类欧几里得（模板题推导）

类欧几里得

设三个函数 $\sum\limits_{i = 0} ^{n} \frac{a \times i + b}{c}, g(a, b, c, n) = \sum\limits_{i = 0} ^{n} i \times \frac{a \times i + b}{c}, h(a, b, c, n) = \sum\limits_{i = 0} ^{n} \left(\frac{a \times i + b}{c} \right) ^ 2$ 。

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

下面除法均为向下取整：

$\ge c$

设 $\frac{b}{c} \times c$ ：
$\sum_{i = 0} ^{n} \frac{a \times i + b'}{c} + \frac{b}{c}\\ = \frac{b}{c} \times n + \sum_{i = 0} ^{n} \frac{a \times i + b'}{c}\\ = \frac{b}{c} \times n + f(a, b', c, n)$

$\ge c$

设 $\frac{a}{c} \times c$ ：
$\sum_{i = 0} ^{n} \frac{a' \times i + b}{c} + \frac{a}{c} \times i\\ = \frac{a}{c} \times \frac{n \times (n + 1)}{2} + \sum_{i = 0} ^{n} \frac{a' \times i + b}{c}\\ = \frac{a}{c} \times \frac{n \times (n + 1)}{2} + f(a', b, c, n)\\$

综上，如果 $\ge c,\ b \ge c$ ：
$\frac{b}{c} \times n + \frac{a}{c} \times \frac{n \times (n + 1)}{2}$

考虑 $c,\ b < c$ ：
$∑i=0na×i+bc∑i=0n∑j=1a×i+bc1\sum_{i = 0} ^{n} \frac{a \times i + b}{c}\\ \sum_{i = 0} ^{n} \sum_{j = 1} ^{\frac{a \times i + b}{c}}1\\$
令 $\frac{a \times n + b}{c}$ ，
$∑i=0n∑j=1m[a×i+bc≥j]∑i=0n∑j=0m−1[a×i+bc≥j+1]∑i=0n∑j=0m−1[a×i+b≥jc+c]∑i=0n∑j=0m−1[a×i≥jc+c−b]∑i=0n∑j=0m−1[a×i>jc+c−b−1]∑i=0n∑j=0m−1[i>jc+c−b−1]a∑j=0m−1∑i=0n[i>jc+c−b−1]a∑j=0m−1n−jc+c−b−1an×m−∑i=0m−1ic+c−b−1a\sum_{i = 0} ^{n} \sum_{j = 1} ^{m} [\frac{a \times i + b}{c} \ge j]\\ \sum_{i = 0} ^{n} \sum_{j = 0} ^{m - 1}[\frac{a \times i + b}{c} \ge j + 1]\\ \sum_{i = 0} ^{n} \sum_{j = 0} ^{m - 1}[a \times i + b \ge jc + c]\\ \sum_{i = 0} ^{n} \sum_{j = 0} ^{m - 1}[a \times i \ge jc + c - b]\\ \sum_{i = 0} ^{n} \sum_{j = 0} ^{m - 1}[a \times i > jc + c - b - 1]\\ \sum_{i = 0} ^{n} \sum_{j = 0} ^{m - 1}[i > \frac{jc + c - b - 1]}{a}\\ \sum_{j = 0} ^{m - 1} \sum_{i = 0} ^{n}[i > \frac{jc + c - b - 1]}{a}\\ \sum_{j = 0} ^{m - 1} n - \frac{jc + c - b - 1}{a}\\ n \times m - \sum_{i = 0} ^{m - 1} \frac{ic + c - b - 1}{a}\\$
$a^{'} = c, b^{'} = c - b - 1, c^{'} = a$ ， $\times m - f(a', b', c', m - 1)$ 。

过程中 $\% c, c = a$ ，发现跟 $gcd⁡\gcd$ 的求解过程是一样的，最后 $a$ 会变成 $0$ ，也就是递归出口。

long long f(long long a, long long b, long long c, long long n) {if (!a) {return (b / c) * (n + 1);}if (a >= c || b >= c) {return f(a % c, b % c, c, n) + (b / c) * (n + 1) + (a / c) * n * (n + 1) / 2;}long long m = (a * n + b) / c;return n * m - f(c, c - b - 1, a, m - 1);
}

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

$\ge c, b' = b\ \%\ c, d = \frac{b}{c}$ :
$∑i=0ni×a×i+b′c+d×id×n×(n+1)2+∑i=0ni×a×i+b′cg(a,b,c,n)=d×n×(n+1)2+g(a,b′,c,n)\sum_{i = 0} ^{n} i \times \frac{a \times i + b'}{c} + d \times i\\ d \times \frac{n \times (n + 1)}{2} + \sum_{i = 0} ^{n} i \times \frac{a \times i + b'}{c}\\ g(a, b, c, n) = d \times \frac{n \times (n + 1)}{2} + g(a, b', c, n)\\$
$\ge c, a' = a\ \%\ c, d = \frac{a}{c}$ ：
$∑i=0ni×a′×i+bc+d×i2d×n×(n+1)×(2n+1)6+∑i=0ni×a′×i+bcd×n×(n+1)×(2n+1)6+g(a′,b,c,n)\sum_{i = 0} ^{n} i \times \frac{a' \times i + b}{c} + d \times i ^ 2\\ d \times \frac{n \times (n + 1) \times (2n + 1)}{6} + \sum_{i = 0} ^{n} i \times \frac{a' \times i + b}{c}\\ d \times \frac{n \times (n + 1) \times (2 n + 1)}{6} + g(a', b, c, n)\\$
综上， $\ge c, \ b \ge c$ ：
$\frac{b}{c} \times \frac{n \times (n + 1)}{2} + \frac{a}{c} \times \frac{n \times (n + 1) \times (2n + 1)}{6} + g(a', b', c, n)\\$

$\frac{a \times n + b}{c}, a' = c, b' = c - b - 1, c'= a$ ：
$∑i=0ni×∑j=1a×i+bc1∑i=0ni×∑j=1m[a×i+bc≥j]∑i=0ni×∑j=0m−1[a×i+bc≥j+1]∑i=0ni×∑j=0m−1[i>jc+c−b−1a]∑j=0m−1∑i=0ni[i>jc+c−b−1a]calc(n)=n×(n+1)2∑j=0m−1calc(n)−calc(jc+c−b−1a)m×n×(n+1)−f(a′,b′,c′,m−1)−h(a′,b′,c′,m−1)2\sum_{i = 0} ^{n} i \times \sum_{j = 1} ^{\frac{a \times i + b}{c}}1\\ \sum_{i = 0} ^{n} i \times \sum_{j = 1} ^{m} [\frac{a \times i + b}{c} \ge j]\\ \sum_{i = 0} ^{n} i \times \sum_{j = 0} ^{m - 1} [\frac{a \times i + b}{c} \ge j + 1]\\ \sum_{i = 0} ^{n} i \times \sum_{j = 0} ^{m - 1} [i > \frac{jc + c - b - 1}{a}]\\ \sum_{j = 0} ^{m - 1} \sum_{i = 0} ^{n} i [i > \frac{jc + c - b - 1}{a}]\\ calc(n) = \frac{n \times (n + 1)}{2}\\ \sum_{j = 0} ^{m - 1} calc(n) - calc(\frac{jc + c - b - 1}{a})\\ \frac{m \times n \times (n + 1) - f(a', b', c', m- 1) - h(a', b', c', m - 1)}{2}\\$

$\sum\limits_{i = 0} ^{n} \left(\frac{a \times i + b}{c} \right) ^ 2, (0 \le a, b, c, n)$

$\ge c,b' = b \ \% \ c, d = \frac{b}{c}$
$∑i=0n(a×i+b′c+d)2∑i=0n(a×i+b′c)2+d2+2×d×a×i+b′cd2×(n+1)+2×d×f(a,b′,c,n)+h(a,b′,c,n)\sum_{i = 0} ^{n} \left(\frac{a \times i + b'}{c} + d \right) ^2\\ \sum_{i = 0} ^{n} \left(\frac{a \times i + b'}{c} \right) ^ 2 + d ^ 2 + 2 \times d \times \frac{a \times i + b'}{c}\\ d ^ 2 \times (n + 1) + 2 \times d \times f(a, b', c, n) + h(a, b', c, n)\\$
$\ge c, a' = a\ \%\ c, d = \frac{a}{c}$
$∑i=0n(d×i+a′×i+bc)2∑i=0nd2×i2+(a′×i+bc)2+2×d×i×a′×i+bcd2×n×(n+1)×(2n+1)6+h(a′,b,c,n)+2×d×g(a′,b,c,n)\sum_{i = 0} ^{n} \left(d \times i + \frac{a' \times i + b}{c} \right) ^ 2\\ \sum_{i = 0} ^{n} d ^ 2 \times i ^ 2 + \left(\frac{a' \times i + b}{c} \right) ^ 2 + 2 \times d \times i \times \frac{a' \times i + b}{c}\\ d ^ 2 \times \frac{n \times (n + 1) \times (2n + 1)}{6} + h(a', b, c, n) + 2 \times d \times g(a', b, c, n)\\$
综上， $\ge c,\ b \ge c$ ：

$\frac{a}{c} \times \frac{b}{c} \times n \times (n + 1) + (\frac{b}{c}) ^ 2 \times (n + 1) + (\frac{a}{c}) ^ 2 \times \frac{n \times (n + 1) \times (2n + 1)}{6}+\\ 2 \times \frac{b}{c} \times f(a', b', c, n) + h(a', b', c, n) + 2 \times \frac{a}{c} \times g(a', b', c, n)\\$

$c,\ b < c, m = \frac{a \times n + b}{c}$ ：

$\times \frac{n \times (n + 1)}{2} - n = 2 \times \sum_{i = 0} ^{n} i - n\\ \sum_{i = 0} ^{n} (\frac{a \times i + b}{c}) ^ 2\\ \sum_{i = 0} ^{n} \left(2 \times \sum_{j = 0} ^{\frac{a \times i + b}{c}}j - \frac{a \times i + b}{c} \right)\\ h(a, b, c, n) = 2 \times \sum_{i = 0} ^{n} \sum_{j = 0} ^{\frac{a \times i + b}{c}} j - f(a, b, c, n)\\ \sum_{i = 0} ^{n} \sum_{j = 0} ^ \frac{a \times i + b}{c}j\\ \sum_{j = 1} ^{m} j\sum_{i = 0} ^{n} [\frac{a \times i + b}{c} \ge j]\\ \sum_{j = 0} ^{m - 1} (j + 1) \sum_{i = 0} ^{n} [\frac{a \times i + b}{c} \ge j + 1]\\ \sum_{j = 0} ^{m - 1} (j + 1) \sum_{i = 0} ^{n}[a > \frac{jc + c - b - 1}{a}]\\ \sum_{j = 0} ^{m - 1} (j + 1) (n - \frac{jc + c - b - 1}{a})\\ \frac{m \times (m + 1)}{2} \times n - g(a', b', c', n) - f(a', b', c')\\ h(a, b, c, n) = m \times (m + 1) \times n - f(a, b, c, n) - 2 \times g(a', b', c', n) - 2 \times f(a', b', c', n)\\$

P5170 【模板】类欧几里得算法

#include <bits/stdc++.h>using namespace std;const int mod = 998244353, inv2 = 499122177, inv6 = 166374059;struct Res {int f, g, h;
};inline int add(int x, int y) {return x + y < mod ? x + y : x + y - mod;
}inline int sub(int x, int y) {return x >= y ? x - y : x - y + mod;
}Res calc(int a, int b, int c, int n) {if (!a) {return {1ll * (b / c) * (n + 1) % mod, 1ll * (b / c) * n % mod * (n + 1) % mod * inv2 % mod, 1ll * (b / c) * (b / c) % mod * (n + 1) % mod};}if (a >= c || b >= c) {Res cur= calc(a % c, b % c, c, n);int f = add(add(cur.f, 1ll * (b / c) * (n + 1) % mod), 1ll * (a / c) * n % mod * (n + 1) % mod * inv2 % mod);int g = add(add(1ll * (b / c) * n % mod * (n + 1) % mod * inv2 % mod, 1ll * (a / c) * n % mod * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod), cur.g);int h = add(add(add(1ll * (a / c) * (b / c) % mod * n % mod * (n + 1) % mod, 1ll * (b / c) * (b / c) % mod * (n + 1) % mod), add(1ll * (a / c) * (a / c) % mod * n % mod * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod, 1ll * 2 * (b / c) % mod * cur.f % mod)), add(cur.h, 1ll * 2 * (a / c) * cur.g % mod));return {f, g, h};}int m = (1ll * a * n + b) / c;Res cur = calc(c, c - b - 1, a, m - 1);int f = sub(1ll * n * m % mod, cur.f);int g = 1ll * sub(sub(1ll * m * n % mod * (n + 1) % mod, cur.f), cur.h) * inv2 % mod;int h = sub(sub(sub(1ll * m * (m + 1) % mod * n % mod, f), 2ll * cur.g % mod), 2ll * cur.f % mod);return {f, g, h};
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);int T, a, b, c, n;scanf("%d", &T);while (T--) {scanf("%d %d %d %d", &n, &a, &b, &c);Res ans = calc(a, b, c, n);printf("%d %d %d\n", ans.f, ans.h, ans.g);}return 0;
}