类欧几里得算法

时间复杂度O(logn)O(logn) 求f(a,b,c,n)=∑i=0n⌊ai+bc⌋g(a,b,c,n)=∑i=0ni⌊ai+bc⌋h(a,b,c,n)=∑i=0n⌊ai+bc⌋2f(a,b,c,n)=\sum_{i=0}^{n} \lfloor \frac{ai+b}{c} \rfloor \quad g(a,b,c,n)=\sum_{i=0}^{n} i\lfloor \frac{ai+b}{c} \rfloor \quad h(a,b,c,n)=\sum_{i=0}^{n} \lfloor \frac{ai+b}{c} \rfloor^2 这三个函数的值

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#include <bits/stdc++.h>
#define int long long
using namespace std;
const int P=998244353;
int i2=499122177,i6=166374059;// i2,i6分别是2,6在(mod p)意义下的逆元
struct data {
  data() { f = g = h = 0; }
  int f, g, h;
};  // 三个函数打包
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) {  // 迭代到最底层
    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;
}