[模板] 多项式全家桶

包含: 多项式乘法, 多项式求逆, 多项式 ln, 多项式 exp, 多项式快速幂.

exp 的 (O(n log n)) 做法常数太大, 实际表现还不如 (O(n log^2 n)) (当然也有可能是我写丑了), 所以就放了 (O(n log^2 n)) 的做法.

学习笔记什么的会找时间补上 (尽量不咕).

#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 - 2), 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;
}
原文地址:https://www.cnblogs.com/BruceW/p/13892271.html