結果

問題 No.526 フィボナッチ数列の第N項をMで割った余りを求める
コンテスト
ユーザー ppp-skyblue
提出日時 2022-07-07 00:00:45
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 4,277 bytes
コンパイル時間 4,688 ms
コンパイル使用メモリ 256,728 KB
最終ジャッジ日時 2025-01-30 05:07:26
ジャッジサーバーID
（参考情報） 		judge5 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 12
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
using namespace atcoder;
// using mint = modint1000000007;
// using mint = modint998244353;
typedef long long ll;
#define int long long
#define rep(i,n) for (int i = 0; i < (int)(n); ++i)
#define nrep(i,n) for (int i = 1; i <= (int)(n); ++i)
#define all(x) (x).begin(),(x).end()
#define bit(n,k) ((n>>k)&1) /*nのk bit目*/
#define bit_count(x) __builtin_popcountll(x)
#define debug(x) cout << #x << ": " << x << endl;
using P = pair<int,int>;
#define INF 1001001001
#define LINF (1LL<<60)
template<class T> inline bool chmax(T& a, T b){ if(a<b){ a=b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b){ if(a>b){ a=b; return 1; } return 0; }

using mint = modint;

template <typename T>
struct Matrix {
  vector<vector<T>> A;
  Matrix() {}
  Matrix(size_t n, size_t m) : A(n, vector<T>(m, 0)) {}
  Matrix(size_t n) : A(n, vector<T>(n, 0)) {}
  size_t height() const { return A.size(); }
  size_t width() const {
    assert(height() > 0);
    return A[0].size();
  }
  inline const vector<T> &operator[](int k) const { return A.at(k); }
  inline vector<T> &operator[](int k) { return A.at(k); }
  static Matrix I(size_t n) {
    Matrix mat(n);
    for (int i = 0; i < n; ++i) mat[i][i] = 1;
    return mat;
  }
  Matrix &operator+=(const Matrix &B) {
    size_t n = height(), m = width();
    assert(n == B.height() and m == B.width());
    for (int i = 0; i < n; ++i) for (int j = 0; j < m; ++j) (*this)[i][j] += B[i][j];
    return *this;
  }
  Matrix &operator-=(const Matrix &B) {
    size_t n = height(), m = width();
    assert(n == B.height() and m == B.width());
    for (int i = 0; i < n; ++i) for (int j = 0; j < m; ++j) (*this)[i][j] -= B[i][j];
    return *this;
  }
  Matrix &operator*=(const Matrix &B) {
    size_t n = height(), m = B.width(), p = width();
    assert(p == B.height());
    vector<vector<T>> C(n, vector<T>(m, 0));
    for (int i = 0; i < n; ++i) for (int j = 0; j < m; ++j) for (int k = 0; k < p; ++k) C[i][j] += (*this)[i][k] * B[k][j];
    A.swap(C);
    return *this;
  }
  Matrix &operator^=(long long k) {
    Matrix B = Matrix::I(height());
    while (k > 0) {
      if (k & 1) B *= *this;
      *this *= *this;
      k >>= 1LL;
    }
    A.swap(B.A);
    return *this;
  }
  Matrix operator+(const Matrix &B) const { return (Matrix(*this) += B); }
  Matrix operator-(const Matrix &B) const { return (Matrix(*this) -= B); }
  Matrix operator*(const Matrix &B) const { return (Matrix(*this) *= B); }
  Matrix operator^(const long long k) const { return (Matrix(*this) ^= k); }
  friend istream &operator>>(istream &is, Matrix &p) {
    size_t n = p.height(), m = p.width();
    for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { is >> p[i][j]; } }
    return is;
  }
  friend ostream &operator<<(ostream &os, Matrix &p) {
    size_t n = p.height(), m = p.width();
    for (int i = 0; i < n; ++i) {
      os << '[';
      for (int j = 0; j < m; ++j) { os << p[i][j] << (j + 1 == m ? "]\n" : ", "); }
    }
    return os;
  }
  T determinant() {
    Matrix B(*this);
    assert(width() == height());
    T ret = 1;
    for (int i = 0; i < width(); ++i) {
      int idx = -1;
      for (int j = i; j < width(); ++j) if (B[j][i] != 0) idx = j;
      if (idx == -1) return T(0);
      if (i != idx) {
        ret *= -1;
        swap(B[i], B[idx]);
      }
      ret *= B[i][i];
      T vv = B[i][i];
      for (int j = 0; j < width(); ++j) B[i][j] /= vv;
      for (int j = i + 1; j < width(); ++j) {
        T a = B[j][i];
        for (int k = 0; k < width(); ++k) B[j][k] -= B[i][k] * a;
      }
    }
    return ret;
  }
  Matrix pow(ll b) {
    Matrix<mint> ret = I(height());
    while(b) {
    if (b%2 == 1) ret = ret * (*this);
    b /= 2;
    (*this) = (*this) * (*this);
    }
    return ret;
  }
  void show() {
    for (int i = 0; i < height(); i++) {
      for (int j = 0; j < width(); j++) {
        if (j != 0) cout << " ";
        cout << A[i][j];
      }
      cout << endl;
    }
  }
};

int32_t main() {
  int n, m; cin >> n >> m;
  mint::set_mod(m);
  Matrix<mint> a(2);
  a[0][0] = 0, a[0][1] = 1;
  a[1][0] = 1, a[1][1] = 1;
  Matrix<mint> b(2,1);
  b[0][0] = 0, b[1][0] = 1;
  cout << (a.pow(n-2)*b)[1][0].val() << endl;
  
}
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