結果

問題 No.541 3 x N グリッド上のサイクルの個数
コンテスト
ユーザー paruki
提出日時 2017-07-03 15:48:20
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 5,008 bytes
コンパイル時間 1,580 ms
コンパイル使用メモリ 176,804 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-10-05 12:22:44
合計ジャッジ時間 2,851 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge5
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ファイルパターン 結果
other AC * 62
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 FOR(i,j,k) for(int (i)=(j);(i)<(int)(k);++(i))
#define rep(i,j) FOR(i,0,j)
#define each(x,y) for(auto &(x):(y))
#define mp make_pair
#define MT make_tuple
#define all(x) (x).begin(),(x).end()
#define debug(x) cout<<#x<<": "<<(x)<<endl
#define smax(x,y) (x)=max((x),(y))
#define smin(x,y) (x)=min((x),(y))
#define MEM(x,y) memset((x),(y),sizeof (x))
#define sz(x) (int)(x).size()
#define rt return
using dbl = double;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;

template<int MOD>
class ModInt {
public:
    ModInt() :value(0) {}
    ModInt(long long val) :value((int)(val<0 ? MOD + val%MOD : val%MOD)) { }

    ModInt& operator+=(ModInt that) {
        value = value + that.value;
        if (value >= MOD)value -= MOD;
        return *this;
    }
    ModInt& operator-=(ModInt that) {
        value -= that.value;
        if (value<0)value += MOD;
        return *this;
    }
    ModInt& operator*=(ModInt that) {
        value = (int)((long long)value * that.value % MOD);
        return *this;
    }
    ModInt &operator/=(ModInt that) {
        return *this *= that.inverse();
    }
    ModInt operator+(ModInt that) const {
        return ModInt(*this) += that;
    }
    ModInt operator-(ModInt that) const {
        return ModInt(*this) -= that;
    }
    ModInt operator*(ModInt that) const {
        return ModInt(*this) *= that;
    }
    ModInt operator/(ModInt that) const {
        return ModInt(*this) /= that;
    }
    ModInt pow(long long k) const {
        if (value == 0)return 0;
        ModInt n = *this, res = 1;
        while (k) {
            if (k & 1)res *= n;
            n *= n;
            k >>= 1;
        }
        return res;
    }
    ModInt inverse() const {
        long long a = value, 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);
        }
        return ModInt(u);
    }
    int toi() const { return value; }

private:
    int value;
};
typedef ModInt<1000000007> mint;
ostream& operator<<(ostream& os, const mint& x) {
    os << x.toi();
    return os;
}

template<class Val>
class Matrix {
public:
    int n, m;
    Matrix() { }
    Matrix(int n_, int m_) :n(n_), m(m_), A(n, vector<Val>(m)) { }
#define ITER(a,b) for(int i=0;i<a;++i)for(int j=0;j<b;++j)
    Matrix operator +(Matrix &B) {
        assert(n == B.n && m == B.m);
        Matrix C(n, m);
        ITER(n, m) C(i, j) = A[i][j] + B(i, j);
        return C;
    }
    Matrix operator -(Matrix &B) {
        assert(n == B.n && m == B.m);
        Matrix C(n, m);
        ITER(n, m) C(i, j) = A[i][j] - B(i, j);
        return C;
    }
    Matrix* operator +=(Matrix &B) {
        assert(n == B.n && m == B.m);
        ITER(n, m) A[i][j] += B(i, j);
        return this;
    }
    Matrix* operator -=(Matrix &B) {
        assert(n == B.n && m == B.m);
        ITER(n, m) A[i][j] -= B(i, j);
        return this;
    }
    Matrix operator *(Matrix &B) {
        assert(m == B.n);
        Matrix C(n, B.m);
        ITER(C.n, B.m)
            for (int k = 0; k < m; ++k)
                C(i, j) += A[i][k] * B(k, j);
        return C;
    }
    Matrix* operator *=(Matrix &B) {
        return &(*this = *this*B);
    }
    Matrix* operator ^=(long long k) {
        return &(*this = *this^k);
    }
#undef ITER
    Matrix operator ^(long long k) {
        assert(n == m);
        Matrix C(n, n), D = *this;
        for (int i = 0; i < n; ++i)C(i, i) = 1;
        while (k > 0) {
            if (k & 1) C *= D;
            D *= D;
            k >>= 1;
        }
        return C;
    }
    Val& operator()(int i, int j) {
        return A[i][j];
    }
    vector<Val>& operator[](int i) {
        return A[i];
    }
    vector<Val> mulVec(const vector<Val> & u) {
        assert((int)u.size() == m);
        Matrix v(m, 1);
        for (int i = 0; i < m; ++i)v[i][0] = u[i];
        v = (*this)*v;
        vector<Val> w(n);
        for (int i = 0; i < n; ++i)w[i] = v[i][0];
        return w;
    }
private:
    vector<vector<Val>> A;
};
typedef Matrix<mint> mat;

/*
多分行列の累乗みたいな

10 x 10
(0,0,0,0,0,0,0,0,0,1)

条件が複雑なので有効グラフ書く
*/

// 完,(1,2),(1,3),(1,4),(2,3),(2,4),(3,4),{(1,2),(3,4)},{(1,4),(2,3)},1
int a[10][10] = {
    {1,1,1,1,1,1,1,0,1,0},//完
    {0,1,1,1,0,0,0,0,1,1},//(1,2)
    {0,1,1,1,1,1,0,0,0,1},//(1,3)
    {0,1,1,1,1,1,1,1,0,1},//(1,4)
    {0,0,1,1,1,1,0,0,0,1},//(2,3)
    {0,0,1,1,1,1,1,0,0,1},//(2,4)
    {0,0,0,1,0,1,1,0,1,1},//(3,4)
    {0,1,0,0,0,0,1,1,0,1},//(1,2),(3,4)
    {0,0,0,1,0,0,0,0,1,0},//(1,4),(2,3)
    {0,0,0,0,0,0,0,0,0,1},//1
};

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);

    ll N;
    cin >> N;
    mat A(10, 10);
    rep(i, 10)rep(j, 10)A[i][j] = a[i][j];
    vector<mint> v(10);
    v.back() = 1;
    A ^= (N + 1);
    v=A.mulVec(v);
    cout << v[0] << endl;
}
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