Hello Codeforcers! As a student of department of mathematics, I'd like to demonstrate an application of Extended Field.

Extended Field

The idea

First we should include the idea of splitting field, a splitting field F of a (may assumed) monic polynomial P is the smallest field such that

I use the word "smallest" with the sense that if P split over E, .

In the other word, P decomposes into linear factor under it's splitting field.

In most of the problem, we need not find the exact splitting field... Just a field that P split over it is enough.

Example

1. P = x² + 1 does not split over but split over where which is widely used as complex.

2. Let p = 998244353 = 7 * 17 * 2²³ + 1, P = x²²⁴ - 1 does not split over but split over where . The multiplication order ord(w²) = 2²³.

3. Let p = 10⁹ + 7, Fibonacci number can be easily written as . Where φ, ψ are both root of (Remark that x² = 5 has no root in Z_p)

Problem Solving

718C - Саша и массив

The problem to be demonstrated corresponds to the third example. This may be not the best solution for the problem. I've got several TL on this problem for large constant on module. But I do think that it's a nice try to solve some problem in different way :)

The problem contains several operations: - shift all element of a segment to the next x Fibonacci term. - query for the sum of an segment

In the problem, we should handle Fibonacci number 1, 1, 2, 3, 5, 8, ... with the formula . Once we have decompose the Fibonacci into two sequence of number powers. The problem becomes easier:

• multiply a segment by φ^x and ψ^x
• query for the sum of an segment

It can be easily handled by two segment tree.

My implementation

We can implement Extended Field using C++ class, just like complex:

template<int sq = 5, int MOD = 1000000007>
class EX {
  int re, im;
  static int trim(int a) {
    if (a >= MOD) a -= MOD;
    if (a < 0) a += MOD;
    return a;
  }
  static int inv(const int a) {
    int ans = 1;
    for (int cur = a, p = MOD - 2; p; p >>= 1, cur = 1ll * cur * cur % MOD) {
      if (p&1) ans = 1ll * ans * cur % MOD;
    }
    return ans;
  };
public:
  EX(int re = 0, int im = 0) : re(re), im(im) {}
  EX& operator=(EX oth) { return re = oth.re, im = oth.im, *this; }
  int norm() const { 
    return trim((1ll * re * re - 1ll * sq * im % MOD * im) % MOD);
  }
  EX conj() const { 
    return EX(re, trim(MOD - im)); 
  }
  EX operator*(EX oth) const {
    return EX((1ll * re * oth.re + 1ll * sq * im % MOD * oth.im) % MOD,
              (1ll * re * oth.im + 1ll * im * oth.re) % MOD);
  };
  EX operator/(int n) const { 
    return EX(1ll * re * inv(n) % MOD, 1ll * im * inv(n) % MOD);
  }
  EX operator/(EX oth) const { return *this * oth.conj() / oth.norm(); }
  EX operator+(EX oth) const { return EX(trim(re + oth.re), trim(im + oth.im)); }
  EX operator-(EX oth) const {
    return EX(trim(re - oth.re), trim(im - oth.im));
  }
  EX pow(long long n) const {
    EX ans(1);
    for (EX a = *this; n; n >>= 1, a = a * a) {
      if (n&1) ans = a * ans;
    }
    return ans;
  }
  bool operator==(EX oth) const { return re == oth.re and im == oth.im; }
  bool operator!=(EX oth) const { return not (*this == oth); }
  int real() const& { return re; }
  int imag() const& { return im; }
};

FFT ------------------ Once a splitting field implementation is done, we can FFT under Z_p more than. I haven't implemented it yet.