Submission #1253805

#	Time	Username	Problem	Language	Result	Execution time	Memory
1253805		abdullah_	Feast (NOI19_feast)	C++20	4 / 100	166 ms	9816 KiB

feast

#include <bits/stdc++.h>

using namespace std;

#define int long long
typedef pair<int, int> pp;
#define all(a) (a).begin(), (a).end()
#define what_is(x) cerr << #x << " is " << x << '\n';
#define print(a) for (const auto &x : (a)) cout << x << ' '; cout << '\n';
const int N = 2e5 + 7;
const int MOD = 1e9 + 7;
const int INF = 1e15;


void solve() {
    int n, k; cin >> n >> k;
    vector<int> a(n + 1);
    vector<int> pre(n + 1);

    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        pre[i] = pre[i - 1] + a[i];
    }

    // The maximum sum when taking 3 segments and some segments may be big

    // we will add some random value into the

    auto check = [&] (int val)->pp  {
        vector<int> dp(n + 1), g(n + 1);
        int mxDp = 0, mxG = 0;
        for (int i = 1; i <= n; ++i) {
            int sum = pre[i] + mxDp - val;
            dp[i] = dp[i - 1];
            g[i] = g[i - 1];
            if (sum > dp[i]) {
                dp[i] = sum;
                g[i] = mxG + 1;
            }

            if (dp[i] - pre[i - 1] > mxDp) {
                mxDp = dp[i] - pre[i];
                mxG = g[i];
            } else if (dp[i] - pre[i - 1] == mxDp) {
                mxG = min(mxG, g[i]);
            }
        }

        return {dp[n], g[n]};
    };


    int l = 1, r = 1e18, pos = -1;

    int res = 0;



    while (l <= r) {
        int mid = (l + r) / 2;
        auto [ans, count]  = check(mid);
        // cout << ans << ' ' << count << '\n';
        if (count <= k) {
            res = max(res, ans + mid * count);
            r = mid - 1;
        } else {
            l = mid + 1;
        }
    }

    cout << res << endl;

}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

// #ifndef ONLINE_JUDGE
//     freopen("input.txt", "r", stdin);
//     freopen("output.txt", "w", stdout);
// #endif

    int tt = 1;
    // cin >> tt;
    while (tt--) {
        solve();
    }

    return 0;
}

• Subtask 1: 4 / 4
• Subtask 2: 0 / 8
• Subtask 3: 0 / 18
• Subtask 4: 0 / 10
• Subtask 5: 0 / 11
• Subtask 6: 0 / 20
• Subtask 7: 0 / 29