結果

問題 No.1073 無限すごろく
ユーザー au7777
提出日時 2021-12-10 22:19:53
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 4,810 bytes
コンパイル時間 4,025 ms
コンパイル使用メモリ 231,512 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-07-18 16:42:16
合計ジャッジ時間 5,411 ms
ジャッジサーバーID
（参考情報） 		judge4 / 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 * 30
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 #
プレゼンテーションモードにする

#include <bits/stdc++.h>
#include <atcoder/all>
typedef long long int ll;
typedef long long int ull;
#define MP make_pair
using namespace std;
using namespace atcoder;
typedef pair<ll, ll> P;
// const ll MOD = 998244353;
const ll MOD = 1000000007;
// using mint = modint998244353;
using mint = modint1000000007;
const double pi = 3.1415926536;
const int MAX = 2000003;
long long fac[MAX], finv[MAX], inv[MAX];
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
ll gcd (ll x, ll y) {
   if (y == 0) return x;
   else if (y > x) {
       return gcd (y, x); 
   }
   else return gcd(x % y, y);
}
ll lcm (ll x, ll y) {
   return x / gcd(x, y) * y;
}
ll my_sqrt(ll x) {
    ll m = 0;
    ll M = 1000000001;
    while (M - m > 1) {
        ll now = (M + m) / 2;
        if (now * now <= x) {
            m = now;
        }
        else {
            M = now;
        }
    }
    return m;
}
ll keta(ll n) {
    ll ret = 0;
    while (n) {
        n /= 10;
        ret++;
    }
    return ret;
}
ll ceil(ll n, ll m) {
    ll ret = n / m;
    if (n % m) ret++;
    return ret;
}
ll pow_ll(ll x, ll n) {
    if (n == 0) return 1;
    if (n % 2) {
        return pow_ll(x, n - 1) * x;
    }
    else {
        ll tmp = pow_ll(x, n / 2);
        return tmp * tmp;
    }
}
template <typename T, int H, int W>
struct Matrix {
  using Array = array<array<T, W>, H>;
  Array A;
 
  Matrix() : A() {
    for (int i = 0; i < H; i++)
      for (int j = 0; j < W; j++) (*this)[i][j] = T();
  }
 
  int height() const { return H; }
 
  int width() const { return W; }
 
  inline const array<T, W> &operator[](int k) const { return A[k]; }
 
  inline array<T, W> &operator[](int k) { return A[k]; }
 
  static Matrix I() {
    assert(H == W);
    Matrix mat;
    for (int i = 0; i < H; i++) mat[i][i] = 1;
    return (mat);
  }
 
  Matrix &operator+=(const Matrix &B) {
    for (int i = 0; i < H; i++)
      for (int j = 0; j < W; j++) A[i][j] += B[i][j];
    return (*this);
  }
 
  Matrix &operator-=(const Matrix &B) {
    for (int i = 0; i < H; i++)
      for (int j = 0; j < W; j++) A[i][j] -= B[i][j];
    return (*this);
  }
 
  Matrix &operator*=(const Matrix &B) {
    assert(H == W);
    Matrix C;
    for (int i = 0; i < H; i++)
      for (int k = 0; k < H; k++)
        for (int j = 0; j < H; j++) C[i][j] += A[i][k] * B[k][j];
    A.swap(C.A);
    return (*this);
  }
 
  Matrix &operator^=(long long k) {
    Matrix B = Matrix::I();
    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) {
    for (int i = 0; i < H; i++) {
      os << "[";
      for (int j = 0; j < W; j++) {
        os << p[i][j] << (j + 1 == W ? "]\n" : ",");
      }
    }
    return (os);
  }
 
  T determinant(int n = -1) {
    if (n == -1) n = H;
    Matrix B(*this);
    T ret = 1;
    for (int i = 0; i < n; i++) {
      int idx = -1;
      for (int j = i; j < n; j++) {
        if (B[j][i] != 0) {
          idx = j;
          break;
        }
      }
      if (idx == -1) return 0;
      if (i != idx) {
        ret *= T(-1);
        swap(B[i], B[idx]);
      }
      ret *= B[i][i];
      T inv = T(1) / B[i][i];
      for (int j = 0; j < n; j++) {
        B[i][j] *= inv;
      }
      for (int j = i + 1; j < n; j++) {
        T a = B[j][i];
        if (a == 0) continue;
        for (int k = i; k < n; k++) {
          B[j][k] -= B[i][k] * a;
        }
      }
    }
    return (ret);
  }
};
Matrix<mint, 6, 6> f(Matrix<mint, 6, 6> m, ll n) {
    if (n == 0) {
        return m.I();
    }
    if (n % 2 == 0) {
        Matrix<mint, 6, 6> tmp = f(m, n / 2);
        return tmp * tmp; 
    }
    else {
        return m * f(m, n - 1);
    }
}
int main() {
    Matrix<mint, 6, 6> m;
    for (int i = 0; i < 6; i++) {
        for (int j = 0; j < 6; j++) {
            if (i + 1 == j) m[i][j] = 1;
            else if (i == 5) m[i][j] = 166666668;
            else m[i][j] = 0;
        } 
    }
    ll n;
    cin >> n;
    cout << f(m,n)[5][5].val() << endl;
    return 0;
}
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