結果

問題 No.3082 Make Palindromic Multiple(Judge)
ユーザー zurr
提出日時 2025-04-05 09:07:46
言語 C++17(clang)
(17.0.6 + boost 1.87.0)
結果
AC  
実行時間 344 ms / 3,500 ms
コード長 4,058 bytes
コンパイル時間 4,038 ms
コンパイル使用メモリ 190,948 KB
実行使用メモリ 11,076 KB
最終ジャッジ日時 2025-04-16 13:14:23
合計ジャッジ時間 8,255 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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ファイルパターン 結果
sample AC * 4
other AC * 73
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);}

random_device rd;
mt19937 mt(rd());

int random_prime(){
  int pr;
  do {
    pr = static_cast<int>(mt() >> 2 | 1U << 29);
  } while(!atcoder::internal::is_prime_constexpr(pr)
       || !atcoder::internal::is_prime_constexpr(pr*2+1));
  return pr;
}

using ull = unsigned long long;

using mint1 = atcoder::dynamic_modint<1>;

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

template<typename T>
pair<T, T> pow_mod(T A, T N, T MOD){
  assert(N >= 0);
  if(N == 0) return {1 % MOD, 0};
  if(N == 1) return {A % MOD, 1 % MOD};
  A %= MOD;
  if(N % 2 == 0){
    auto [p, s] = pow_mod(A, N / 2, MOD);
    T p2 = (p * p) % MOD;
    T s2 = (s + s * p) % MOD;
    return {p2, s2};
  }else{
    auto [p, s] = pow_mod(A, N - 1, MOD);
    T p2 = (p * A) % MOD;
    T s2 = (s + p) % MOD;
    return {p2, s2};
  }
}

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(mt() % (mint1::mod() - 3) + 2);

  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;
}
// https://yukicoder.me/submissions/1046297
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