[模板] 類歐幾里得演算法
阿新 • • 發佈：2021-08-05
類歐幾里得演算法
\[\rm
f(a,b,c,n)=\sum\limits_{i=0}^n \lfloor \dfrac{ai+b}{c} \rfloor\\
g(a,b,c,n)=\sum\limits_{i=0}^n i\lfloor \dfrac{ai+b}{c} \rfloor\\
h(a,b,c,n)={\sum\limits_{i=0}^n \lfloor \dfrac{ai+b}{c} \rfloor }^2
\]
抄了一份板子
未經許可，禁止搬運。#include <bits/stdc++.h> #define int long long using namespace std; const int P = 998244353; int i2 = 499122177, i6 = 166374059; struct data { data() { f = g = h = 0; } int f, g, h; }; // 三個函式打包 data calc(int n, int a, int b, int c) { int ac = a / c, bc = b / c, m = (a * n + b) / c, n1 = n + 1, n21 = n * 2 + 1; data d; if (a == 0) { // 迭代到最底層 d.f = bc * n1 % P; d.g = bc * n % P * n1 % P * i2 % P; d.h = bc * bc % P * n1 % P; return d; } if (a >= c || b >= c) { // 取模 d.f = n * n1 % P * i2 % P * ac % P + bc * n1 % P; d.g = ac * n % P * n1 % P * n21 % P * i6 % P + bc * n % P * n1 % P * i2 % P; d.h = ac * ac % P * n % P * n1 % P * n21 % P * i6 % P + bc * bc % P * n1 % P + ac * bc % P * n % P * n1 % P; d.f %= P, d.g %= P, d.h %= P; data e = calc(n, a % c, b % c, c); // 迭代 d.h += e.h + 2 * bc % P * e.f % P + 2 * ac % P * e.g % P; d.g += e.g, d.f += e.f; d.f %= P, d.g %= P, d.h %= P; return d; } data e = calc(m - 1, c, c - b - 1, a); d.f = n * m % P - e.f, d.f = (d.f % P + P) % P; d.g = m * n % P * n1 % P - e.h - e.f, d.g = (d.g * i2 % P + P) % P; d.h = n * m % P * (m + 1) % P - 2 * e.g - 2 * e.f - d.f; d.h = (d.h % P + P) % P; return d; } int T, n, a, b, c; signed main() { scanf("%lld", &T); while (T--) { scanf("%lld%lld%lld%lld", &n, &a, &b, &c); data ans = calc(n, a, b, c); printf("%lld %lld %lld\n", ans.f, ans.h, ans.g); } return 0; }