結果

問題 No.12 限定された素数
コンテスト
ユーザー fumiphys
提出日時 2020-04-05 16:50:19
言語 C++14
(gcc 13.3.0 + boost 1.89.0)
結果
AC  
実行時間 40 ms / 5,000 ms
コード長 3,977 bytes
記録
記録タグの例:
初AC ショートコード 純ショートコード 純主流ショートコード 最速実行時間
コンパイル時間 1,659 ms
コンパイル使用メモリ 172,004 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-11-24 10:26:47
合計ジャッジ時間 3,340 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge1
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
other AC * 26
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 #
raw source code 

// 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 whole(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define bit(n) (1LL<<(n))
// functions
template <typename T> void unique(T& c){c.erase(std::unique(c.begin(), c.end()), c.end());}
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(nullptr); cout << fixed << setprecision(20);}
} fast_io_;

template <typename T>
vector<T> list_prime(T n){
  vector<T> res;
  vector<bool> i_prime = vector<bool>(n+1, true);
  i_prime[0] = i_prime[1] = false;
  for(ll i = 2; i <= n; i++){
    if(i_prime[i]){
      res.push_back(i);
      for(ll j = 2; j * i <= n; j++){
        i_prime[i * j] = false;
      }
    }
  }
  return res;
}


int main(int argc, char const* argv[])
{
  int n; cin >> n;
  vector<int> a(n); cin >> a;
  auto p = list_prime(5001000);
  int x = 0;
  rep(i, n)x ^= (1 << a[i]);
  int ng = (1<<10) - 1; ng ^= x;
  int res = -1;
  int prev = 1;
  int curr = 0;
  rep(j, sz(p)){
    int y = 0;
    int tmp = p[j];
    while(tmp){
      y |= (1 << (tmp % 10));
      tmp /= 10;
    }
    if((ng & y) != 0 || p[j] > 5000000){
      if(p[j] > 5000000){
        if(prev != 5000000)chmax(res, 5000000 - prev);
        break;
      }
      if(curr == x){
        chmax(res, p[j] - 1 - prev);
      }
      curr = 0;
      prev = p[j] + 1;
    }else{
      curr |= y;
    }
  }
  cout << res << endl;
  return 0;
}
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