类欧几里得算法推导

初识

给出三种形式：

• $f(a,b,c,n)={\sum }_{i=0}^n\lfloor \frac{ai+b}{c}\rfloor$
• $g(a,b,c,n)={\sum }_{i=0}^{n}i\lfloor \frac{ai+b}{c}\rfloor$
• $h(a,b,c,n)={\sum }_{i=1}^n\lfloor \frac{ai+b}{c}{\rfloor }^2$

接下来我们进入正题，讲解这三个式子的求解

f(a, b, c, n)

$$\begin{gathered} f(a,b,c,n)=\sum _{i=1}^n\lfloor \frac{ai+b}{c}\rfloor \\ ifa\ge c或者b\ge c \\ f(a,b,c,n)=f(a\%c,b\%c,c,n)+\frac{n(n+1)}{2}\lfloor \frac{a}{c}\rfloor +(n+1)\lfloor \frac{b}{c}\rfloor \\ ifa<c并且b<c \\ 我们假定m=\lfloor \frac{an+b}{c}\rfloor \\ 有f(a,b,c,n)=\sum _{i=0}^n\sum _{j=1}^m(\lfloor \frac{ai+b}{c}\rfloor \ge j) \\ =\sum _{i=0}^n\sum _{j=0}^{m-1}(\lfloor \frac{ai+b}{c}\rfloor \ge j+1) \\ =\sum _{i=0}^n\sum _{j=0}^{m-1}(i>\lfloor \frac{jc+c-b-1}{a})\rfloor \\ =\sum _{j=0}^{m-1}n-\lfloor \frac{jc+c-b-1}{a}\rfloor \\ =nm-f(c,c-b-1,a,m-1) \\ 也就是a=c,c=a\%c \\ 最后的递归边界就是a=0 \end{gathered}$$
最后我们就可以通过递归求得其函数值。

g(a, b, c, n)

$$\begin{gathered} g(a,b,c,n)=\sum _{i=0}^{n}i\lfloor \frac{ai+b}{c}\rfloor \\ ifa\ge c或者b\ge c \\ g(a,b,c,n)=g(a\%c,b\%c,c,n)+\frac{n(n+1)(2n+1)}{6}\lfloor \frac{a}{c}\rfloor +\frac{n(n+1)}{2}\lfloor \frac{b}{c}\rfloor \\ ifa<c并且b<c \\ 同上面式子一样m=\lfloor \frac{an+b}{c}\rfloor \\ g(a,b,c,n)=\sum _{i=0}^{n}i\sum _{j=1}^m(\lfloor \frac{ai+b}{c}\rfloor \ge j) \\ =\sum _{i=0}^{n}i\sum _{j=0}^{m-1}(i>\lfloor \frac{jc+c-b-1}{a})\rfloor \\ 显然有i的下界为\lfloor \frac{jc+c-b-1}{a}\rfloor +1，上界是n，所以构成了一个等差数列 \\ 最后得到g(a,b,c,n)=\sum _{j=0}^{m-1}\lfloor \frac{(\lfloor \frac{jc+c-b-1}{a}\rfloor +1+n)(n-\lfloor \frac{jc+c-b-1}{a}\rfloor )}{2}\rfloor \\ =\lfloor \frac{mn(n+1)-f(c,c-b-1,a,m-1)-h(c,c-b-1,a,m-1)}{2}\rfloor \\ 所以我们先推导h(a,b,c,n) \end{gathered}$$

h(a, b, c, n)

$$\begin{gathered} h(a,b,c,n)=\sum _{i=0}^n\lfloor \frac{ai+b}{c}{\rfloor }^2 \\ ifa\ge c或者b\ge c \\ h(a,b,c,n)=\sum _{i=0}^n{\left(\frac{i(a\%c)+b\%c}{c}+\lfloor \frac{a}{c}\rfloor i+\lfloor \frac{b}{c}\rfloor \right)}^2 \\ =\sum _{i=0}^n{\left(\frac{i(a\%c)+b\%c}{c}\right)}^2+\sum _{i=0}^n\lfloor \frac{a}{c}{\rfloor }^2{i}^2+\sum _{i=0}^n\lfloor \frac{b}{c}{\rfloor }^2+2\lfloor \frac{a}{c}\rfloor \sum _{i=0}^{n}i\frac{i(a\%c)+b\%c}{c}+2\lfloor \frac{b}{c}\rfloor \sum _{i=0}^n\frac{i(a\%c)+b\%c}{c}+2\lfloor \frac{a}{c}\rfloor \lfloor \frac{b}{c}\rfloor \sum _{i=0}^{n}i \\ =h(a\%c,b\%c,c,n)+2\lfloor \frac{b}{c}\rfloor f(a\%c,b\%c,c,n)+2\lfloor \frac{a}{c}\rfloor g(a\%c,b\%c,c,n)+\frac{n(n+1)(2n+1)}{6}\lfloor \frac{a}{c}{\rfloor }^2+n(n+1)\lfloor \frac{a}{c}\rfloor \lfloor \frac{b}{c}\rfloor +(n+1)\lfloor \frac{b}{c}{\rfloor }^2 \\ ifa<c并且b<c \\ {n}^2=2\frac{n(n+1)}{2}-n=(2\sum _{i=0}^{n}i)-n \\ h(a,b,c,n)=\sum _{i=0}^n(2\sum _{j=0}^{\lfloor \frac{ai+b}{c}\rfloor }j)-\lfloor \frac{ai+b}{c}\rfloor \\ m=\lfloor \frac{an+b}{c}\rfloor \\ h(a,b,c,n)=2\sum _{j=0}^{m-1}(j+1)\sum _{i=0}^n(\lfloor \frac{ai+b}{c}\rfloor \ge j+1)-f(a,b,c,n) \\ =2\sum _{j=0}^m(j+1)\sum _{i=0}^{n}i>\lfloor \frac{cj+c-b-1}{a}\rfloor =\sum _{j=0}^n(j+1)(n-\lfloor \frac{cj+c-b-1}{a})\rfloor )-f(a,b,c,n) \\ =2\left(\sum _{j=1}^{m}jn-\sum _{j=0}^{m-1}(j+1)\lfloor \frac{cj+c-b-1}{a})\rfloor \right)-f(a,b,c,n) \\ 最后得到h(a,b,c,n)=nm(m+1)-2f(c,c-b-1,a,m-1)-2g(c,c-b-1,a,m-1)-f(a,b,c,n) \end{gathered}$$

总结

$f(a,b,c,n)$是可以单独求的，但是$g(a,b,c,n),h(a,b,c,n)$是得互相得到并且要通过$f(a,b,c,n)$得到，所以我们可以通过一个结构体一次求得三个函数的值，然后一起返回。

P5170 【模板】类欧几里得算法

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int mod = 998244353, inv2 = 499122177, inv6 = 166374059;struct Node {ll f, g, h;
};ll quick_pow(ll a, ll n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans;
}Node solve(ll a, ll b, ll c, ll n) {Node ans, temp;if(!a) {ans.f = (n + 1) * (b / c) % mod;ans.g = (n + 1) * n % mod * inv2 % mod * (b / c) % mod;ans.h = (n + 1) * (b / c) % mod * (b / c) % mod;return ans;}if(a >= c || b >= c) {temp = solve(a % c, b % c, c, n);ans.f = (temp.f + n * (n + 1) % mod * inv2 % mod * (a / c) % mod + (n + 1) * (b / c) % mod) % mod;ans.g = (temp.g + n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod * (a / c) % mod + n * (n + 1) % mod * inv2 % mod * (b / c) % mod) % mod;ans.h = (temp.h + 2 * (b / c) % mod * temp.f % mod + 2 * (a / c) % mod * temp.g % mod +  n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod * (a / c) % mod * (a / c) % mod + n * (n + 1) % mod * (a / c) % mod * (b / c) % mod + (n + 1) * (b / c) % mod * (b / c) % mod) % mod;return ans;}ll m = (a * n + b) / c;temp = solve(c, c - b - 1, a, m - 1);ans.f = (n * m % mod - temp.f + mod) % mod;ans.g = ((m * n % mod * (n + 1) % mod - temp.f - temp.h) % mod * inv2 % mod + mod) % mod;ans.h = ((n * m % mod * (m + 1) % mod - 2 * temp.f % mod - 2 * temp.g % mod - ans.f) % mod + mod) % mod;return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T = read();while(T--) {ll n = read(), a = read(), b = read(), c = read();Node ans = solve(a, b, c, n);printf("%lld %lld %lld\n", ans.f, ans.h, ans.g);}return 0;
}