形式的冪級数(Formal-Power-Series)

説明

えーこれはなんでしょうかねー

形式的べき級数(Formal power series)を扱う。

$P(x) = \sum_{i=0}^{\infty} c_{i} x^{i}$

実用的には多項式 $P(x), Q(x)$ の最初のいくつかの項を与えたときに以下の演算を行う。基本的には演算を畳み込みに落とし込んで FFT を用いて高速化できる。

• $O(n)$ : $P(x)+Q(x), P(x)-Q(x), -P(x), P’(x), \int P(x) dx, kP(x)$
• $O(n \log n)$ : $P(x)Q(x)$
• $O(n \log n)$ : $\frac {1} {P(x)}$
• $O(n \log n)$ : $\sqrt {P(x)}$
• $O(n \log n)$ : $\exp(P(x))$
• $O(n \log n)$ : $log(P(x)) = \int \frac {P’(x)} {P(x)}$
• $O(n \log n)$ : $P(x)^{k} = \exp(k \log P(x))$

計算量

上に示した

実装例

依存ライブラリ Mod-Int

テンプレート引数としてMod-Intが渡されることを想定している。

TODO 使い方

template< typename T >
struct FormalPowerSeries : vector< T > {
  using vector< T >::vector;
  using P = FormalPowerSeries;

  using MULT = function< P(P, P) >;

  static MULT &get_mult() {
    static MULT mult = nullptr;
    return mult;
  }

  static void set_fft(MULT f) {
    get_mult() = f;
  }

  void shrink() {
    while(this->size() && this->back() == T(0)) this->pop_back();
  }

  P operator+(const P &r) const { return P(*this) += r; }

  P operator+(const T &v) const { return P(*this) += v; }

  P operator-(const P &r) const { return P(*this) -= r; }

  P operator-(const T &v) const { return P(*this) -= v; }

  P operator*(const P &r) const { return P(*this) *= r; }

  P operator*(const T &v) const { return P(*this) *= v; }

  P operator/(const P &r) const { return P(*this) /= r; }

  P operator%(const P &r) const { return P(*this) %= r; }

  P &operator+=(const P &r) {
    if(r.size() > this->size()) this->resize(r.size());
    for(int i = 0; i < r.size(); i++) (*this)[i] += r[i];
    return *this;
  }

  P &operator+=(const T &r) {
    if(this->empty()) this->resize(1);
    (*this)[0] += r;
    return *this;
  }

  P &operator-=(const P &r) {
    if(r.size() > this->size()) this->resize(r.size());
    for(int i = 0; i < r.size(); i++) (*this)[i] -= r[i];
    shrink();
    return *this;
  }

  P &operator-=(const T &r) {
    if(this->empty()) this->resize(1);
    (*this)[0] -= r;
    shrink();
    return *this;
  }

  P &operator*=(const T &v) {
    const int n = (int) this->size();
    for(int k = 0; k < n; k++) (*this)[k] *= v;
    return *this;
  }

  P &operator*=(const P &r) {
    if(this->empty() || r.empty()) {
      this->clear();
      return *this;
    }
    assert(get_mult() != nullptr);
    return *this = get_mult()(*this, r);
  }

  P &operator%=(const P &r) {
    return *this -= *this / r * r;
  }

  P operator-() const {
    P ret(this->size());
    for(int i = 0; i < this->size(); i++) ret[i] = -(*this)[i];
    return ret;
  }

  P &operator/=(const P &r) {
    if(this->size() < r.size()) {
      this->clear();
      return *this;
    }
    int n = this->size() - r.size() + 1;
    return *this = (rev().pre(n) * r.rev().inv(n)).pre(n).rev(n);
  }

  P pre(int sz) const {
    return P(begin(*this), begin(*this) + min((int) this->size(), sz));
  }

  P operator>>(int sz) const {
    if(this->size() <= sz) return {};
    P ret(*this);
    ret.erase(ret.begin(), ret.begin() + sz);
    return ret;
  }

  P operator<<(int sz) const {
    P ret(*this);
    ret.insert(ret.begin(), sz, T(0));
    return ret;
  }

  P rev(int deg = -1) const {
    P ret(*this);
    if(deg != -1) ret.resize(deg, T(0));
    reverse(begin(ret), end(ret));
    return ret;
  }

  P diff() const {
    const int n = (int) this->size();
    P ret(max(0, n - 1));
    for(int i = 1; i < n; i++) ret[i - 1] = (*this)[i] * T(i);
    return ret;
  }

  P integral() const {
    const int n = (int) this->size();
    P ret(n + 1);
    ret[0] = T(0);
    for(int i = 0; i < n; i++) ret[i + 1] = (*this)[i] / T(i + 1);
    return ret;
  }

  // F(0) must not be 0
  P inv(int deg = -1) const {
    assert(((*this)[0]) != T(0));
    const int n = (int) this->size();
    if(deg == -1) deg = n;
    P ret({T(1) / (*this)[0]});
    for(int i = 1; i < deg; i <<= 1) {
      ret = (ret + ret - ret * ret * pre(i << 1)).pre(i << 1);
    }
    return ret.pre(deg);
  }

  // F(0) must be 1
  P log(int deg = -1) const {
    assert((*this)[0] == 1);
    const int n = (int) this->size();
    if(deg == -1) deg = n;
    return (this->diff() * this->inv(deg)).pre(deg - 1).integral();
  }

  P sqrt(int deg = -1) const {
    const int n = (int) this->size();
    if(deg == -1) deg = n;

    if((*this)[0] == T(0)) {
      for(int i = 1; i < n; i++) {
        if((*this)[i] != T(0)) {
          if(i & 1) return {};
          if(deg - i / 2 <= 0) break;
          auto ret = (*this >> i).sqrt(deg - i / 2) << (i / 2);
          if(ret.size() < deg) ret.resize(deg, T(0));
          return ret;
        }
      }
      return P(deg, 0);
    }

    P ret({T(1)});
    T inv2 = T(1) / T(2);
    for(int i = 1; i < deg; i <<= 1) {
      ret = (ret + pre(i << 1) * ret.inv(i << 1)) * inv2;
    }
    return ret.pre(deg);
  }

  // F(0) must be 0
  P exp(int deg = -1) const {
    assert((*this)[0] == T(0));
    const int n = (int) this->size();
    if(deg == -1) deg = n;
    P ret({T(1)});
    for(int i = 1; i < deg; i <<= 1) {
      ret = (ret * (pre(i << 1) + T(1) - ret.log(i << 1))).pre(i << 1);
    }
    return ret.pre(deg);
  }

  P pow(int64_t k, int deg = -1) const {
    const int n = (int) this->size();
    if(deg == -1) deg = n;
    for(int i = 0; i < n; i++) {
      if((*this)[i] != T(0)) {
        T rev = T(1) / (*this)[i];
        P C(*this * rev);
        P D(n - i);
        for(int j = i; j < n; j++) D[j - i] = C[j];
        D = (D.log() * k).exp() * (*this)[i].pow(k);
        P E(deg);
        if(i * k > deg) return E;
        auto S = i * k;
        for(int j = 0; j + S < deg && j < D.size(); j++) E[j + S] = D[j];
        return E;
      }
    }
    return *this;
  }


  T eval(T x) const {
    T r = 0, w = 1;
    for(auto &v : *this) {
      r += w * v;
      w *= x;
    }
    return r;
  }
};

検証

yukicoder No.3046 yukicoderの過去問 inverse()

int main() {
  int K, N;
  cin >> K >> N;
  PolynominalMod< modint > X(K + 1);
  X[0] = 1;
  for(int i = 0; i < N; i++) {
    int x;
    cin >> x;
    if(x <= K) X[x] = -1;
  }
  cout << X.inverse()[K] << endl;
}

Codeforces Round #250 (Div. 1) E - The Child and Binary Tree inverse(), sqrt()

int main() {
  int N, M;
  cin >> N >> M;
  PolynominalMod< modint > C(M + 1);
  C[0] = 1;
  for(int i = 0; i < N; i++) {
    int k;
    cin >> k;
    if(k <= M) C[k] = mod - 4;
  }
  C = C.sqrt();
  C[0] += 1;
  C = C.inverse();
  for(int i = 1; i <= M; i++) {
    cout << C[i] * 2 << "\n";
  }
}

参考

Operations on Formal Power Series