結果

問題 No.93 ペガサス
ユーザー fumiphysfumiphys
提出日時 2019-09-04 23:16:00
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 40 ms / 5,000 ms
コード長 4,342 bytes
コンパイル時間 1,680 ms
コンパイル使用メモリ 169,068 KB
実行使用メモリ 22,784 KB
最終ジャッジ日時 2024-12-30 00:00:24
合計ジャッジ時間 2,764 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge3
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 * 4
other AC * 16
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 # 

// includes
#include <bits/stdc++.h>
using namespace std;

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define bit(n) (1LL<<(n))
// functions
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(20);}
} fast_io_;

ll dp[1010][1010][2][2];

int main(int argc, char const* argv[])
{
  int n;
  cin >> n;
  dp[0][0][0][0] = 1;
  rep(i, n){
    rep(j, i + 1){
      // k = 0, l = 0
      if(j > 0)(dp[i+1][j-1][0][0] += j * dp[i][j][0][0] % mod) %= mod;
      if(i > 1){
        (dp[i+1][j+1][0][1] += 2 * dp[i][j][0][0] % mod) %= mod;
        if(i-1-j>=0)(dp[i+1][j][0][0] += (i-1-j) * dp[i][j][0][0] % mod) %= mod;
      }else{
        (dp[i+1][j][0][0] += (i+1-j) * dp[i][j][0][0] % mod) %= mod;
      }
      // k = 0, l = 1
      if(j > 0){
        (dp[i+1][j-1][1][0] += (j - 1) * dp[i][j][0][1] % mod) %= mod;
        (dp[i+1][j-1][0][0] += dp[i][j][0][1]) %= mod;
      }
      (dp[i+1][j+1][1][1] += 2 * dp[i][j][0][1] % mod) %= mod;
      if(i-1-j>=0)(dp[i+1][j][1][0] += (i-1-j) * dp[i][j][0][1] % mod) %= mod;
      // k = 1, l = 0
      (dp[i+1][j][0][1] += dp[i][j][1][0]) %= mod;
      if(j > 0)(dp[i+1][j-1][0][0] += (j - 1) * dp[i][j][1][0] % mod) % mod;
      (dp[i+1][j+1][0][1] += dp[i][j][1][0]) %= mod;
      (dp[i+1][j][0][0] += (i-j) * dp[i][j][1][0] % mod) %= mod;
      // k = 1, l = 1
      (dp[i+1][j][1][1] += dp[i][j][1][1]) %= mod;
      if(j > 0){
        (dp[i+1][j-1][0][0] += dp[i][j][1][1]) %= mod;
        if(j > 1)(dp[i+1][j-1][1][0] += (j - 2) * dp[i][j][1][1]) %= mod;
      }
      (dp[i+1][j+1][1][1] += dp[i][j][1][1]) %= mod;
      (dp[i+1][j][1][0] += (i-j) * dp[i][j][1][1]) %= mod;
    }
  }
  cout << dp[n][0][0][0] << endl;
  return 0;
}
0