結果

問題 No.2911 位相の公理
コンテスト
ユーザー 👑 binap
提出日時 2024-10-04 22:02:47
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,173 bytes
コンパイル時間 4,023 ms
コンパイル使用メモリ 254,948 KB
最終ジャッジ日時 2025-02-24 15:17:48
ジャッジサーバーID
（参考情報） 		judge4 / judge2
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
other WA * 19 TLE * 1 -- * 4
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);}

int main(){
	int n, m;
	cin >> n >> m;
	vector<string> s(m);
	cin >> s;
	
	auto f = [&](string s){
		int res = 0;
		rep(i, n){
			res <<= 1;
			if(s[i] == '1') res++;
		}
		return res;
	};
	
	vector<int> memo(1 << n);
	rep(i, m){
		memo[f(s[i])] = 1;
		cout << f(s[i]) << "\n";
	}
	cout << memo;
	
	{
		string t;
		rep(k, n) t += '0';
		if(memo[f(t)] != 1){
			cout << "No\n";
			return 0;
		}
	}
	{
		string t;
		rep(k, n) t += '1';
		if(memo[f(t)] != 1){
			cout << "No\n";
			return 0;
		}
	}
	rep(i, m) rep(j, m){
		{
			string t;
			rep(k, n){
				if(s[i][k] == '1' and s[j][k] == '1') t += '1';
				else t += '0';
			}
			if(memo[f(t)] != 1){
				cout << "No\n";
				return 0;
			}
		}
		{
			string t;
			rep(k, n){
				if(s[i][k] == '1' or s[j][k] == '1') t += '1';
				else t += '0';
			}
			if(memo[f(t)] != 1){
				cout << "No\n";
				return 0;
			}
		}
	}
	cout << "Yes\n";
	
	return 0;
}
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