| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 1302315 | tripadva | 삶의 질 (IOI10_quality) | C++20 | 0 ms | 0 KiB |
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }
void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }
template <class T> using indexed_set = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#ifdef LOCAL
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif
#define ar array
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
#define sza(x) ((int)x.size())
#define all(a) (a).begin(), (a).end()
#define sort(a) sort(all(a))
#define mp make_pair
#define pb push_back
#define f first
#define s second
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template<class T> using pq = priority_queue<T>;
template<class T> using pqg = priority_queue<T, vector<T>, greater<T>>;
const int MAX_N = 1e5 + 5;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;
const ld EPS = 1e-9;
const char nl = '\n';
const long long int ll_MAX=9223372036854775807;
const string deb = "------------";
//[x,y],[a,b]
int stepenDvojke(int n) {
int s = 1;
while (s < n)
s <<= 1;
return s;
}
int a[3001][3001];
int A[3001][3001];
int h,w,N,M;
bool check(int k){
memset(A, 0, sizeof(A));
for(int i=1;i<=N;i++)
for(int j=1;j<=M;j++)
A[i][j] = (a[i-1][j-1] <= k ? 1 : -1);
for(int i =1;i<=N;i++){
for(int j = 1;j<=M;j++){
A[i][j] = A[i][j] + A[i-1][j] + A[i][j-1] - A[i-1][j-1];
}
}
for(int i =h;i<=N;i++){
for(int j = w;j<=M;j++){
if(A[i][j]-A[i-h][j]-A[i][j-w]+A[i-h][j-w]>=0)return true;
}
}
return false;
}
void solve() {
cin>>N>>M>>h>>w;
for(int i =0;i<N;i++){
for(int j = 0;j<M;j++)cin>>a[i][j];
}
int l = 1,r= N*M,rez = N*M;
while(l<=r){
int mid = (l+r)/2;
if(check(mid))r=mid-1,rez=mid;
else l=mid+1;
}
cout<<rez;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int tc = 1;
// cin >> tc;
for (int t = 1; t <= tc; t++) {
// cout << "Case #" << t << ": ";
solve();
}
}