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>
typedef long long ll;
using namespace std;
#define f first
#define s second
#define pb push_back
#define ep emplace
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define uniquev(v) sort(all(v)), (v).resize(unique(all(v)) - (v).begin())
#define mem(f,x) memset(f , x , sizeof(f))
#define sz(x) (int)(x).size()
#define __lcm(a, b) (1ll * ((a) / __gcd((a), (b))) * (b))
#define mxx *max_element
#define mnn *min_element
#define cntbit(x) __builtin_popcountll(x)
#define len(x) (int)(x.length())
const int N = 2e5 + 100;
pair <int, int> A[N];
int B[N], pr[N], ans[N];
ll solve() {
int n;
cin >> n;
for (int i = 1; i <= n + 1; i++) {
cin >> A[i].f;
A[i].s = i;
}
for (int i = 1; i <= n; i++) {
cin >> B[i];
}
sort(B + 1, B + 1 + n);
sort(A + 1, A + 2 + n);
for (int i = 1; i <= n; i++) {
pr[i] = max(pr[i - 1], A[i].f - B[i]);
}
int cur = 0;
for (int i = n + 1; i >= 1; i--) {
ans[A[i].s] = max(cur, pr[i - 1]);
cur = max(cur, A[i].f - B[i - 1]);
}
for (int i = 1; i <= n + 1; i++)
cout << ans[i] << " ";
return 0;
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t = 1;
//cin >> t;
while (t--) {
solve();
//cout << solve() << '\n';
}
return 0;
}
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