結果

問題 No.243 出席番号(2)
ユーザー fumiphysfumiphys
提出日時 2019-07-18 01:27:14
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 5,443 bytes
コンパイル時間 1,760 ms
コンパイル使用メモリ 174,592 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-06-06 08:04:46
合計ジャッジ時間 25,518 ms
ジャッジサーバーID
(参考情報)
judge3 / judge2
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
6,816 KB
testcase_01 AC 2 ms
6,940 KB
testcase_02 AC 2 ms
6,944 KB
testcase_03 WA -
testcase_04 WA -
testcase_05 WA -
testcase_06 WA -
testcase_07 WA -
testcase_08 WA -
testcase_09 WA -
testcase_10 WA -
testcase_11 WA -
testcase_12 WA -
testcase_13 WA -
testcase_14 WA -
testcase_15 WA -
testcase_16 WA -
testcase_17 WA -
testcase_18 WA -
testcase_19 WA -
testcase_20 WA -
testcase_21 WA -
testcase_22 WA -
testcase_23 TLE -
testcase_24 TLE -
testcase_25 TLE -
testcase_26 TLE -
testcase_27 TLE -
testcase_28 WA -
testcase_29 WA -
testcase_30 WA -
testcase_31 WA -
testcase_32 WA -
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>

// macros
#define ll long long int
#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 dump(a, n) for(int i = 0; i < n; i++)cout << a[i] << "\n "[i + 1 != n];
#define dump2(a, n, m) for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)cout << a[i][j] << "\n "[j + 1 != m];
#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
typedef pair<int, int> P;
typedef pair<ll, int> Pl;
typedef pair<ll, ll> Pll;
typedef pair<double, double> Pd;
typedef complex<double> cd;

// 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
template <typename T>
T power(T a, T n, T mod) {
  T res = 1;
  T tmp = n;
  T curr = a;
  while(tmp){
    if(tmp % 2 == 1){
      res = (T)(res * curr % mod);
    }
    curr = (T)(curr * curr % mod);
    tmp >>= 1;
  }

  return res;
}

struct Factorial{
  int n;
  const int MOD = 1e9 + 7;
  vector<long long> fac;
  vector<long long> inv_;
  Factorial(int n): n(n){
    fac.resize(n + 1);
    inv_.resize(n + 1);
    calc_factorial();
    calc_inv();
  }
  void calc_factorial(){
    fac[0] = 1;
    for(int i = 1; i <= n; i++){
      fac[i] = i * fac[i-1] % MOD;
    }
  }
  void calc_inv(){
    inv_[n] = power<long long>(fac[n], MOD - 2, MOD);
    for(int i = n - 1; i >= 0; i--){
      inv_[i] = (i + 1) * inv_[i+1] % MOD;
    }
  }
  long long& operator[](size_t i){
    if(i < 0 || i > n){
      cerr << "list index out of range" << endl;
      abort();
    }
    return fac[i];
  }
  long long inv(size_t i){
    if(i < 0 || i > n){
      cerr << "list index out of range" << endl;
      abort();
    }
    return inv_[i];
  }
  long long comb(int n, int k){
    if(n < k)return 0;
    long long res = fac[n];
    res = res * inv_[n-k] % MOD;
    res = res * inv_[k] % MOD;
    return res;
  }
  long long perm(int n, int k){
    if(n < k)return 0;
    long long res = fac[n];
    res = res * inv_[n-k] % MOD;
    return res;
  }
  long long h(int n, int k){
    if(n == 0 && k == 0)return 1;
    return comb(n + k - 1, k);
  }
};
ll dp[2][5001];
int c[5001];
int sum[5001];

int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  INT(n); vector<int> a(n); cin >> a;
  rep(i, n)c[a[i]]++;
  rep(i, n)sum[i] = (i > 0 ? sum[i-1]: 0) + c[i];
  dp[0][0] = 1;
  Factorial fac(n);
  rep(i, n - 1){
    rep(j, sum[i] + 1){
      int rem = sum[i] - j;
      for(int k = 0; k <= min(rem, c[i+1]); k++){
        (dp[(i+1)&1][j+k+1] += (fac.perm(rem, k) * fac.comb(c[i+1], k) % mod) * (dp[i&1][j] * rem % mod) % mod) %= mod;
        (dp[(i+1)&1][j+k] += (fac.perm(rem, k) * fac.comb(c[i+1], k) % mod) * (dp[i&1][j] * 1 % mod) % mod) %= mod;
      }
    }
  }
  cout << dp[(n-1)&1][n] << endl;
  return 0;
}
0