[模板] 多項式全家桶
阿新 • • 發佈:2020-10-28
包含: 多項式乘法, 多項式求逆, 多項式 ln, 多項式 exp, 多項式快速冪.
exp 的 \(O(n \log n)\) 做法常數太大, 實際表現還不如 \(O(n \log^2 n)\) (當然也有可能是我寫醜了), 所以就放了 \(O(n \log^2 n)\) 的做法.
學習筆記什麼的會找時間補上 (儘量不咕).
#include <cstdio> #include <cstring> #include <iostream> using namespace std; typedef long long ll; const int _ = (1 << 18) + 7; const int mod = 998244353, rt = 3; int n, K1, K2, g[_], f[_]; bool flag; int Pw(int a, int p) { int res = 1; while (p) { if (p & 1) res = (ll)res * a % mod; a = (ll)a * a % mod; p >>= 1; } return res; } namespace POLY { int tot, num[_], pwrt[2][_], inv[_], tmp[5][_]; void Clear(int *f, int L) { memset(f, 0, L << 2); } void Cpy(int *h, int *f, int L) { memcpy(h, f, L << 2); } void Init() { tot = 1; while (tot <= n + n) tot <<= 1; inv[1] = 1; for (int i = 2; i <= tot; ++i) inv[i] = (ll)inv[mod % i] * (mod - mod / i) % mod; pwrt[0][tot] = Pw(rt, (mod - 1) / tot); pwrt[1][tot] = Pw(pwrt[0][tot], mod - 2); for (int len = (tot >> 1); len; len >>= 1) { pwrt[0][len] = (ll)pwrt[0][len << 1] * pwrt[0][len << 1] % mod; pwrt[1][len] = (ll)pwrt[1][len << 1] * pwrt[1][len << 1] % mod; } } void NTT(int *f, int t, bool ty) { for (int i = 1; i < t; ++i) { num[i] = (num[i >> 1] >> 1) | ((i & 1) ? t >> 1 : 0); if (i < num[i]) swap(f[i], f[num[i]]); } for (int len = 2; len <= t; len <<= 1) { int gap = len >> 1, w1 = pwrt[ty][len]; for (int i = 0, w = 1, tmp; i < t; i += len, w = 1) for (int j = i; j < i + gap; ++j) { tmp = (ll)w * f[j + gap] % mod; f[j + gap] = (f[j] - tmp + mod) % mod; f[j] = (f[j] + tmp) % mod; w = (ll)w * w1 % mod; } } if (ty) for (int i = 0; i < t; ++i) f[i] = (ll)f[i] * inv[t] % mod; } void Inv(int *h, int *f, int L) { int a[_]; Clear(h, L >> 1), Clear(a, L >> 1); h[0] = Pw(f[0], mod - 1), a[0] = f[0], a[1] = f[1]; for (int len = 2, t = 4; len <= L; len <<= 1, t <<= 1) { NTT(h, t, 0), NTT(a, t, 0); for (int i = 0; i < t; ++i) h[i] = (ll)h[i] * (2 - (ll)a[i] * h[i] % mod + mod) % mod; NTT(h, t, 1), NTT(a, t, 1); for (int i = len; i < t; i++) a[i] = f[i], h[i] = 0; } } void Deriv(int *h, int *f, int L) { for (int i = 0; i < L - 1; ++i) h[i] = (ll)f[i + 1] * (i + 1) % mod; } void Integ(int *h, int *f, int L) { for (int i = L - 1; i; --i) h[i] = (ll)f[i - 1] * inv[i] % mod; h[0] = 0; } void Ln(int *h, int *f, int L) { Clear(h, L << 1); int a[(L << 1) + 7], b[(L << 1) + 7]; Clear(a, L << 1), Clear(b, L << 1); Deriv(a, f, L), Inv(b, f, L); NTT(a, L << 1, 0), NTT(b, L << 1, 0); for (int i = 0; i < (L << 1); ++i) h[i] = (ll)a[i] * b[i] % mod; NTT(h, L << 1, 1); Integ(h, h, L); } void dcExp(int *f, int *g, int t, int l, int r) { if (t == 1) { f[0] = l ? (ll)f[0] * inv[l] % mod : f[0]; return; } dcExp(f, g, t >> 1, l, (l + r) >> 1); int a[t + 7], b[t + 7]; Clear(a, t), Clear(b, t); Cpy(a, f, t >> 1), Cpy(b, g, t); NTT(a, t, 0), NTT(b, t, 0); for (int i = 0; i < t; ++i) a[i] = (ll)a[i] * b[i] % mod; NTT(a, t, 1); for (int i = (t >> 1); i < t; ++i) f[i] = (f[i] + a[i - 1]) % mod; dcExp(f + (t >> 1), g, t >> 1, (l + r) >> 1, r); } void Exp(int *f, int *g, int L) { Deriv(g, g, L), f[0] = 1; dcExp(f, g, L, 1, L); } void Pow(int *h, int *f, int K1, int K2, int L) { int st = 0; while (st < n and !f[st]) ++st; if ((flag and st) || (ll)st * K1 >= (ll)n) return; int inv = Pw(f[st], mod - 2), tmp = f[st]; for (int i = 0; i < n; ++i) f[i] = (ll)f[i + st] * inv % mod; int a[(L << 1) + 7]; Clear(a, L << 1); Ln(a, f, L); for (int i = 0; i < n; ++i) a[i] = (ll)a[i] * K1 % mod; Exp(g, a, L); st *= K1, tmp = Pw(tmp, K2); for (int i = n - 1; i >= st; --i) g[i] = (ll)g[i - st] * tmp % mod; for (int i = 0; i < st; ++i) g[i] = 0; } } void Gi(int &K1, int &K2) { ll t1 = 0, t2 = 0; char c = getchar(); while (!isdigit(c)) c = getchar(); while (isdigit(c)) { t1 = t1 * 10 + c - '0', t2 = t2 * 10 + c - '0'; if (t1 >= mod) flag = 1, t1 %= mod; t2 %= (mod - 1); c = getchar(); } K1 = t1, K2 = t2; }