This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC target ("avx2")
#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("unroll-loops")
#define f first
#define s second
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((int) (x).size())
#define pb push_back
#define mp make_pair
#define int long long
using namespace std;
using namespace __gnu_pbds;
template <typename T> using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T> inline bool umin(T &a, const T &b) { if(a > b) { a = b; return 1; } return 0; }
template <typename T> inline bool umax(T &a, const T &b) { if(a < b) { a = b; return 1; } return 0; }
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const ll mod = 998244353;
const ll base = 1e6 + 9;
const ll inf = 1e18;
const int MAX = 3e5 + 42;
const int LG = 20;
random_device rd;
mt19937 gen(rd());
uniform_int_distribution<ll> dis(1, inf);
void solve() {
int n, m, k;
cin >> n >> m >> k;
vector<string> a(n);
for(auto &i : a) {
cin >> i;
for(auto &j : i) j -= 'A';
}
if(n * n * m <= 1e7) {
for(int i = 0; i < n; i++) {
bool ok = 1;
for(int j = 0; j < n; j++) {
if(j == i) continue;
int cnt = 0;
for(int p = 0; p < m && cnt <= k; p++) {
cnt += a[i][p] != a[j][p];
}
if(cnt != k) {
ok = 0;
break;
}
}
if(ok) {
cout << i + 1 << '\n';
return;
}
}
assert(0);
}
const int BUBEN = 40;
vector<vector<vector<int>>> have(BUBEN, vector<vector<int>>(m, vector<int>(4)));
vector<int> cnt(BUBEN);
vector<int> idx(n);
for(int i = 0; i < n; i++) {
idx[i] = dis(gen) % BUBEN;
cnt[idx[i]]++;
for(int j = 0; j < m; j++) {
for(int c = 0; c < 4; c++) {
have[idx[i]][j][c]++;
}
}
}
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
have[idx[i]][j][a[i][j]]--;
}
}
vector<int> p(n); iota(all(p), 0); shuffle(all(p), gen);
for(auto i : p) {
int need = (cnt[idx[i]] - 1) * k;
for(int j = 0; j < m && need >= 0; j++) {
need -= have[idx[i]][j][a[i][j]];
}
if(need == 0) {
cout << i + 1 << '\n';
return;
}
}
assert(0);
}
signed main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int ttt = 1;
// cin >> ttt;
while(ttt--) {
solve();
}
}
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |