이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
//#pragma GCC optimize("Ofast")
//#pragma GCC target ("avx2")
#define ll long long
#define ff first
#define ss second
#define int ll
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define pb push_back
#define pii pair <int, int>
#define pdd pair <double, double>
#define vi vector <int>
using namespace std;
using namespace __gnu_pbds;
template <class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
const int inf = 1e15;
const int mod = 1e9+7;
const int N = 2e5+100;
int n, a, b;
vector <int> v;
bool check(int k){
vector <int> dp(n+100);
dp[0] = 1;
for (int i = 0, j = 1; i < n; ++i){
if (i > 1) dp[i] += dp[i-1];
if (dp[i] <= 0) continue;
j = max(j, i + 1);
while (j < n && v[j] - v[i] <= k) ++j;
if (i + a <= min(i + b, j)) {
dp[i+a]++;
dp[min(i + b, j) + 1]--;
}
}
dp[n] += dp[n-1];
return dp[n] > 0;
}
void solve(){
cin >> n >> a >> b; v.resize(n);
for (int i = 0; i < n; ++i) cin >> v[i];
sort(all(v));
int l = -1, r = 1e9;
for (; r - l > 1; ){
int m = (l + r) / 2;
if (check(m)) r = m;
else l = m;
}
cout << r;
}
signed main(){
ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
int tt = 1;
while (tt--){
solve();
cout << '\n';
}
}
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