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/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;
//Dijkstra->set
//set.find_by_order(x) x-position value
//set.order_of_key(x) number of strictly less elements don't need *set.??
#define N 100005
#define wr cout << "Continue debugging\n";
#define all(x) (x).begin(), (x).end()
#define ll long long int
#define pii pair <int, int>
#define pb push_back
#define ff first
#define ss second
int binary(vector<int> vec, int x, int n){
int l = 0, r = n, pos = 0;
while(l <= r){
int md = (l+r)/2;
if (vec[md] >= x) r = md - 1;
else{
pos = md;
l = md + 1;
}
}
return pos;
}
int LIS(int n, vector<int> a){
vector<int> vec(n+1, 1e9);
vec[0] = 0;
int mx = 0;
for (int i = 1; i <= n; i++){
int pos = binary(vec, a[i], n);
vec[pos+1] = min(vec[pos+1], a[i]);
mx = max(mx, pos+1);
}
return mx;
}
int main ()
{
ios::sync_with_stdio(false);
cin.tie(0);
int n, x;
cin >> n >> x;
vector<int> a(n+1);
for (int i = 1; i <= n; i++) cin >> a[i];
if (!x) return cout << LIS(n, a), 0;
assert(x);
int mx = 0;
for (int i = 1; i <= n; i++){
a[i] -= x;
mx = max(mx, LIS(n, a));
}
cout << mx;
}
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