This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
/** MIT License Copyright (c) 2018 Vasilyev Daniil **/
#include <bits/stdc++.h>
using namespace std;
template<typename T> using v = vector<T>;
typedef long long longlong;
#define int longlong
typedef long double ld;
typedef string str;
typedef vector<int> vint;
#define rep(a, l, r) for(int a = (l); a < (r); a++)
#define pb push_back
#define fs first
#define sc second
#define sz(a) ((int) a.size())
const long long inf = 4611686018427387903; //2^62 - 1
#if 0 //FileIO
const string fileName = "";
ifstream fin ((fileName == "" ? "input.txt" : fileName + ".in" ));
ofstream fout((fileName == "" ? "output.txt" : fileName + ".out"));
#define get fin>>
#define put fout<<
#else
#define get cin>>
#define put cout<<
#endif
#define eol put endl
void read() {} template<typename Arg,typename... Args> void read (Arg& arg,Args&... args){get (arg) ;read(args...) ;}
void print(){} template<typename Arg,typename... Args> void print(Arg arg,Args... args){put (arg)<<" ";print(args...);}
int getInt(){int a; get a; return a;}
//code goes here
inline int lsb(int a) { return a & -a; }
const int N = 262144;
int segTree[N * 2];
void update(int p, int val, int cur = 1, int ll = 1, int rr = N) {
if (ll == rr) {
segTree[cur] = max(segTree[cur], val);
return;
}
int mid = (ll + rr) / 2;
if (p <= mid)
update(p, val, cur * 2, ll, mid);
else
update(p, val, cur * 2 + 1, mid + 1, rr);
segTree[cur] = max(segTree[cur * 2], segTree[cur * 2 + 1]);
}
int query(int l, int r, int cur = 1, int ll = 1, int rr = N) {
if (l > r)
return 0;
else if (l == ll && r == rr)
return segTree[cur];
else {
int mid = (ll + rr) / 2;
int res = max(query(l, min(r, mid), cur * 2, ll, mid),
query(max(l, mid + 1), r, cur * 2 + 1, mid + 1, rr));
return res;
}
}
void run() {
memset(segTree, 0, sizeof segTree);
int n, k;
read(n, k);
int a[n];
set<int> d;
rep(i, 0, n) {
get a[i];
d.insert(a[i]);
d.insert(a[i] - k);
}
map<int, int> inv;
int p = 2;
for (int j : d)
inv[j] = p++;
int v[n];
rep(i, 0, n)
v[i] = inf;
int dp[n][2];
rep(i, 0, n) {
dp[i][0] = (lower_bound(v, v + n, a[i]) - v) + 1;
v[dp[i][0] - 1] = a[i];
dp[i][1] = query(1, inv[a[i]] - 1) + 1;
update(inv[a[i] - k], dp[i][0]);
update(inv[a[i]], dp[i][1]);
}
int ans = 0;
rep(i, 0, n)
ans = max(ans, dp[i][1]);
put ans;
}
signed main() {srand(time(0)); ios::sync_with_stdio(0); cin.tie(0); put fixed << setprecision(12); run(); return 0;}
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