Submission #765940

#	Time	Username	Problem	Language	Result	Execution time	Memory
765940		Shreyan_Paliwal	Split the sequence (APIO14_sequence)	C++17	0 / 100	46 ms	131072 KiB

sequence

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

#define int long long

template<typename T>
string tostr(const T& value) {
    ostringstream oss;
    oss << value;
    return oss.str();
}

template<typename... Args>
string fstr(const string& format, Args... args) {
    string result = format;
    size_t pos = 0;
    size_t argIndex = 0;

    auto replaceArg = [&](const auto& arg) {
        pos = result.find("{}", pos);
        if (pos != string::npos) {
            result.replace(pos, 2, tostr(arg));
            ++argIndex;
        }
    };

    (replaceArg(args), ...);

    return result;
}

const int maxn = 100000;
const int maxk = 200;
const int INF = 2000000000000000;

int n, k;
int v[maxn+1];
int dp[maxn+1][2];
int p[maxn+1][maxk+1];

struct F {
    int a, b;
    bool operator<(F o) const { return a*o.b<b*o.a; }
};
struct L {
    int m, b, i;
    int f(int x) { return m*x+b; }
    F operator^(L o) { return F{b-o.b,o.m-m}; }
};
struct CHT {
    deque<L> d;
    void add(L l) {
        while (d.size() >= 2 && (d.end()[-2]^l) < (d.end()[-2]^d.back())) d.pop_back();
        d.push_back(l);
    }
    L qry(int x) {
        while (d.size() >= 2 && d[1].f(x) > d[0].f(x)) d.pop_front();
        return d[0];
    }
};

signed main() {
    cin.tie(nullptr) -> ios::sync_with_stdio(false);
    // freopen("main.in", "r", stdin);

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

    fill(dp[0], dp[0] + 2*(maxn+1), -INF);
    // dp[0][0 & 1] = 0;

    for (int i = 1; i <= k; i++) {
        CHT h;
        for (int j = 1; j <= n; j++) {
            h.add(L{v[j-1], dp[j-1][(i-1)&1] - v[j-1] * v[j-1], j-1});
            L l = h.qry(v[j]);
            dp[j][i&1] = max(l.f(v[j]), -INF);
            p[j][i] = l.i;
            // cout << fstr("{} {} {} {} {}", i, j, i&1, j&1, (i-1)&1) << endl;
            // cout << fstr("(i {}) (j {}) (dp[j-1][(i-1)&1] {}) (dp[j][i&1] {})", i, j, dp[j-1][(i-1)&1], dp[j][i&1]) << endl;
        }
        // cout << endl;
    }

    int ans = -INF;
    for (int i = 0; i <= n; i++)
        ans = max(ans, (v[n] - v[i]) * v[i] + dp[i][k & 1]);

    cout << ans << endl;

    for (int i = 0; i <= n; i++) 
        if ((v[n] - v[i]) * v[i] + dp[i][k&1] == ans) {
            while (k) {
                cout << i << ' ';
                i = p[i][k--];
            }
        }


    return 0;
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 11
• Subtask 3: 0 / 11
• Subtask 4: 0 / 17
• Subtask 5: 0 / 21
• Subtask 6: 0 / 29