結果

問題 No.541 3 x N グリッド上のサイクルの個数
ユーザー はむこ
提出日時 2017-05-30 15:32:20
言語 C++11
(gcc 13.3.0)
結果
AC  
実行時間 3 ms / 2,000 ms
コード長 12,256 bytes
コンパイル時間 1,945 ms
コンパイル使用メモリ 181,144 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-09-22 12:40:45
合計ジャッジ時間 3,461 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge2
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ファイルパターン 結果
other AC * 62
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ソースコード

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プレゼンテーションモードにする

#include <bits/stdc++.h>
#include <sys/time.h>
using namespace std;
#define rep(i,n) for(long long i = 0; i < (long long)(n); i++)
#define repi(i,a,b) for(long long i = (long long)(a); i < (long long)(b); i++)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mt make_tuple
#define mp make_pair
template<class T1, class T2> bool chmin(T1 &a, T2 b) { return b < a && (a = b, true); }
template<class T1, class T2> bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
#define exists find_if
#define forall all_of
using ll = long long; using vll = vector<ll>; using vvll = vector<vll>; using P = pair<ll, ll>;
using ld = long double;  using vld = vector<ld>; 
using vi = vector<int>; using vvi = vector<vi>; vll conv(vi& v) { vll r(v.size()); rep(i, v.size()) r[i] = v[i]; return r; }
using Pos = complex<double>;
template <typename T, typename U> ostream &operator<<(ostream &o, const pair<T, U> &v) {  o << "(" << v.first << ", " << v.second << ")"; return o; }
template<size_t...> struct seq{}; template<size_t N, size_t... Is> struct gen_seq : gen_seq<N-1, N-1, Is...>{}; template<size_t... Is> struct gen_seq
    <0, Is...> : seq<Is...>{};
template<class Ch, class Tr, class Tuple, size_t... Is>
void print_tuple(basic_ostream<Ch,Tr>& os, Tuple const& t, seq<Is...>){ using s = int[]; (void)s{0, (void(os << (Is == 0? "" : ", ") << get<Is>(t)), 
    0)...}; }
template<class Ch, class Tr, class... Args> 
auto operator<<(basic_ostream<Ch, Tr>& os, tuple<Args...> const& t) -> basic_ostream<Ch, Tr>& { os << "("; print_tuple(os, t, gen_seq<sizeof...(Args
    )>()); return os << ")"; }
ostream &operator<<(ostream &o, const vvll &v) { rep(i, v.size()) { rep(j, v[i].size()) o << v[i][j] << " "; o << endl; } return o; }
template <typename T> ostream &operator<<(ostream &o, const vector<T> &v) { o << '['; rep(i, v.size()) o << v[i] << (i != v.size()-1 ? ", " : ""); o 
    << "]";  return o; }
template <typename T>  ostream &operator<<(ostream &o, const set<T> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next
    (it) != m.end() ? ", " : ""); o << "]";  return o; }
template <typename T, typename U>  ostream &operator<<(ostream &o, const map<T, U> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o 
    << *it << (next(it) != m.end() ? ", " : ""); o << "]";  return o; }
template <typename T, typename U, typename V>  ostream &operator<<(ostream &o, const unordered_map<T, U, V> &m) { o << '['; for (auto it = m.begin(); 
    it != m.end(); it++) o << *it; o << "]";  return o; }
vector<int> range(const int x, const int y) { vector<int> v(y - x + 1); iota(v.begin(), v.end(), x); return v; }
template <typename T> istream& operator>>(istream& i, vector<T>& o) { rep(j, o.size()) i >> o[j]; return i;}
string bits_to_string(ll input, ll n=64) { string s; rep(i, n) s += '0' + !!(input & (1ll << i)); reverse(all(s)); return s; }
template <typename T> unordered_map<T, ll> counter(vector<T> vec){unordered_map<T, ll> ret; for (auto&& x : vec) ret[x]++; return ret;};
string substr(string s, P x) {return s.substr(x.fi, x.se - x.fi); }
struct ci : public iterator<forward_iterator_tag, ll> { ll n; ci(const ll n) : n(n) { } bool operator==(const ci& x) { return n == x.n; } bool 
    operator!=(const ci& x) { return !(*this == x); } ci &operator++() { n++; return *this; } ll operator*() const { return n; } };
size_t random_seed; namespace std { using argument_type = P; template<> struct hash<argument_type> { size_t operator()(argument_type const& x) const 
    { size_t seed = random_seed; seed ^= hash<ll>{}(x.fi); seed ^= (hash<ll>{}(x.se) << 1); return seed; } }; }; // hash for various class
namespace myhash{ const int Bsizes[]={3,9,13,17,21,25,29,33,37,41,45,49,53,57,61,65,69,73,77,81}; const int xor_nums[]={0x100007d1,0x5ff049c9
    ,0x14560859,0x07087fef,0x3e277d49,0x4dba1f17,0x709c5988,0x05904258,0x1aa71872,0x238819b3,0x7b002bb7,0x1cf91302,0x0012290a,0x1083576b,0x76473e49
    ,0x3d86295b,0x20536814,0x08634f4d,0x115405e8,0x0e6359f2}; const int hash_key=xor_nums[rand()%20]; const int mod_key=xor_nums[rand()%20]; template 
    <typename T> struct myhash{ std::size_t operator()(const T& val) const { return (hash<T>{}(val)%mod_key)^hash_key; } }; };
template <typename T> class uset:public std::unordered_set<T,myhash::myhash<T>> { using SET=std::unordered_set<T,myhash::myhash<T>>; public: uset
    ():SET(){SET::rehash(myhash::Bsizes[rand()%20]);} };
template <typename T,typename U> class umap:public std::unordered_map<T,U,myhash::myhash<T>> { public: using MAP=std::unordered_map<T,U,myhash
    ::myhash<T>>; umap():MAP(){MAP::rehash(myhash::Bsizes[rand()%20]);} };    
struct timeval start; double sec() { struct timeval tv; gettimeofday(&tv, NULL); return (tv.tv_sec - start.tv_sec) + (tv.tv_usec - start.tv_usec) * 
    1e-6; }
struct init_{init_(){ gettimeofday(&start, NULL); ios::sync_with_stdio(false); cin.tie(0); srand((unsigned int)time(NULL)); random_seed = RAND_MAX / 
    2 + rand() / 2; }} init__;
static const long long mo = 1e9+7;
class Mod {
    public:
        ll num;
        Mod() : Mod(0) {}
        Mod(long long int n) : num(n) { }
        Mod(const string &s){ long long int tmp = 0; for(auto &c:s) tmp = (c-'0'+tmp*10) % mo; num = tmp; }
        Mod(int n) : Mod(static_cast<long long int>(n)) {}
        operator int() { return num; }
};
istream &operator>>(istream &is, Mod &x) { long long int n; is >> n; x = n; return is; }
ostream &operator<<(ostream &o, const Mod &x) { o << x.num; return o; }
Mod operator+(const Mod a, const Mod b) { return Mod((a.num + b.num) % mo); }
Mod operator+(const long long int a, const Mod b) { return Mod(a) + b; }
Mod operator+(const Mod a, const long long int b) { return b + a; }
Mod operator++(Mod &a) { return a + Mod(1); }
Mod operator-(const Mod a, const Mod b) { return Mod((mo + a.num - b.num) % mo); }
Mod operator-(const long long int a, const Mod b) { return Mod(a) - b; }
Mod operator--(Mod &a) { return a - Mod(1); }
Mod operator*(const Mod a, const Mod b) { return Mod(((long long)a.num * b.num) % mo); }
Mod operator*(const long long int a, const Mod b) { return Mod(a)*b; }
Mod operator*(const Mod a, const long long int b) { return Mod(b)*a; }
Mod operator*(const Mod a, const int b) { return Mod(b)*a; }
Mod operator+=(Mod &a, const Mod b) { return a = a + b; }
Mod operator+=(long long int &a, const Mod b) { return a = a + b; }
Mod operator-=(Mod &a, const Mod b) { return a = a - b; }
Mod operator-=(long long int &a, const Mod b) { return a = a - b; }
Mod operator*=(Mod &a, const Mod b) { return a = a * b; }
Mod operator*=(long long int &a, const Mod b) { return a = a * b; }
Mod operator*=(Mod& a, const long long int &b) { return a = a * b; }
Mod factorial(const long long n) {
    if (n < 0) return 0;
    Mod ret = 1;
    for (int i = 1; i <= n; i++) {
        ret *= i;
    }
    return ret;
}
Mod operator^(const Mod a, const long long n) {
    if (n == 0) return Mod(1);
    Mod res = (a * a) ^ (n / 2);
    if (n % 2) res = res * a;
    return res;
}
Mod modpowsum(const Mod a, const long long b) {
    if (b == 0) return 0;
    if (b % 2 == 1) return modpowsum(a, b - 1) * a + Mod(1);
    Mod result = modpowsum(a, b / 2);
    return result * (a ^ (b / 2)) + result;
}
/*************************************/
// 以下、modは素数でなくてはならない！
/*************************************/
Mod inv(const Mod a) { return a ^ (mo - 2); }
/*************************************/
// GF(p)の行列演算
/*************************************/
using number = Mod;
using arr = vector<number>;
using matrix = vector<vector<Mod>>;
ostream &operator<<(ostream &o, const arr &v) { rep(i, v.size()) cout << v[i] << " "; return o; }
ostream &operator<<(ostream &o, const matrix &v) { rep(i, v.size()) cout << v[i]; return o; }
matrix zero(int n) { return matrix(n, arr(n, 0)); } // O(n^2)
matrix identity(int n) { matrix A(n, arr(n, 0)); rep(i, n) A[i][i] = 1; return A; } // O(n^2)
// O(n^2)
arr mul(const matrix &A, const arr &x) { 
    arr y(A.size(), 0); 
    rep(i, A.size()) rep(j, A[0].size()) y[i] += A[i][j] * x[j]; 
    return y; 
} 
// O(n^3)
matrix mul(const matrix &A, const matrix &B) {
    matrix C(A.size(), arr(B[0].size(), 0));
    rep(i, C.size())
        rep(j, C[i].size())
        rep(k, A[i].size())
        C[i][j] += A[i][k] * B[k][j];
    return C;
}
// O(n^2)
matrix plu(const matrix &A, const matrix &B) {
    matrix C(A.size(), arr(B[0].size(), 0));
    rep(i, C.size())
        rep(j, C[i].size())
            C[i][j] += A[i][j] + B[i][j];
    return C;
}
// O(n^2)
matrix sub(const matrix &A, const matrix &B) {
    matrix C(A.size(), arr(B[0].size(), 0));
    rep(i, C.size())
        rep(j, C[i].size())
            C[i][j] += A[i][j] - B[i][j];
    return C;
}
// O(n^2)
arr plu(const arr &A, const arr &B) {
    arr C(A.size());
    rep(i, A.size())
        C[i] += A[i] + B[i];
    return C;
}
// O(n)
arr sub(const arr &A, const arr &B) {
    arr C(A.size());
    rep(i, A.size())
        C[i] += A[i] - B[i];
    return C;
}
// 構築なし累乗
//
// O(n^3 log e)
matrix pow(const matrix &A, long long e) {
    return e == 0 ? identity(A.size())  :
        e % 2 == 0 ? pow(mul(A, A), e/2) : mul(A, pow(A, e-1));
}
// input : a, b
// output : x, y  s.t. ax + by = （符号付き）gcd(a, b)
int extGcd( int a, int b, int& x, int& y ) {
  if ( b == 0 ) {
    x = 1; y = 0; return a;
  }
  int g = extGcd( b, a % b, y, x );
  y -= (a / b) * x;
  return g;
}
// xn = 1 (mod p)
int invMod(int n, int p) {
  int x, y, g = extGcd ( n, p, x, y );
  if (g == 1) return x;
  else if (g == -1) return -x;
  else return 0; // gcd(n, p) != 1，解なし
}
// 有限体上の線型方程式系 Ax = b (mod q)を解く
// a = [A | b]: m × n の係数行列
// x: 解を記録するベクトル
// 計算量： O(min(m, n) * m * n)
bool gauss(matrix a, arr& x, int m, int n, int q) {
  int rank = 0;
  vll pivot(n);
  // 前進消去
  for (int i = 0, j = 0; i < m && j < n-1; ++j) {
    int p = -1;
    ll tmp = 0;
    // ピボットを探す
    for (int k = i; p < 0 && k < m; ++k) {
      if (a[k][j] != 0) p = k;  // 有限体上なので非零で十分
    }
    // ランク落ち対策
    if (p == -1) continue;
    // 第i行と第p行を入れ替える
    for (int k = j; k < n; ++k)
      tmp = a[i][k], a[i][k] = a[p][k], a[p][k] = tmp;
    // 第i行を使って掃き出す
    for (int k = i+1; k < m; ++k) {
      tmp = -(ll)a[k][j] * invMod(a[i][j], q) % q;
      for (int l = j; l < n; ++l)
        a[k][l] += tmp * a[i][l];
    }
    // 第i行を正規化： a[i][j] = 1 にする
    tmp = invMod(a[i][j], q);
    for (int k = j; k < n; ++k)
      a[i][k] = a[i][k] * tmp % q;
    pivot[i++] = j, rank++;
  }
  // 解の存在のチェック
  for (int i = rank; i < m; ++i)
    if (a[i][n-1] != 0) return false;
  // 解をxに代入（後退代入）
  for (int i = 0; i < rank; ++i)
    x[i] = a[i][n-1];
  for (int i = rank-1; i >= 0; --i) {
    for (int j = pivot[i] + 1; j < n-1; ++j)
      x[i] -= a[i][j] * x[j];
      x[i] -= (ll)x[i] / q * q, x[i] = ((ll)x[i] + q) % q;  // 0 <= x[i] < q に調整
  }
  rep(i, x.size()) x[i] = (x[i] + mo) % mo;
  return true;
}
arr solve(matrix a, arr b) {
    int m = a.size();
    arr ret(a.size());
    rep(i, a.size()) {
        a[i].pb(b[i]);
    }
    gauss(a, ret, m, m+1, mo);
    return ret;
}
int main(void) {
    ll n; cin >> n;
    matrix A = {
        {1,1,1,1,1,1,1,0},
        {1,1,1,1,1,0,0,0},
        {1,1,1,0,0,0,0,1},
        {1,1,0,1,1,0,0,0},
        {1,1,0,1,1,1,0,0},
        {1,0,0,0,1,1,0,1},
        {0,0,1,0,0,1,1,0},
        {1,0,0,0,0,0,0,1},
    };
    arr x = 
        {1,1,1,1,1,1,1,0};
    ll m = 8;
    auto an = pow(A, n);
    auto y = x;
    // s = (n-1)s
    arr s = solve(sub(identity(m), A), mul(sub(identity(m), an), x));
    // (I-A)t  = s - x - (n-1)A^n x
    arr annm1x = mul(an, x);
    rep(i, m) annm1x[i] = annm1x[i] * ((n-1) % mo);
    annm1x = sub(s, annm1x);
    annm1x = sub(annm1x, x);
    arr t = solve(sub(identity(m), A), annm1x);
    arr r(m);
    rep(i, m) r[i] = (n % mo * s[i] % mo - t[i]) % mo;
    Mod ret = 0;
    rep(i, m) if (i != 6) {
        ret = (ret + r[i]) % mo;
    }
    cout << ret % mo << endl;
    return 0;
}
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