結果

問題 No.3226 2×2行列累乗
ユーザー pia019
提出日時 2025-08-08 21:48:42
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 3 ms / 2,000 ms
コード長 4,324 bytes
コンパイル時間 7,264 ms
コンパイル使用メモリ 355,884 KB
実行使用メモリ 7,716 KB
最終ジャッジ日時 2025-08-08 21:48:59
合計ジャッジ時間 8,401 ms
ジャッジサーバーID
(参考情報)
judge1 / 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 * 27
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 #

#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
typedef long long ll;
const int INF = 1<<30;
const ll INFLL = 1LL<<60;
const ll MOD = 998244353;
const double INFD = 1.0E10;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, -1, 0, 1};
//const int dx[8] = {1, 1, 0, -1, -1, -1, 0, 1};
//const int dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
using Pair = pair<ll, ll>;
using Graph = vector<vector<ll>>;
using mint = atcoder::modint;


template< class 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 {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() && 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() && 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] = (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 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 (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);
  }
};

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(15);
    
    //const int p = 67;
    //mint::set_mod(p);
    Matrix<ll> A(2, 2);
    for (int i = 0; i < 2; i++){
        for (int j = 0; j < 2; j++){
            ll x; cin >> x;
            A[i][j] = x;
        }
    }

    ll s, t, n, k; cin >> s >> t >> n >> k;
    mint::set_mod(k);
    Matrix<mint> B(2, 2);
    for (int i = 0; i < 2; i++){
        for (int j = 0; j < 2; j++){
            B[i][j] = A[i][j];
        }
    }
    
    vector<Matrix<mint>> pows(60, Matrix<mint>(2,2));
    pows[0] = B;
    for (int i = 1; i < 60; i++) pows[i] = pows[i - 1] * pows[i - 1];
    
    
    Matrix<mint> ans(2, 2); ans[0][0] = 1, ans[1][1] = 1;
    for (int i = 0; i < 60; i++){
        if ((n >> i) & 1) ans *= pows[i];
    }
    
    mint a = ans[0][0] * s + ans[0][1] * t;
    mint b = ans[1][0] * s + ans[1][1] * t;
    cout << a.val() << " " << b.val() << endl;
    

    return 0;
}
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