\})$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$f(a,b,c,n)=\sum _{i=0}^n\lfloor \frac{a\cdot i+b}{c}\rfloor$

• case 1：$a=0$

  $$f(a,b,c,n)=\lfloor \frac{b}{c}\rfloor \cdot (n+1)$$

• case 2：$a\ge c||b\ge c$

  $$\begin{array}{rl}f(a,b,c,n)& =\sum _{i=0}^n\lfloor \frac{a\cdot i+b}{c}\rfloor \\ & =\sum _{i=0}^n\lfloor \frac{(\lfloor \frac{a}{c}\rfloor \cdot c+a\mathrm{\%}c)\cdot i+\lfloor \frac{b}{c}\rfloor \cdot c+b\mathrm{\%}c}{c}\rfloor \\ & =\sum _{i=0}^n\lfloor \frac{a}{c}\rfloor \cdot i+\sum _{i=0}^n\lfloor \frac{b}{c}\rfloor +\sum _{i=0}^n\lfloor \frac{(a\mathrm{\%}c)\cdot i+b\mathrm{\%}c}{c}\rfloor \\ & =\lfloor \frac{a}{c}\rfloor \cdot \frac{n\cdot (n+1)}{2}+\lfloor \frac{b}{c}\rfloor \cdot (n+1)+f(a\mathrm{\%}c,b\mathrm{\%}c,c,n)\end{array}$$

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

  $$\begin{array}{rl}& k\le \frac{a\cdot i+b}{c}<k+1\\ & \iff k\cdot c-b\le a\cdot i<c\cdot (k+1)-b\\ & \iff \lfloor \frac{k\cdot c-b+a-1}{a}\rfloor \le i\le \lfloor \frac{c\cdot (k+1)-b+a-1}{a}\rfloor -1\end{array}$$

  我们变换求和的方法，枚举$k$

  $$\begin{array}{rl}f(a,b,c,n)& =\sum _{i=1}^n\lfloor \frac{a\cdot i+b}{c}\rfloor \\ & =\sum _{k=0}^{\lfloor \frac{a\cdot n+b}{c}\rfloor -1}(\lfloor \frac{c\cdot (k+1)-b+a-1}{a}\rfloor -1-\lfloor \frac{k\cdot c-b+a-1}{a}\rfloor +1)\cdot k+\\ & (n-\lfloor \frac{\lfloor \frac{a\cdot n+b}{c}\rfloor \cdot c-b+a-1}{a}\rfloor +1)\cdot \lfloor \frac{a\cdot n+b}{c}\rfloor \\ & =\sum _{k=1}^{\lfloor \frac{a\cdot n+b}{c}\rfloor }\lfloor \frac{c\cdot k-b-1}{a}\rfloor \cdot k-\sum _{k=0}^{\lfloor \frac{a\cdot n+b}{c}\rfloor -1}\lfloor \frac{c\cdot k-b-1}{a}\rfloor \cdot k-\sum _{k=0}^{\lfloor \frac{a\cdot n+b}{c}\rfloor -1}\lfloor \frac{c\cdot (k+1)-b-1}{a}\rfloor +\\ & (n-\lfloor \frac{\lfloor \frac{a\cdot n+b}{c}\rfloor \cdot c-b-1}{a}\rfloor )\cdot \lfloor \frac{a\cdot n+b}{c}\rfloor \\ & =\lfloor \frac{a\cdot n+b}{c}\rfloor \cdot n-\sum _{k=0}^{\lfloor \frac{a\cdot n+b}{c}\rfloor -1}\lfloor \frac{c\cdot (k+1)-b-1}{a}\rfloor \\ & =\lfloor \frac{a\cdot n+b}{c}\rfloor \cdot n-f(c,c-b-1,a,\lfloor \frac{a\cdot n+b}{c}\rfloor -1)\end{array}$$

我们发现上面的计算过程与欧几里得算法类似，复杂度是$O(lo{g}_{2}max\{a,b\})$

进阶款

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

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

等会我们会看到这两个式子具有关联性。

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

• case 2：$a\ge c||b\ge c$

  $$\begin{array}{rl}g(a,b,c,n)& =\sum _{i=0}^n{\lfloor \frac{a\cdot i+b}{c}\rfloor }^2\\ & =\sum _{i=0}^n{\lfloor \frac{(\lfloor \frac{a}{c}\rfloor \cdot c+a\mathrm{\%}c)\cdot i+\lfloor \frac{b}{c}\rfloor \cdot c+b\mathrm{\%}c}{c}\rfloor }^2\\ & =\sum _{i=0}^n{(\lfloor \frac{a}{c}\rfloor \cdot i+\lfloor \frac{b}{c}\rfloor +\lfloor \frac{(a\mathrm{\%}c)\cdot i+b\mathrm{\%}c}{c}\rfloor )}^2\\ & ={\lfloor \frac{a}{c}\rfloor }^2\cdot \frac{n\cdot (n+1)\cdot (2n+1)}{6}+\lfloor \frac{a}{c}\rfloor \cdot \lfloor \frac{b}{c}\rfloor \cdot n\cdot (n+1)+{\lfloor \frac{b}{c}\rfloor }^2\cdot (n+1)+\\ & g(a\mathrm{\%}c,b\mathrm{\%}c,c,n)+2\cdot \lfloor \frac{a}{c}\rfloor \cdot h(a\mathrm{\%}c,b\mathrm{\%}c,c,n)+2\cdot \lfloor \frac{b}{c}\rfloor \cdot f(a\mathrm{\%}c,b\mathrm{\%}c,c,n)\\ h(a,b,c,n)& =\sum _{i=0}^n\lfloor \frac{a\cdot i+b}{c}\rfloor \cdot i\\ & =\sum _{i=0}^n\lfloor \frac{(\lfloor \frac{a}{c}\rfloor \cdot c+a\mathrm{\%}c)\cdot i+\lfloor \frac{b}{c}\rfloor \cdot c+b\mathrm{\%}c}{c}\rfloor \cdot i\\ & =\lfloor \frac{a}{c}\rfloor \cdot \frac{n\cdot (n+1)\cdot (2n+1)}{6}+\lfloor \frac{b}{c}\rfloor \cdot \frac{n\cdot (n+1)}{2}+h(a\mathrm{\%}c,b\mathrm{\%}c,c,n)\end{array}$$

• case 3：$a<c\mathrm{\&}\mathrm{\&}b<c$

  $$\begin{array}{rl}g(a,b,c,n)& =\sum _{i=1}^n{\lfloor \frac{a\cdot i+b}{c}\rfloor }^2\\ & =\sum _{k=0}^{\lfloor \frac{a\cdot n+b}{c}\rfloor -1}(\lfloor \frac{c\cdot (k+1)-b+a-1}{a}\rfloor -1-\lfloor \frac{k\cdot c-b+a-1}{a}\rfloor +1)\cdot k^2+\\ & (n-\lfloor \frac{\lfloor \frac{a\cdot n+b}{c}\rfloor \cdot c-b+a-1}{a}\rfloor +1)\cdot {\lfloor \frac{a\cdot n+b}{c}\rfloor }^2\\ & =\sum _{k=1}^{\lfloor \frac{a\cdot n+b}{c}\rfloor }\lfloor \frac{c\cdot k-b-1}{a}\rfloor \cdot k^2-\sum _{k=0}^{\lfloor \frac{a\cdot n+b}{c}\rfloor -1}\lfloor \frac{c\cdot k-b-1}{a}\rfloor \cdot k^2-\sum _{k=0}^{\lfloor \frac{a\cdot n+b}{c}\rfloor -1}\lfloor \frac{c\cdot (k+1)-b-1}{a}\rfloor -\\ & \sum _{k=0}^{\lfloor \frac{a\cdot n+b}{c}\rfloor -1}2k\cdot \lfloor \frac{c\cdot (k+1)-b-1}{a}\rfloor +(n-\lfloor \frac{\lfloor \frac{a\cdot n+b}{c}\rfloor \cdot c-b-1}{a}\rfloor )\cdot {\lfloor \frac{a\cdot n+b}{c}\rfloor }^2\\ & ={\lfloor \frac{a\cdot n+b}{c}\rfloor }^2\cdot n-f(c,c-b-1,a,\lfloor \frac{a\cdot n+b}{c}\rfloor -1)-2\cdot h(c,c-b-1,a,\lfloor \frac{a\cdot n+b}{c}\rfloor -1)\\ h(a,b,c,n)& =\sum _{i=0}^n\lfloor \frac{a\cdot i+b}{c}\rfloor \cdot i\\ & =\sum _{k=0}^{\lfloor \frac{a\cdot n+b}{c}\rfloor -1}\frac{k}{2}\cdot (\lfloor \frac{c\cdot (k+1)-b-1}{a}\rfloor -\lfloor \frac{c\cdot k-b-1}{a}\rfloor )\cdot (\lfloor \frac{c\cdot (k+1)-b-1}{a}\rfloor +\lfloor \frac{c\cdot k-b-1}{a}\rfloor +1)+\\ & \frac{1}{2}\cdot \lfloor \frac{n\cdot a+b}{c}\rfloor \cdot (n-\lfloor \frac{\lfloor \frac{n\cdot a+b}{c}\rfloor \cdot c-b-1}{a}\rfloor )\cdot (n+\lfloor \frac{\lfloor \frac{n\cdot a+b}{c}\rfloor \cdot c-b-1}{a}\rfloor +1)\\ & =\frac{1}{2}\cdot \sum _{i=0}^{\lfloor \frac{n\cdot a+b}{c}\rfloor -1}i\cdot ({\lfloor \frac{c\cdot (i+1)-b-1}{a}\rfloor }^2-{\lfloor \frac{c\cdot i-b-1}{a}\rfloor }^2)+\frac{1}{2}\cdot \sum _{i=0}^{\lfloor \frac{n\cdot a+b}{c}\rfloor -1}i\cdot (\lfloor \frac{c\cdot (i+1)-b-1}{a}\rfloor -\lfloor \frac{c\cdot i-b-1}{a}\rfloor )\\ & \frac{1}{2}\cdot \lfloor \frac{n\cdot a+b}{c}\rfloor \cdot (n-\lfloor \frac{\lfloor \frac{n\cdot a+b}{c}\rfloor \cdot c-b-1}{a}\rfloor )\cdot (n+\lfloor \frac{\lfloor \frac{n\cdot a+b}{c}\rfloor \cdot c-b-1}{a}\rfloor +1)\\ & =\lfloor \frac{n\cdot a+b}{c}\rfloor \cdot \frac{n\cdot (n+1)}{2}-\frac{1}{2}\cdot g(c,c-b-1,a,\lfloor \frac{a\cdot n+b}{c}\rfloor -1)-\frac{1}{2}\cdot f(c,c-b-1,a,\lfloor \frac{a\cdot n+b}{c}\rfloor -1)\end{array}$$

综上，

$$\begin{array}{rl}f(a,b,c,n)& =\{\begin{array}{ll}\lfloor \frac{b}{c}\rfloor \cdot (n+1),& a=0\\ \lfloor \frac{a}{c}\rfloor \cdot \frac{n\cdot (n+1)}{2}+\lfloor \frac{b}{c}\rfloor \cdot (n+1)+f(a\mathrm{\%}c,b\mathrm{\%}c,c,n),& a\ge c||b\ge c\\ \lfloor \frac{a\cdot n+b}{c}\rfloor \cdot n-f(c,c-b-1,a,\lfloor \frac{a\cdot n+b}{c}\rfloor -1),& a<c\mathrm{\&}\mathrm{\&}b<c\end{array}\\ g(a,b,c,n)& =\{\begin{array}{ll}{\lfloor \frac{b}{c}\rfloor }^2\cdot (n+1),& a=0\\ {\lfloor \frac{a}{c}\rfloor }^2\cdot \frac{n\cdot (n+1)\cdot (2n+1)}{6}+\lfloor \frac{a}{c}\rfloor \cdot \lfloor \frac{b}{c}\rfloor \cdot n\cdot (n+1)+{\lfloor \frac{b}{c}\rfloor }^2\cdot (n+1)+\\ g(a\mathrm{\%}c,b\mathrm{\%}c,c,n)+2\cdot \lfloor \frac{a}{c}\rfloor \cdot h(a\mathrm{\%}c,b\mathrm{\%}c,c,n)+2\cdot \lfloor \frac{b}{c}\rfloor \cdot f(a\mathrm{\%}c,b\mathrm{\%}c,c,n),& a\ge c||b\ge c\\ {\lfloor \frac{a\cdot n+b}{c}\rfloor }^2\cdot n-f(c,c-b-1,a,\lfloor \frac{a\cdot n+b}{c}\rfloor -1)-2\cdot h(c,c-b-1,a,\lfloor \frac{a\cdot n+b}{c}\rfloor -1),& a<c\mathrm{\&}\mathrm{\&}b<c\end{array}\\ h(a,b,c,n)& =\{\begin{array}{ll}\lfloor \frac{b}{c}\rfloor \cdot \frac{n\cdot (n+1)}{2},& a=0\\ \lfloor \frac{a}{c}\rfloor \cdot \frac{n\cdot (n+1)\cdot (2n+1)}{6}+\lfloor \frac{b}{c}\rfloor \cdot \frac{n\cdot (n+1)}{2}+h(a\mathrm{\%}c,b\mathrm{\%}c,c,n),& a\ge c||b\ge c\\ \lfloor \frac{n\cdot a+b}{c}\rfloor \cdot \frac{n\cdot (n+1)}{2}-\frac{1}{2}\cdot g(c,c-b-1,a,\lfloor \frac{a\cdot n+b}{c}\rfloor -1)-\frac{1}{2}\cdot f(c,c-b-1,a,\lfloor \frac{a\cdot n+b}{c}\rfloor -1),& a<c\mathrm{\&}\mathrm{\&}b<c\end{array}\end{array}$$

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;
}