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;
const int INF = 1e9;
int main()
{
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int N,X; cin >> N >> X;
vector<int> arr(N);
for(int i = 0;i<N;++i){
cin >> arr[i];
}
vector<int> dp(N+1,INF), L(N,1);
dp[0] = -INF; int ans = 1;
for(int i = 0;i<N;++i){
int ind = lower_bound(dp.begin(),dp.end(),arr[i])-dp.begin(); //first that is >= arr[i]
dp[ind] = arr[i];
L[i] = ind+1;
ans = max(ans,L[i]);
}
dp = vector<int>(N+1,INF);
dp[0] = -INF;
for(int i = N-1;i>=0;--i){ //1 to i have been decreased by X and we want to find the LDS from N-1 to i -> this is equivalent to increasing i by X
int ind = lower_bound(dp.begin(),dp.end(),-(arr[i]+X))-dp.begin(); //we want to find the negative version so that we can still lower bound
ans = max(ans,L[i]+ind); //no -1 because in ind, i was not counted
ind = lower_bound(dp.begin(),dp.end(),-arr[i])-dp.begin();
dp[ind] = -arr[i];
}
cout << ans << "\n";
return 0;
}
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