結果

問題 No.212 素数サイコロと合成数サイコロ (2)
ユーザー fumiphysfumiphys
提出日時 2019-10-05 17:12:59
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 38 ms / 5,000 ms
コード長 3,592 bytes
コンパイル時間 1,716 ms
コンパイル使用メモリ 169,720 KB
実行使用メモリ 13,040 KB
最終ジャッジ日時 2024-10-06 09:56:37
合計ジャッジ時間 2,220 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge4
このコードへのチャレンジ
（要ログイン）

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 11 ms
11,012 KB
testcase_01 AC 22 ms
11,068 KB
testcase_02 AC 11 ms
11,616 KB
testcase_03 AC 22 ms
11,228 KB
testcase_04 AC 31 ms
13,040 KB
testcase_05 AC 38 ms
12,984 KB
testcase_06 AC 35 ms
13,036 KB
testcase_07 AC 35 ms
12,956 KB
testcase_08 AC 22 ms
12,932 KB
testcase_09 AC 23 ms
12,820 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 a[6] = {2, 3, 5, 7, 11, 13};
int b[6] = {4, 6, 8, 9, 10, 12};

double p1[400001], p2[400001];
double tmp[400001];

int main(int argc, char const* argv[])
{
  int p, c; cin >> p >> c;
  p1[1] = 1.;
  int m = 400000;
  rep(i, p){
    rep(j, m+1)tmp[j] = 0.;
    for(int j = m; j > 0; j--){
      rep(k, 6){
        if(j * a[k] <= m)tmp[j * a[k]] += p1[j] / 6.;
      }
    }
    swap(p1, tmp);
  }
  p2[1] = 1.;
  rep(i, c){
    rep(j, m+1)tmp[j] = 0.;
    for(int j = m; j > 0; j--){
      rep(k, 6){
        if(j * b[k] <= m)tmp[j * b[k]] += p2[j] / 6.;
      }
    }
    swap(p2, tmp);
  }
  double r1 = 0., r2 = 0.;
  rep(i, m)r1 += i * p1[i];
  rep(i, m)r2 += i * p2[i];
  cout << r1 * r2 << endl;
  return 0;
}
0