Submission #1280987

#TimeUsernameProblemLanguageResultExecution timeMemory
1280987ionut27Rabbit Carrot (LMIO19_triusis)C++20
100 / 100
21 ms3520 KiB
#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];

int solve(int j, ll hAct) {
    if (j == n + 1) {
        return 0;
    }

    int ans = solve(j + 1, hAct + k) + 1;

    if (hAct + k >= a[j]) {
        ans = min(ans, solve(j + 1, a[j]));
    }
    return ans;
}

int main() {
    read();

    fin >> n >> k;
    for (int i = 1; i <= n; ++i) {
        fin >> a[i];
    }
    // fill(cache, cache + N, -1);
    dbg(solve(1, 0));

    vector<int> b;
    for (int i = 1; i <= n; ++i) {
        if (i * k >= a[i]) {
            b.push_back(i * k - a[i]);
        }
    }
    vector<int> lis;
    for (int i : b) {
        int p = upper_bound(all(lis), i) - lis.begin();
        if (p == sz(lis)) {
            lis.push_back(i);
        } else {
            lis[p] = min(lis[p], i);
        }
    }
    dbg(lis);
    cout << n - sz(lis) << '\n';


    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 timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...