結果

問題 No.108 トリプルカードコンプ
ユーザー fumiphysfumiphys
提出日時 2019-09-07 19:42:28
言語 C++14
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 11 ms / 5,000 ms
コード長 3,763 bytes
コンパイル時間 1,529 ms
コンパイル使用メモリ 167,188 KB
実行使用メモリ 11,660 KB
最終ジャッジ日時 2023-09-09 09:17:52
合計ジャッジ時間 3,076 ms
ジャッジサーバーID
（参考情報） 		judge12 / judge13
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 1 ms
4,376 KB
testcase_01 AC 2 ms
4,380 KB
testcase_02 AC 2 ms
4,380 KB
testcase_03 AC 1 ms
4,376 KB
testcase_04 AC 1 ms
4,380 KB
testcase_05 AC 1 ms
4,376 KB
testcase_06 AC 1 ms
4,376 KB
testcase_07 AC 11 ms
11,540 KB
testcase_08 AC 6 ms
11,660 KB
testcase_09 AC 5 ms
11,520 KB
testcase_10 AC 5 ms
11,620 KB
testcase_11 AC 6 ms
11,616 KB
testcase_12 AC 1 ms
4,376 KB
testcase_13 AC 3 ms
6,712 KB
testcase_14 AC 3 ms
6,656 KB
testcase_15 AC 3 ms
6,748 KB
testcase_16 AC 3 ms
6,616 KB
testcase_17 AC 3 ms
6,684 KB
testcase_18 AC 9 ms
11,220 KB
testcase_19 AC 6 ms
11,136 KB
testcase_20 AC 5 ms
11,444 KB
testcase_21 AC 8 ms
11,524 KB
testcase_22 AC 5 ms
11,216 KB
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_;

int n;
double N;
double dp[101][101][101];

double calc(int i0, int i1, int i2){
  if(dp[i0][i1][i2] >= 0)return dp[i0][i1][i2];
  if(i0 == 0 && i1 == 0 && i2 == 0)return dp[i0][i1][i2] = 0.;
  // dp[i0][i1][i2] = 1 + ((i0)/n * dp[i0-1][i1+1][i2] + (i1)/n * dp[i0][i1-1][i2+1] + (i1)/n * dp[i0][i1][i2-1] + (n-i0-i1-i2)/n*dp[i0][i1][i2]);
  double dp1 = (i0 > 0 ? calc(i0-1, i1+1, i2): 0) * (i0) / N;
  double dp2 = (i1 > 0 ? calc(i0, i1-1, i2+1): 0) * (i1) / N;
  double dp3 = (i2 > 0 ? calc(i0, i1, i2-1): 0) * (i2) / N;
  dp[i0][i1][i2] = (1. + dp1 + dp2 + dp3) * N / (i0 + i1 + i2);
  return dp[i0][i1][i2];
}

int main(int argc, char const* argv[])
{
  cin >> n;
  N = n;
  vector<int> a(n);
  cin >> a;
  vector<int> cnt(11, 0);
  rep(i, n){
    cnt[a[i]]++;
  }
  rep(i, n + 1)rep(j, n + 1)rep(k, n + 1)dp[i][j][k] = -1;
  cout << calc(cnt[0], cnt[1], cnt[2]) << endl;
  return 0;
}
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