結果

問題 No.3082 Make Palindromic Multiple(Judge)
ユーザー 👑 binap
提出日時 2025-02-26 01:53:08
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 660 ms / 3,500 ms
コード長 5,570 bytes
コンパイル時間 5,006 ms
コンパイル使用メモリ 262,852 KB
実行使用メモリ 11,036 KB
最終ジャッジ日時 2025-04-16 13:11:05
合計ジャッジ時間 11,403 ms
ジャッジサーバーID
（参考情報） 		judge5 / 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(){
    while (true){
        int pr = mt() % 100000000 + 900000000;
        if (atcoder::internal::is_prime_constexpr(pr)){
            return pr;
        }
    }
}

using ull = unsigned long long;

using mint1 = atcoder::dynamic_modint<1>;
using mint2 = atcoder::dynamic_modint<2>;

struct mints{
	mint1 d1;
	mint2 d2;
	mints(long long val = 0) : d1(val), d2(val) {}
	mints(mint1 d1, mint2 d2) : d1(d1), d2(d2) {}
	mints& operator++() {
		d1++; d2++;
		return *this;
	}
	mints& operator--() {
		d1--; d2--;
		return *this;
	}
	mints& operator+=(const mints& a) {
		d1 += a.d1;	d2 += a.d2;
		return *this;
	}
	mints& operator-=(const mints& a) {
		d1 -= a.d1;	d2 -= a.d2;
		return *this;
	}
	mints& operator*=(const mints& a) {
		d1 *= a.d1;	d2 *= a.d2;
		return *this;
	}
	mints& operator/=(const mints& a) { return *this = *this * a.inv(); }
	mints operator+() const { return *this; }
	mints operator-() const { return mints(0) - *this; }
	
	mints pow(long long n) const {
		return mints(d1.pow(n), d2.pow(n));
	}
	mints inv() const {
		return mints(d1.inv(), d2.inv());
	}
	
	mints operator+(const mints& a) const {
		return mints(d1 + a.d1, d2 + a.d2);
	}
	mints operator-(const mints& a) const {
		return mints(d1 - a.d1, d2 - a.d2);
	}
	mints operator*(const mints& a) const {
		return mints(d1 * a.d1, d2 * a.d2);
	}
	mints operator/(const mints& a) const {
		return mints(d1 / a.d1, d2 / a.d2);
	}
	bool operator==(const mints& a) const {
		return d1 == a.d1 and d2 == a.d2;
	}
	bool operator!=(const mints& a) const {
		return d1 != a.d1 or d2 != a.d2;
	}
    ull val(){
        return (ull(d1.val()) << 32) + ull(d2.val());
    };
};
ostream& operator<<(ostream& os, const mints& a) {os << a.d1 << '&' << a.d2; return os;}

void random_mod(){
	const int p1 = random_prime();
	const int p2 = random_prime();
	mint1::set_mod(p1);
	mint2::set_mod(p2);
}

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];
}

template<typename T>
T primitive(){
	T g = 0;
	int p = T::mod();
	
	auto factor = [&](int x){
		assert(x >= 1);
		vector<int> res;
		for(int y = 1; y * y <= x; y++){
			if(x % y == 0){
				res.push_back(y);
				res.push_back(x / y);
			}
		}
		return res;
	};
	
	vector<int> factor_list = factor(p- 1);
	
	while(true){
		g = mt() % (p - 1) + 1;
		bool success = true;
		for(int y : factor_list){
			if(y == p - 1) continue;
			T h = g.pow(y);
			if(h == T(1)) success = false;
		}
		if(success) break;
	}
	return g;
}

bool judge_palindrome(){
	
	
	using F = mints;
	vector<F> hash(2, 0);
	F base(primitive<mint1>(), primitive<mint2>());
	
	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;
}
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