#include "bits/stdc++.h"
#include <type_traits>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
// ============ Macros starts here ============
int recur_depth = 0;
#ifdef DEBUG
#define dbg(x) {++recur_depth; auto x_=x; --recur_depth; cerr<<string(recur_depth, '\t')<<"\e[91m"<<__func__<<":"<<__LINE__<<"\t"<<#x<<" = "<<x_<<"\e[39m"<<endl;}
#else
#define dbg(x)
#endif // DEBUG
template<typename Ostream, typename Cont>
typename enable_if<is_same<Ostream, ostream>::value, Ostream&>::type operator<<(Ostream& os, const Cont& v) {
os << "[";
for (auto& x : v) { os << x << ", "; }
return os << "]";
}
template<typename Ostream, typename ...Ts>
Ostream& operator<<(Ostream& os, const pair<Ts...>& p) {
return os << "{" << p.first << ", " << p.second << "}";
}
#define readFast \
ios_base::sync_with_stdio(false); \
cin.tie(0); \
cout.tie(0);
#ifdef LOCAL
#define read() ifstream fin("date.in.txt")
#else
#define read() readFast
#endif // LOCAL
// ============ Macros ends here ============
#define fin cin
#define ll long long
#define sz(x) (int)(x).size()
#define all(v) v.begin(), v.end()
#define output(x) (((int)(x) && cout << "YES\n") || cout << "NO\n")
#define LSB(x) (x & (-x))
#define test cout << "WORKS\n";
const int N = 2e5 + 15;
const int MOD = 998244353;
int n, k;
int a[N];
unordered_map<int, unordered_map<int, int>> cache;
int solve(int j, ll hAct) {
if (j == n + 1) {
return 0;
}
if (cache.find(j) != cache.end() && cache[j].find(hAct) != cache[j].end()) {
return cache[j][hAct];
}
int ans = min(solve(j + 1, hAct + k), solve(j + 1, max(hAct - k, 0ll))) + 1;
if (hAct + k >= a[j]) {
ans = min(ans, solve(j + 1, a[j]));
}
return cache[j][hAct] = ans;
}
int main() {
read();
fin >> n >> k;
for (int i = 1; i <= n; ++i) {
fin >> a[i];
}
// fill(cache, cache + N, -1);
cout << solve(1, 0);
// vector<vector<int>> dp(n + 1, vector<int>(3, INT_MAX));
// dp[0][0] = 0;
// for (int i = 1; i <= n; ++i) {
// // 0 - no change
// if (a[i - 1] + k >= a[i]) {
// dp[i][0] = min({ dp[i - 1][0], dp[i - 1][1], dp[i - 1][2] });
// }
// // 1 - increase
// else {
// dp[i][1]
// }
// }
return 0;
} /*stuff you should look for !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
* test the solution with the given example
* int overflow, array bounds, matrix bounds
* special cases (n=1?)
* do smth instead of nothing and stay organized
* WRITE STUFF DOWN
* DON'T GET STUCK ON ONE APPROACH
~Benq~*/
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |