結果

問題 No.130 XOR Minimax
ユーザー fumiphysfumiphys
提出日時 2019-09-14 23:39:31
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 100 ms / 5,000 ms
コード長 3,328 bytes
コンパイル時間 2,469 ms
コンパイル使用メモリ 172,484 KB
実行使用メモリ 29,912 KB
最終ジャッジ日時 2024-09-12 22:53:11
合計ジャッジ時間 4,478 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge1
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 53 ms
6,812 KB
testcase_01 AC 2 ms
6,812 KB
testcase_02 AC 2 ms
6,940 KB
testcase_03 AC 2 ms
6,944 KB
testcase_04 AC 48 ms
29,788 KB
testcase_05 AC 53 ms
29,912 KB
testcase_06 AC 47 ms
17,500 KB
testcase_07 AC 50 ms
17,480 KB
testcase_08 AC 100 ms
7,432 KB
testcase_09 AC 11 ms
6,940 KB
testcase_10 AC 17 ms
6,944 KB
testcase_11 AC 52 ms
15,444 KB
testcase_12 AC 7 ms
6,940 KB
testcase_13 AC 44 ms
12,892 KB
testcase_14 AC 100 ms
7,048 KB
testcase_15 AC 3 ms
6,944 KB
testcase_16 AC 72 ms
6,944 KB
testcase_17 AC 75 ms
6,940 KB
testcase_18 AC 81 ms
6,944 KB
testcase_19 AC 95 ms
6,940 KB
testcase_20 AC 49 ms
6,944 KB
testcase_21 AC 100 ms
6,944 KB
testcase_22 AC 24 ms
6,944 KB
testcase_23 AC 9 ms
6,940 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_;

ll calc(vector<ll> &a, int k){
  if(k < 0)return 0;
  ll res = 0;
  vector<ll> a1, a2;
  rep(i, sz(a)){
    if(a[i] & bit(k))a1.pb(a[i]);
    else a2.pb(a[i]);
  }
  if(sz(a1) == 0 || sz(a2) == 0){
    res = calc(a, k - 1);
  }else{
    res = min(bit(k) + calc(a1, k-1), bit(k) + calc(a2, k-1));
  }
  return res;
}

int main(int argc, char const* argv[])
{
  int n;
  cin >> n;
  vector<ll> a(n);
  cin >> a;
  cout << calc(a, 31) << endl;
  return 0;
}
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