제출 #1162113

#제출 시각아이디문제언어결과실행 시간메모리
1162113NonozeGenetics (BOI18_genetics)C++20
0 / 100
10 ms8520 KiB
/*
*	Author: Nonoze
*	Created: Wednesday 05/03/2025
*/
#include <bits/stdc++.h>
using namespace std;

#ifndef IN_LOCAL
	#define dbg(...)
#endif

// #define cout cerr << "OUT: "
#define endl '\n'
#define endlfl '\n' << flush
#define quit(x) return (void)(cout << x << endl)

template<typename T> void read(T& x) { cin >> x;}
template<typename T1, typename T2> void read(pair<T1, T2>& p) { read(p.first), read(p.second);}
template<typename T> void read(vector<T>& v) { for (auto& x : v) read(x); }
template<typename T1, typename T2> void read(T1& x, T2& y) { read(x), read(y); }
template<typename T1, typename T2, typename T3> void read(T1& x, T2& y, T3& z) { read(x), read(y), read(z); }
template<typename T1, typename T2, typename T3, typename T4> void read(T1& x, T2& y, T3& z, T4& zz) { read(x), read(y), read(z), read(zz); }
template<typename T> void print(vector<T>& v) { for (auto& x : v) cout << x << ' '; cout << endl; }

#define sz(x) (int)(x.size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define make_unique(v) sort(all(v)), v.erase(unique(all(v)), (v).end())
#define pb push_back
#define mp(a, b) make_pair(a, b)
#define fi first
#define se second
#define cmin(a, b) a = min(a, b)
#define cmax(a, b) a = max(a, b)
#define YES cout << "YES" << endl
#define NO cout << "NO" << endl
#define QYES quit("YES")
#define QNO quit("NO")

#define int long long
#define double long double
const int inf = numeric_limits<int>::max() / 4;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 1e9+7, LOG=20;



void solve();

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int tt=1;
	// cin >> tt;
	while(tt--) solve();
	return 0;
}




int n, k, m, q;
vector<string> a;


void solve() {
	read(n, m, k);
	a.clear(), a.resize(n); read(a);
	map<string, int> nb;
	for (auto u: a) nb[u]++;
	queue<string> pos;
	for (int i=0; i<2*n; i++) {
		while (!pos.empty()) {
			int cnt=0;
			for (int j=0; j<m; j++) cnt+=(pos.front()[j]!=a[i%n][j]);
			if (cnt==k || cnt==0) break;
			pos.pop();
		}
		if (i<n && nb[a[i]]==1) pos.push(a[i]);
		// dbg(pos);
	}
	assert(!pos.empty());
	for (int i=0; i<n; i++) {
		if (a[i]==pos.front()) quit(i+1);
	}
	assert(0);
}
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