結果
問題 | No.683 Two Operations No.3 |
ユーザー | fumiphys |
提出日時 | 2019-05-26 13:00:59 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
MLE
|
実行時間 | - |
コード長 | 3,642 bytes |
コンパイル時間 | 1,625 ms |
コンパイル使用メモリ | 168,248 KB |
実行使用メモリ | 814,592 KB |
最終ジャッジ日時 | 2024-09-17 15:12:12 |
合計ジャッジ時間 | 3,444 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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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 | 2 ms
6,812 KB |
testcase_01 | AC | 2 ms
6,944 KB |
testcase_02 | AC | 1 ms
6,944 KB |
testcase_03 | AC | 1 ms
6,940 KB |
testcase_04 | AC | 1 ms
6,940 KB |
testcase_05 | AC | 2 ms
6,944 KB |
testcase_06 | MLE | - |
testcase_07 | -- | - |
testcase_08 | -- | - |
testcase_09 | -- | - |
testcase_10 | -- | - |
testcase_11 | -- | - |
testcase_12 | -- | - |
testcase_13 | -- | - |
testcase_14 | -- | - |
testcase_15 | -- | - |
ソースコード
// includes #include <bits/stdc++.h> // macros #define ll long long int #define pb emplace_back #define mk make_pair #define pq priority_queue #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 vrep(v, i) for(int i = 0; i < (v).size(); i++) #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 FI first #define SE second #define dump(a, n) for(int i = 0; i < n; i++)cout << a[i] << "\n "[i + 1 != n]; #define dump2(a, n, m) for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)cout << a[i][j] << "\n "[j + 1 != m]; #define bit(n) (1LL<<(n)) #define INT(n) int n; cin >> n; #define LL(n) ll n; cin >> n; #define DOUBLE(n) double n; cin >> n; using namespace std; // types typedef pair<int, int> P; typedef pair<ll, int> Pl; typedef pair<ll, ll> Pll; typedef pair<double, double> Pd; typedef complex<double> cd; // constants const int inf = 1e9; const ll linf = 1LL << 50; const double EPS = 1e-10; const int mod = 1e9 + 7; const int dx[4] = {-1, 0, 1, 0}; const int dy[4] = {0, -1, 0, 1}; // solve 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;} bool ok(ll a, ll b){ if(a == 0 && b == 0)return true; if(a % 2 == 0 && b > 0 && ok(a / 2, b - 1))return true; if(b % 2 == 0 && a > 0 && ok(a - 1, b / 2))return true; return false; } int main(int argc, char const* argv[]) { ios_base::sync_with_stdio(false); cin.tie(0); LL(a); LL(b); if(ok(a, b))cout << "Yes" << endl; else cout << "No" << endl; return 0; }