用来快速求解 $\sum\limits_{i=0}^{n}\lfloor \frac{ai+b}{c} \rfloor,\sum\limits_{i=0}^{n}{\lfloor \frac{ai+b}{c} \rfloor}^2,\sum\limits_{i=0}^{n}i\lfloor \frac{ai+b}{c} \rfloor $
有多快呢？据说是log的？反正abc取1e9可以200ms过1e5组询问……
#include <bits/stdc++.h>
typedef long long ll;
constexpr int mod = 998244353;
constexpr ll inv2 = 499122177;
constexpr ll inv6 = 166374059;
ll f(ll a, ll b, ll c, ll n);
ll g(ll a, ll b, ll c, ll n);
ll h(ll a, ll b, ll c, ll n);
struct Query {
ll f, g, h;
};
Query solve(ll a, ll b, ll c, ll n) {
Query ans, tmp;
if(a == 0) {
ans.f = (n + 1) * (b / c) % mod;
ans.g = (b / c) * n % mod * (n + 1) % mod * inv2 % mod;
ans.h = (n + 1) * (b / c) % mod * (b / c) % mod;
return ans;
}
if(a >= c || b >= c) {
tmp = solve(a % c, b % c, c, n);
ans.f = (tmp.f + (a / c) * n % mod * (n + 1) % mod * inv2 % mod + (b / c) * (n + 1) % mod) % mod;
ans.g = (tmp.g + (a / c) * n % mod * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod + (b / c) * n % mod * (n + 1) % mod * inv2 % mod) % mod;
ans.h = ((a / c) * (a / c) % mod * n % mod * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod +
(b / c) * (b / c) % mod * (n + 1) % mod + (a / c) * (b / c) % mod * n % mod * (n + 1) % mod +
tmp.h + 2 * (a / c) % mod * tmp.g % mod + 2 * (b / c) % mod * tmp.f % mod) % mod;
return ans;
}
ll m = (a * n + b) / c;
tmp = solve(c, c - b - 1, a, m - 1);
ans.f = (n * (m % mod) % mod - tmp.f) % mod;
ans.g = (n * (n + 1) % mod * (m % mod) % mod - tmp.f - tmp.h) % mod * inv2 % mod;
ans.h = (n * (m % mod) % mod * ((m + 1) % mod) % mod - 2 * tmp.g - 2 * tmp.f - ans.f) % mod;
return ans;
}
inline char nc() {
static char buf[1000000], *p1 = buf, *p2 = buf;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2) ? EOF : *p1++;
}
inline ll read() {
ll res = 0;
char ch;
do
ch = nc();
while(ch < 48 || ch > 57);
do
res = res * 10 + ch - 48, ch = nc();
while(ch >= 48 && ch <= 57);
return res;
}
int main() {
ll t = read();
ll n, a, b, c;
while(t--) {
n = read(), a = read(), b = read(), c = read();
Query ans = solve(a, b, c, n);
printf("%lld %lld %lld\n", (ans.f + mod) % mod, (ans.h + mod) % mod, (ans.g + mod) % mod);
}
return 0;
}