結果

問題 No.1073 無限すごろく
ユーザー kissshot7kissshot7
提出日時 2020-11-24 19:29:40
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 4 ms / 2,000 ms
コード長 6,317 bytes
コンパイル時間 1,917 ms
コンパイル使用メモリ 177,576 KB
実行使用メモリ 7,680 KB
最終ジャッジ日時 2024-07-23 18:46:42
合計ジャッジ時間 3,069 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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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 4 ms
7,488 KB
testcase_01 AC 4 ms
7,496 KB
testcase_02 AC 4 ms
7,552 KB
testcase_03 AC 4 ms
7,552 KB
testcase_04 AC 4 ms
7,492 KB
testcase_05 AC 4 ms
7,552 KB
testcase_06 AC 4 ms
7,552 KB
testcase_07 AC 4 ms
7,552 KB
testcase_08 AC 4 ms
7,424 KB
testcase_09 AC 4 ms
7,488 KB
testcase_10 AC 4 ms
7,492 KB
testcase_11 AC 4 ms
7,496 KB
testcase_12 AC 4 ms
7,488 KB
testcase_13 AC 4 ms
7,552 KB
testcase_14 AC 4 ms
7,552 KB
testcase_15 AC 4 ms
7,424 KB
testcase_16 AC 4 ms
7,552 KB
testcase_17 AC 4 ms
7,624 KB
testcase_18 AC 4 ms
7,488 KB
testcase_19 AC 4 ms
7,552 KB
testcase_20 AC 4 ms
7,552 KB
testcase_21 AC 4 ms
7,492 KB
testcase_22 AC 4 ms
7,424 KB
testcase_23 AC 4 ms
7,492 KB
testcase_24 AC 4 ms
7,492 KB
testcase_25 AC 4 ms
7,552 KB
testcase_26 AC 4 ms
7,552 KB
testcase_27 AC 4 ms
7,496 KB
testcase_28 AC 4 ms
7,680 KB
testcase_29 AC 4 ms
7,552 KB
testcase_30 AC 4 ms
7,492 KB
testcase_31 AC 4 ms
7,492 KB
testcase_32 AC 4 ms
7,488 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;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
// const ll mod = 998244353;
const ll INF = mod * mod;
const int INF_N = 1e+9;
typedef pair<int, int> P;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

//繰り返し二乗法
ll mod_pow(ll a, ll n, ll m) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%m;
		a = a * a%m; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

//逆元(Eucledean algorithm)
ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 18;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}
using mP = pair<modint, modint>;

int dx[4] = { 0,1,0,-1 };
int dy[4] = { 1,0,-1,0 };


template< class T >
struct Matrix {
    std::vector< std::vector< T > > A;
    Matrix() {}
    Matrix(size_t n, size_t m) : A(n, std::vector< T >(m, 0)) {}
    Matrix(size_t n) : A(n, std::vector< T >(n, 0)) {};
    size_t height() const {
        return (A.size());
    }
    size_t width() const {
        return (A[0].size());
    }
    inline const std::vector< T > &operator[](int k) const {
        return (A.at(k));
    }
    inline std::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());
        std::vector< std::vector< T > > C(n, std::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+(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);
    }
    friend std::ostream &operator<<(std::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);
    }
    
    
    Matrix pow(int64_t k) const {
        auto res = I(A.size());
        auto M = *this;
        while (k > 0) {
            if (k & 1) {
                res *= M;
            }
            M *= M;
            k >>= 1;
        }
        return res;
    }
};

void solve() {
    ll n; cin >> n;
    Matrix<modint> mat(6);
    rep(i, 5) mat[i][i+1] = 1;
    rep(i, 6) mat[5][i] = modint(1)/modint(6);

    Matrix<modint> m1(6, 1);
    m1[0][0] = 1;
    rep(i, 6){
        Rep(j, i+1, 6){
            m1[j][0] += m1[i][0]/modint(6);
        }
    }

    auto mt = mat.pow(n);
    mt *= m1;

    cout << mt[0][0].n << endl;
}

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  //cout << fixed << setprecision(10);
  //init_f();
  //init();
  //int t; cin >> t; rep(i, t)solve();
  solve();
//   stop
    return 0;
}
0