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>
using namespace std;
#define ll long long int
#define F first
#define S second
#define pb push_back
#define md ((b + e) >> 1)
#define lc (ind << 1)
#define rc (lc | 1)
const ll N = 2e5 + 10;
const ll LOG = 50;
const ll MOD = 1e9 + 7;
const ll INF = 1e9 + 10;
ll n, add, a[N], L[N], R[N], ans;
pair <ll, ll> mxl[N], mxr[N];
int main() {
cin >> n >> add;
for (ll i = 0; i < n; i++){
cin >> a[i];
}
ll lenl = 0;
for (ll i = 0; i < n; i++){
ll p = lower_bound(L, L + lenl, a[i]) - L;
L[p] = a[i];
lenl = max(lenl, p + 1ll);
if (i == 0) mxl[i] = {p + 1ll, -a[0]};
else mxl[i] = max(mxl[i - 1], {p + 1ll, -a[i]});
}
ll lenr = 0;
for (ll i = n - 1; i >= 0; i--){
ll p = lower_bound(R, R + lenr, -(a[i] + add)) - R;
R[p] = -(a[i] + add);
lenr = max(lenr, p + 1ll);
if (i == n - 1) mxr[i] = {p + 1ll, a[i] + add};
else mxr[i] = max(mxr[i + 1], {p + 1ll, a[i] + add});
}
ans = max({ans, lenl, lenr});
for (int i = 1; i < n; i++){
if (-mxl[i - 1].S < mxr[i].S){
ans = max(ans, mxl[i - 1].F + mxr[i].F);
cout << "i = " << i << ' ' << mxl[i - 1].F << ' ' << mxr[i].F << '\n';
}
}
cout << ans << '\n';
return 0;
}
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