結果

問題 No.1105 Many Triplets
ユーザー batsumarubatsumaru
提出日時 2020-07-03 23:22:25
言語 C++14
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 5,923 bytes
コンパイル時間 1,848 ms
コンパイル使用メモリ 178,012 KB
実行使用メモリ 4,348 KB
最終ジャッジ日時 2023-10-17 06:21:17
合計ジャッジ時間 2,768 ms
ジャッジサーバーID
(参考情報)
judge11 / judge14
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
4,348 KB
testcase_01 AC 2 ms
4,348 KB
testcase_02 AC 2 ms
4,348 KB
testcase_03 AC 1 ms
4,348 KB
testcase_04 AC 2 ms
4,348 KB
testcase_05 AC 1 ms
4,348 KB
testcase_06 AC 2 ms
4,348 KB
testcase_07 AC 2 ms
4,348 KB
testcase_08 AC 2 ms
4,348 KB
testcase_09 AC 1 ms
4,348 KB
testcase_10 AC 2 ms
4,348 KB
testcase_11 AC 2 ms
4,348 KB
testcase_12 AC 2 ms
4,348 KB
testcase_13 AC 1 ms
4,348 KB
testcase_14 AC 2 ms
4,348 KB
testcase_15 AC 1 ms
4,348 KB
testcase_16 AC 2 ms
4,348 KB
testcase_17 AC 2 ms
4,348 KB
testcase_18 AC 2 ms
4,348 KB
testcase_19 AC 1 ms
4,348 KB
testcase_20 AC 1 ms
4,348 KB
testcase_21 AC 2 ms
4,348 KB
testcase_22 AC 2 ms
4,348 KB
testcase_23 AC 2 ms
4,348 KB
testcase_24 AC 2 ms
4,348 KB
testcase_25 AC 1 ms
4,348 KB
testcase_26 AC 2 ms
4,348 KB
testcase_27 AC 2 ms
4,348 KB
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;
typedef long long ll;
typedef pair<ll, ll> P;
#define DUMP(x) cout << #x << " = " << (x) << endl;
#define FOR(i, m, n) for (ll i = m; i < n; i++)
#define IFOR(i, m, n) for (ll i = n - 1; i >= m; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)
#define FOREACH(x, a) for (auto&(x) : (a))
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) ll(x.size())

// modint
template <int MOD>
struct Fp {
  long long val;
  constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
    if (val < 0) val += MOD;
  }
  constexpr int getmod() { return MOD; }
  constexpr Fp operator-() const noexcept { return val ? MOD - val : 0; }
  constexpr Fp operator+(const Fp& r) const noexcept { return Fp(*this) += r; }
  constexpr Fp operator-(const Fp& r) const noexcept { return Fp(*this) -= r; }
  constexpr Fp operator*(const Fp& r) const noexcept { return Fp(*this) *= r; }
  constexpr Fp operator/(const Fp& r) const noexcept { return Fp(*this) /= r; }
  constexpr Fp& operator+=(const Fp& r) noexcept {
    val += r.val;
    if (val >= MOD) val -= MOD;
    return *this;
  }
  constexpr Fp& operator-=(const Fp& r) noexcept {
    val -= r.val;
    if (val < 0) val += MOD;
    return *this;
  }
  constexpr Fp& operator*=(const Fp& r) noexcept {
    val = val * r.val % MOD;
    return *this;
  }
  constexpr Fp& operator/=(const Fp& r) noexcept {
    long long a = r.val, b = MOD, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    val = val * u % MOD;
    if (val < 0) val += MOD;
    return *this;
  }
  constexpr bool operator==(const Fp& r) const noexcept {
    return this->val == r.val;
  }
  constexpr bool operator!=(const Fp& r) const noexcept {
    return this->val != r.val;
  }
  friend constexpr ostream& operator<<(ostream& os, const Fp<MOD>& x) noexcept {
    return os << x.val;
  }
  friend constexpr Fp<MOD> modpow(const Fp<MOD>& a, long long n) noexcept {
    if (n == 0) return 1;
    auto t = modpow(a, n / 2);
    t = t * t;
    if (n & 1) t = t * a;
    return t;
  }
};

// binomial coefficient
template <class T>
struct BiCoef {
  vector<T> fact_, inv_, finv_;
  constexpr BiCoef() {}
  constexpr BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
    init(n);
  }
  constexpr void init(int n) noexcept {
    fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
    int MOD = fact_[0].getmod();
    for (int i = 2; i < n; i++) {
      fact_[i] = fact_[i - 1] * i;
      inv_[i] = -inv_[MOD % i] * (MOD / i);
      finv_[i] = finv_[i - 1] * inv_[i];
    }
  }
  constexpr T com(int n, int k) const noexcept {
    if (n < k || n < 0 || k < 0) return 0;
    if (n <= (int)fact_.size() - 1) {
      return fact_[n] * finv_[k] * finv_[n - k];
    }
    // C(n,k), n <= 1e9, k <= 1e6
    T res = 1;
    for (int i = n; i >= n - k + 1; i--) {
      T a = i;
      res *= a;
    }
    res *= finv_[k];
    return res;
  }
  constexpr T fact(int n) const noexcept {
    if (n < 0) return 0;
    return fact_[n];
  }
  constexpr T inv(int n) const noexcept {
    if (n < 0) return 0;
    return inv_[n];
  }
  constexpr T finv(int n) const noexcept {
    if (n < 0) return 0;
    return finv_[n];
  }
};

const int MOD = 1000000007;
// const int MOD = 998244353;
using mint = Fp<MOD>;
BiCoef<mint> bc;

// matrix
template <class T>
struct Matrix {
  vector<vector<T>> val;
  Matrix(int n = 1, int m = 1, T v = 0) : val(n, vector<T>(m, v)) {}
  void init(int n, int m, T v = 0) { val.assign(n, vector<T>(m, v)); }
  void resize(int n, int m) {
    val.resize(n);
    for (int i = 0; i < n; ++i) val[i].resize(m);
  }
  Matrix<T>& operator=(const Matrix<T>& A) {
    val = A.val;
    return *this;
  }
  size_t size() const { return val.size(); }
  vector<T>& operator[](int i) { return val[i]; }
  const vector<T>& operator[](int i) const { return val[i]; }

  // debug
  friend ostream& operator<<(ostream& s, const Matrix<T>& M) {
    for (int i = 0; i < (int)M.size(); ++i)
      for (int j = 0; j < (int)M[i].size(); ++j)
        s << M[i][j]
          << " \n"[i != (int)M.size() - 1 && j == (int)M[i].size() - 1];
    return s;
  }
};

template <class T>
Matrix<T> operator*(const Matrix<T>& A, const Matrix<T>& B) {
  Matrix<T> R(A.size(), B[0].size());
  for (int i = 0; i < A.size(); ++i)
    for (int j = 0; j < B[0].size(); ++j)
      for (int k = 0; k < B.size(); ++k) R[i][j] += A[i][k] * B[k][j];
  return R;
}

template <class T>
Matrix<T> pow(const Matrix<T>& A, long long n) {
  Matrix<T> R(A.size(), A.size());
  auto B = A;
  for (int i = 0; i < A.size(); ++i) R[i][i] = 1;
  while (n > 0) {
    if (n & 1) R = R * B;
    B = B * B;
    n >>= 1;
  }
  return R;
}

template <class T>
vector<T> operator*(const Matrix<T>& A, const vector<T>& B) {
  vector<T> v(A.size());
  for (int i = 0; i < A.size(); ++i)
    for (int k = 0; k < B.size(); ++k) v[i] += A[i][k] * B[k];
  return v;
}

template <class T>
Matrix<T> operator+(const Matrix<T>& A, const Matrix<T>& B) {
  Matrix<T> R(A.size(), A[0].size());
  for (int i = 0; i < A.size(); ++i)
    for (int j = 0; j < A[0].size(); ++j) R[i][j] = A[i][j] + B[i][j];
  return R;
}

template <class T>
Matrix<T> operator-(const Matrix<T>& A, const Matrix<T>& B) {
  Matrix<T> R(A.size(), A[0].size());
  for (int i = 0; i < A.size(); ++i)
    for (int j = 0; j < A[0].size(); ++j) R[i][j] = A[i][j] - B[i][j];
  return R;
}

template <class T>
void printArray(const vector<T>& a) {
  int n = a.size();
  for (int i = 0; i < n; i++) {
    cout << a[i] << " \n"[i == n - 1];
  }
}

int main() {
  ll n;
  cin >> n;
  vector<mint> v(3);
  REP(i, 3) {
    ll a;
    cin >> a;
    v[i] = (mint)a;
  }
  Matrix<mint> X(3, 3, 0);
  X[0][0] = 1, X[0][1] = -1, X[1][1] = 1, X[1][2] = -1, X[2][0] = -1,
  X[2][2] = 1;
  Matrix<mint> Y = pow(X, n - 1);
  vector<mint> z = Y * v;
  printArray(z);
}
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