結果

問題 No.93 ペガサス
ユーザー fumiphys
提出日時 2019-08-01 00:03:23
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 63 ms / 5,000 ms
コード長 4,752 bytes
コンパイル時間 1,607 ms
コンパイル使用メモリ 170,884 KB
実行使用メモリ 35,072 KB
最終ジャッジ日時 2024-07-05 07:21:50
合計ジャッジ時間 2,425 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge5
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 #
プレゼンテーションモードにする

// includes
#include <bits/stdc++.h>
// macros
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#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 vrep(v, i) for(int i = 0; i < (v).size(); i++)
#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 FI first
#define SE second
#define bit(n) (1LL<<(n))
#define INT(n) int n; cin >> n;
#define LL(n) ll n; cin >> n;
#define DOUBLE(n) double n; cin >> n;
using namespace std;
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>;
using Pli = pair<ll, int>;
using Pil = pair<int, ll>;
using Pll = pair<ll, ll>;
using Pdd = pair<double, double>;
using cd = complex<double>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1e9 + 7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// solve
ll dp[1010][1010][2][2];
int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  INT(n);
  dp[0][0][0][0] = 1;
  rep(i, n - 1){
    rep(j, n + 1){
      // 0 0
      if(i>0)(dp[i+1][j+1][0][1] += 2 * dp[i][j][0][0] % mod) %= mod;
      if(j>0)(dp[i+1][j-1][0][0] += j * dp[i][j][0][0] % mod) %= mod;
      if(i-j>=0&&i>0)(dp[i+1][j][0][0] += (i-j) * dp[i][j][0][0] % mod) %= mod;
      else if(i-j+2>=0&&i==0)(dp[i+1][j][0][0] += (i-j+2) * dp[i][j][0][0] % mod) %= mod;
      // 1 0
      (dp[i+1][j][0][1] += dp[i][j][1][0] % mod) %= mod;
      (dp[i+1][j+1][0][1] += dp[i][j][1][0] % mod) %= mod;
      if(j>0)(dp[i+1][j-1][0][0] += (j-1) * dp[i][j][1][0] % mod) %= mod;
      if(i-j+1>=0)(dp[i+1][j][0][0] += (i-j+1) * dp[i][j][1][0] % mod) %= mod;
      // 0 1
      if(i>0)(dp[i+1][j+1][1][1] += 2 * dp[i][j][0][1] % mod) %= mod;
      if(j>0)(dp[i+1][j-1][1][0] += (j-1) * dp[i][j][0][1] % mod) %= mod;
      if(j>0)(dp[i+1][j-1][0][0] += dp[i][j][0][1] % mod) %= mod;
      if(i-j>=0)(dp[i+1][j][1][0] += (i-j) * dp[i][j][0][1] % mod) %= mod;
      // 1 1
      (dp[i+1][j][1][1] += dp[i][j][1][1] % mod) %= mod;
      (dp[i+1][j+1][1][1] += dp[i][j][1][1] % mod) %= mod;
      if(j>1)(dp[i+1][j-1][1][0] += (j-2) * dp[i][j][1][1] % mod) %= mod;
      if(j>0)(dp[i+1][j-1][0][0] += dp[i][j][1][1] % mod) %= mod;
      if(i-j+1>=0)(dp[i+1][j][1][0] += (i-j+1) * dp[i][j][1][1] % mod) %= mod;
    }
  }
  //rep(i, n)rep(j, n)rep(m, 2)rep(k, 2)cout << i << " " << j << " " << m << " " << k << " " << dp[i][j][m][k] << endl;
  cout << dp[n-1][0][0][0] << endl;
  return 0;
}
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