結果
問題 | No.183 たのしい排他的論理和(EASY) |
ユーザー | fumiphys |
提出日時 | 2019-09-26 23:26:49 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 589 ms / 5,000 ms |
コード長 | 3,189 bytes |
コンパイル時間 | 1,948 ms |
コンパイル使用メモリ | 170,488 KB |
実行使用メモリ | 6,948 KB |
最終ジャッジ日時 | 2024-07-02 17:21:58 |
合計ジャッジ時間 | 6,556 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
(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 | 2 ms
6,812 KB |
testcase_01 | AC | 3 ms
6,944 KB |
testcase_02 | AC | 4 ms
6,940 KB |
testcase_03 | AC | 3 ms
6,944 KB |
testcase_04 | AC | 3 ms
6,940 KB |
testcase_05 | AC | 4 ms
6,940 KB |
testcase_06 | AC | 3 ms
6,944 KB |
testcase_07 | AC | 513 ms
6,944 KB |
testcase_08 | AC | 521 ms
6,940 KB |
testcase_09 | AC | 358 ms
6,948 KB |
testcase_10 | AC | 494 ms
6,944 KB |
testcase_11 | AC | 567 ms
6,940 KB |
testcase_12 | AC | 4 ms
6,940 KB |
testcase_13 | AC | 2 ms
6,944 KB |
testcase_14 | AC | 589 ms
6,940 KB |
testcase_15 | AC | 579 ms
6,940 KB |
testcase_16 | AC | 3 ms
6,940 KB |
testcase_17 | AC | 129 ms
6,940 KB |
testcase_18 | AC | 509 ms
6,940 KB |
testcase_19 | AC | 66 ms
6,944 KB |
ソースコード
// 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 main(int argc, char const* argv[]) { int n; cin >> n; vector<int> a(n); cin >> a; vector<bool> dp(bit(15) + 1, false); dp[0] = true; rep(i, n){ rep(j, bit(15) + 1){ dp[j^a[i]] = dp[j^a[i]] | dp[j]; } } int res = 0; rep(i, sz(dp))if(dp[i])++res; cout << res << endl; return 0; }