Submission #64520

#	Time	Username	Problem	Language	Result	Execution time	Memory
64520		forestryks	Studentsko (COCI14_studentsko)	C++14	100 / 100	75 ms	1136 KiB

studentsko

///////////////////////////////////////////////////////////////////////////////////////////////
#include <bits/stdc++.h>
using namespace std;

// #define int long long

#define FAST_IO ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define FILE_IO(x) freopen((string(x) + ".in").c_str(), "r", stdin); freopen((string(x) + ".out").c_str(), "w", stdout)
#define f first
#define s second
#define x1 x1qwer
#define y1 y1qwer
#define right right123
#define left left123
#define foreach(it, v) for (auto it : v)
#define rep(it, n) for (int it = 0; it < n; ++it)
#define forin(it, l, r) for (int it = l; it < r; ++it)
#define all(x) x.begin(), x.end()

typedef long long ll;
typedef unsigned long long ull;
typedef double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const double DINF = numeric_limits<double>::infinity();
const ll MOD = 1e9 + 7;
const double EPS = 1e-7;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
mt19937 mmtw_(MOD);
uniform_int_distribution<ll> rd_;
ll randomll() { return rd_(mmtw_);}
ll rnd(ll x, ll y) { return rd_(mmtw_) % (y - x + 1) + x; }
template <class T> T fact(T n) { if (n == 1) return 1; return n * fact(n - 1); }
////////////////////////////////////////////////////////////////////////////////////////////////

const int MAXN = 5e4 + 5;

int n, k;
int a[MAXN];
int dp[MAXN];

void compress() {
    vector<pii> b(n);
    rep(i, n) {
        b[i] = {a[i], i};
    }
    sort(b.begin(), b.end());

    for (int i = 0; i < n; ++i) {
        a[b[i].s] = i / k;
    }
}

int main() {
    FAST_IO;
    cin >> n >> k;
    rep(i, n) {
        cin >> a[i];
    }

    compress();

    fill(dp, dp + n, 1);
    for (int i = 1; i < n; ++i) {
        for (int j = 0; j < i; ++j) {
            if (a[j] <= a[i] && dp[j] + 1 > dp[i]) {
                dp[i] = dp[j] + 1;
            }
        }
    }

    cout << n - *max_element(dp, dp + n) << endl;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10
• Subtask 8: 10 / 10
• Subtask 9: 10 / 10
• Subtask 10: 10 / 10