Submission #330404

#	Time	Username	Problem	Language	Result	Execution time	Memory
330404		534351	Holding (COCI20_holding)	C++17	110 / 110	167 ms	4900 KiB

holding

#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>

using namespace std;
// using namespace __gnu_pbds;

// template<class T>
// using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template<class T, class U>
void ckmin(T &a, U b)
{
    if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
    if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

const int INF = 1e9 + 7;
const int MAXN = 113;

int N, L, R, K;
int arr[MAXN];
int dp[2][MAXN][MAXN * MAXN];
int ans;

int32_t main()
{
    cout << fixed << setprecision(12);
    cerr << fixed << setprecision(4);
    ios_base::sync_with_stdio(false); cin.tie(0);
    cin >> N >> L >> R >> K; L--; R--;
    FOR(i, 0, N)
    {
        cin >> arr[i];
    }
    FOR(j, 0, R - L + 2)
    {
        FOR(k, 0, K + 1)
        {
            dp[1][j][k] = INF;
        }
    }
    dp[1][0][0] = 0;
    FOR(i, 0, N) //dp[pos][# numbers selected][# moves]
    {
        FOR(j, 0, R - L + 2)
        {
            FOR(k, 0, K + 1)
            {
                dp[0][j][k] = dp[1][j][k];
                dp[1][j][k] = INF;
            }
        }
        FOR(j, 0, R - L + 2)
        {
            FOR(k, 0, K + 1)
            {
                //select another number.
                ckmin(dp[1][j + 1][k + abs(i - (L + j))], dp[0][j][k] + arr[i]);
                //don't select another number.
                ckmin(dp[1][j][k], dp[0][j][k]);
            }
        }
    }
    ans = INF;
    FOR(k, 0, K + 1)
    {
        ckmin(ans, dp[1][R - L + 1][k]);
    }
    cout << ans << '\n';
    return 0;
}

• Subtask 1: 22 / 22
• Subtask 2: 33 / 33
• Subtask 3: 33 / 33
• Subtask 4: 22 / 22