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多项式操作公式记录

不证明了。

未完待续，随学随更。

以下默认 \(A(x)\) 为 \((n - 1)\) 次多项式 \(A(x) = \sum_{i = 0}^{n - 1} a_i x^i\)，且 \(n\) 为 \(2\) 的整数幂。

快速傅里叶变换/数论变换

\[A_(\omega_n^k)~=~A_0(\omega_m^k) + \omega_n^kA_1(\omega_m^k)\]

\[A(\omega_n^{k + m})~=~A_0(\omega_m^k) - \omega_n^kA_1(\omega_m^k)\]

其中 \(m = \frac{n - 1}{2}\)，\(A_0(x) = \sum_{i = 0}^m a_{2i} x^i\)，\(A_1(x) = \sum_{i = 0}^m a^{2i + 1}x^i\)。\(\omega_n\) 代表最小的 \(n\) 次单位根/原根。

求乘法逆

\[B_{2k}(x) \equiv 2 B_{k}(x) - A_{2k}(x) \times B_{k}^{2}(x) \pmod {x^{2k}}\]

其中 \(A_k(x)\) 代表 \(A\) 的前 \(k\) 项；\(B_k(x)~=~A_k^{-1}(x)\)。

注意：\(B_k(x)\) 不一定是 \(A_n(x)\) 乘法逆的前 \(k\) 项，它只是 \(A_n(x)\) 前 \(k\) 项的乘法逆。

求对数函数

设

\[B(x) = \ln A(x)\]

有

\[B'(x)~\equiv~A'(x) A^{-1}(x) \pmod {x^n}\]

则

\[B(x)~=~\int B'(x) \text{d}x\]

求导、积分公式：

\[f(x) = Cx^a~~\Rightarrow~~f'(x) = C(a - 1)x^{a - 1}\]

\[f(x) = Cx^a ~~\Rightarrow~~ \int f(x) \text{d}x = \frac{C}{a + 1} x^{a + 1}\]

其中 \(C\) 是常数。对于多项式，二者均是线性可加的。

封装代码

#include <cstdio>
#include <cstring>
#include <algorithm>

template <typename T>
void qr(T &x);
template <typename T>
void qw(T x, const char Ch, const bool NeedPutChar);

namespace Poly {

const int G = 3;
const int MOD = 998244353;
const int PHI = 998244352;
const int DPHI = 998244351;
const int SzI = sizeof(int);
const int SzL = sizeof(long long int);

void Init(int *const A, const int N) {
  for (int i = 0; i < N; ++i) {
    qr(A[i]);
  }
}

void Print(int *const A, const int N) {
  int DN = N - 1;
  for (int i = 0; i < DN; ++i) {
    qw(A[i], ' ', true);
  }
  qw(A[DN], '\n', true);
}

void GetN(int n, int &N) {
  N = 1;
  while (N < n) N <<= 1;
}

int mpow(int x, int y) {
  int _ret = 1;
  while (y) {
    if (y & 1)  _ret = 1ll * x * _ret % MOD;
    x = 1ll * x * x % MOD;
    y >>= 1;
  }
  return _ret;
}

int tax[maxn], taxlen;

void GetRev(const int N) {
  int p = 1, d = N >> 1;
  for (int w = p; w < N; w = p) {
    for (int i = 0; i < w; ++i) {
      tax[p++] = tax[i] | d;
    }
    d >>= 1;
  }
  taxlen = N;
}

void MakeRev(int *const A, const int N) {
  for (int i = 1; i < N; ++i) if (tax[i] > i) {
    std::swap(A[i], A[tax[i]]);
  }
}

void Modint(int &x) {
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
}

void NTT(int *const A, const int N) {
  if (taxlen != N) {
    GetRev(N);
  }
  MakeRev(A, N);
  for (int w = 2, M = 1; w <= N; w <<= 1) {
    int Wn = mpow(G, PHI / w);
    for (int L = 0; L < N; L += w) {
      ll g = 1;
      for (int i = L, lm = L + M, j = lm; i < lm; ++i, ++j) {
        ll x = A[i], y = A[j];
        A[i] = (x + g * y) % MOD; A[j] = (x - g * y) % MOD;
        (g *= Wn) %= MOD;
      }
    }
    M = w;
  }
  for (int i = 0; i < N; ++i) {
    Modint(A[i]);
  }
}

void GetInv(int *const A, int *const B, const int N) {
  static int C[maxn];
  B[0] = mpow(A[0], DPHI);
  for (int w = 2, M = 4; w <= N; M <<= 1) {
    memcpy(C, A, w * SzI);
    NTT(B, M); NTT(C, M);
    for (int i = 0; i < M; ++i) {
      B[i] = (B[i] << 1) % MOD - 1ll * C[i] * B[i] % MOD * B[i] % MOD;
      Modint(B[i]);
    }
    NTT(B, M);
    std::reverse(B + 1, B + M);
    for (int i = 0, iv = mpow(M, DPHI); i < w; ++i) {
      B[i] = 1ll * B[i] * iv % MOD;
    }
    memset(B + w, 0, w * SzI);
    w = M;
  }
}

void GetDer(int *const A, int *const B, const int N) {
  for (int i = 1; i < N; ++i) {
    B[i - 1] = 1ll * A[i] * i % MOD;
  }
  B[N - 1] = 0;
}

void GetInte(int *const A, int *const B, const int N) {
  B[0] = 0;
  for (int i = 1; i < N; ++i) {
    B[i] = 1ll * A[i - 1] * mpow(i, DPHI) % MOD;
  }
}

void GetLn(int *const A, int *const B, const int N) {
  static int C[maxn];
  GetDer(A, B, N);
  GetInv(A, C, N);
  int M = N << 1;
  NTT(B, M); NTT(C, M);
  for (int i = 0; i < M; ++i) {
    C[i] = 1ll * B[i] * C[i] % MOD;
  }
  NTT(C, M);
  std::reverse(C + 1, C + M);
  for (int i = 0, iv = mpow(M, DPHI); i < N; ++i) {
    C[i] = 1ll * C[i] * iv % MOD;
  }
  GetInte(C, B, N);
}

}

多项式操作公式记录

标签：std   变换   pen   ace   include   play   namespace   make   ever   

原文地址：https://www.cnblogs.com/yifusuyi/p/12157488.html