結果
問題 |
No.3082 Make Palindromic Multiple(Judge)
|
ユーザー |
👑 |
提出日時 | 2025-02-26 02:00:58 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
MLE
|
実行時間 | - |
コード長 | 2,641 bytes |
コンパイル時間 | 4,752 ms |
コンパイル使用メモリ | 254,588 KB |
実行使用メモリ | 823,524 KB |
最終ジャッジ日時 | 2025-03-28 01:21:30 |
合計ジャッジ時間 | 12,099 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 5 MLE * 1 -- * 60 |
ソースコード
#include<bits/stdc++.h> #include<atcoder/all> #define rep(i,n) for(int i=0;i<n;i++) using namespace std; using namespace atcoder; typedef long long ll; typedef vector<int> vi; typedef vector<long long> vl; typedef vector<vector<int>> vvi; typedef vector<vector<long long>> vvl; typedef long double ld; typedef pair<int, int> P; template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;} template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;} template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;} template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;} template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;} template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;} template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;} template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;} template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;} template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;} template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template<typename T> void chmin(T& a, T b){a = min(a, b);} template<typename T> void chmax(T& a, T b){a = max(a, b);} int main(){ int K; cin >> K; vector<string> S(K); vector<long long> T(K); rep(i, K) cin >> S[i] >> T[i]; vector<string> x(2); rep(t, 2){ rep(i, K) rep(j, T[i]) x[t] += S[i]; reverse(S.begin(), S.end()); reverse(T.begin(), T.end()); rep(i, K) reverse(S[i].begin(), S[i].end()); } if(x[0] == x[1]) cout << "Yes\n"; else cout << "No\n"; return 0; }