結果

問題 No.3202 Periodic Alternating Subsequence
ユーザー HoyHoyCharhang
提出日時 2025-07-11 22:36:11
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 357 ms / 2,000 ms
コード長 4,071 bytes
コンパイル時間 3,639 ms
コンパイル使用メモリ 286,484 KB
実行使用メモリ 33,160 KB
最終ジャッジ日時 2025-07-11 22:36:24
合計ジャッジ時間 11,483 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
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 * 2
other AC * 24
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>
#define fi first
#define se second
#define rep(i,s,n) for (int i = (s); i < (n); ++i)
#define rrep(i,n,g) for (int i = (n)-1; i >= (g); --i)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define len(x) (int)(x).size()
#define dup(x,y) (((x)+(y)-1)/(y))
#define pb push_back
#define eb emplace_back
#define Field(T) vector<vector<T>>
using namespace std;
using ll = long long;
using ull = unsigned long long;
template<typename T> using pq = priority_queue<T,vector<T>,greater<T>>;
using P = pair<int,int>;
template<class T>bool chmax(T&a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T>bool chmin(T&a,T b){if(b<a){a=b;return 1;}return 0;}

template< int mod >
struct ModInt {
  int x;
  ModInt() : x(0) {}
  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
  ModInt &operator+=(const ModInt &p) {
    if((x += p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator-=(const ModInt &p) {
    if((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator*=(const ModInt &p) {
    x = (int) (1LL * x * p.x % mod);
    return *this;
  }
  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }
  ModInt operator-() const { return ModInt(-x); }
  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
  bool operator==(const ModInt &p) const { return x == p.x; }
  bool operator!=(const ModInt &p) const { return x != p.x; }

  ModInt inverse() const {
    assert(x);
    int a = x, b = mod, u = 1, v = 0, t;
    while(b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }

  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while(n > 0) {
      if(n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }

  friend ostream &operator<<(ostream &os, const ModInt &p) {
    return os << p.x;
  }

  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt< mod >(t);
    return (is);
  }

  static int get_mod() { return mod; }
};

using mint = ModInt<1000000007>;

template<typename T>
vector<vector<T>> mat_mul(vector<vector<T>> &a, vector<vector<T>> &b) {
  assert(b.size() == a[0].size());
  int n = (int)a.size(), m = (int)b[0].size(), l = (int)b.size();
  vector<vector<T>> c(n, vector<T>(m, T(0)));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      for (int k = 0; k < l; ++k) {
        c[i][j] += a[i][k] * b[k][j];
      }
    }
  }
  return c;
}

template<typename T>
vector<vector<T>> mat_pow(vector<vector<T>> &x, long long n) {
  int m = (int)x.size();
  vector<vector<T>> y(m, vector<T>(m, T(0)));
  for (int i = 0; i < m; ++i) {
    y[i][i] = T(1);
  }
  while (n) {
    if (n & 1) {
      y = mat_mul(x, y);
    }
    x = mat_mul(x, x);
    n >>= 1;
  }
  return y;
}

int main() {
  string s;
  cin >> s;
  ll k;
  cin >> k;
  int n = len(s);
  vector<vector<mint>> a(6, vector<mint>(6));
  rep(p,0,2) rep(q,0,3) {
    vector<vector<vector<mint>>> dp(n+1, vector<vector<mint>>(2, vector<mint>(3, 0)));
    dp[0][p][q] = 1;
    rep(i,0,n) {
      rep(j,0,2) rep(l,0,3) dp[i+1][j][l] += dp[i][j][l];
      if (s[i] == '0') {
        dp[i+1][0][0] += dp[i][1][0];
        dp[i+1][0][1] += dp[i][1][1] + dp[i][1][0];
        dp[i+1][0][2] += dp[i][1][2] + dp[i][1][1]*2 + dp[i][1][0];
      } else {
        dp[i+1][1][0] += dp[i][0][0];
        dp[i+1][1][1] += dp[i][0][1] + dp[i][0][0];
        dp[i+1][1][2] += dp[i][0][2] + dp[i][0][1]*2 + dp[i][0][0];
      }
    }
    rep(np,0,2) rep(nq,0,3) {
      a[3*p+q][3*np+nq] = dp[n][np][nq];
    }
  }
  a = mat_pow(a, k);
  // rep(i,0,6) {
  //   rep(j,0,6) {
  //     cout << a[i][j] << " ";
  //   }
  //   cout << endl;
  // }
  cout << a[0][2]+a[0][5]+a[3][2]+a[3][5] << endl;
  return 0;
}
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