Submission #1162108

#TimeUsernameProblemLanguageResultExecution timeMemory
1162108NonozeGenetics (BOI18_genetics)C++20
0 / 100
11 ms11080 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; pos.push(a[0]); for (int i=1; 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 (nb[a[i%n]]==1) pos.push(a[i%n]); // dbg(pos); } for (int i=0; i<n; i++) { if (a[i]==pos.front()) quit(i+1); } quit(-1); }
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