結果

問題 No.3082 Make Palindromic Multiple(Judge)
ユーザー 👑 binap
提出日時 2025-02-26 12:20:17
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,248 bytes
コンパイル時間 8,739 ms
コンパイル使用メモリ 257,060 KB
実行使用メモリ 11,148 KB
最終ジャッジ日時 2025-03-28 01:21:47
合計ジャッジ時間 11,790 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge5
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 4
other AC * 45 WA * 9 RE * 12
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 # 

#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);}

using ull = unsigned long long;

using mint1 = atcoder::dynamic_modint<1>;

void random_mod(){
	const int p1 = 998244353;
	mint1::set_mod(p1);
}

long long MOD;

int K;
vector<string> S;
vector<long long> T;

void input(){
	cin >> K;
	S.resize(K);
	T.resize(K);
	rep(i, K) cin >> S[i] >> T[i];
}

bool judge_palindrome(){
	
	using F = mint1;
	vector<F> hash(2, 0);
	F base(2524844);
	
	auto get_hash = [&]() -> F{
		F x(0);
		rep(i, K){
			F p = F(base).pow(int(S[i].size()));
			F q = p.pow(T[i]);
			x *= q;
			
			F y(0);
			for(auto c : S[i]){
				y *= base;
				y += c - '0' + 1;
			}
			x += y * (q - 1) / (p - 1);
		}
		return x;
	};
	
	rep(t, 2){
		hash[t] = get_hash();
		
		reverse(S.begin(), S.end());
		reverse(T.begin(), T.end());
		rep(i, K) reverse(S[i].begin(), S[i].end());
	}
	return hash[0] == hash[1];
}

int main() {
	random_mod();
	input();
	bool is_palindrome = judge_palindrome();
	if(is_palindrome) cout << "Yes\n";
	else cout << "No\n";
	return 0;
}
0