Submission #766595

#	Time	Username	Problem	Language	Result	Execution time	Memory
766595		Shreyan_Paliwal	Split the sequence (APIO14_sequence)	C++17	0 / 100	56 ms	131072 KiB

sequence

// #pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;

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;
}

#define int long long

const int maxn = 100000;
const int maxk = 200 + 1;
const int INF = 2000000000'000000;

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

struct F { int a, b; 
F() {}
F(int A, int B) { if (B < 0) B *= -1, A *= -1; int c = abs(__gcd(A, B)); a = A/c, b = B/c; }
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) { if (m==o.m) return F{o.b, 1}; return F{b-o.b,o.m-m}; }
};
struct CHT {
    deque<L> d;
    inline void clear() { d.clear(); }
    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];
    }
};

void solve() {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) { cin >> p[i]; p[i] += p[i-1]; }
    
    fill(dp[0], dp[0] + (maxn+1), 0);
    fill(dp[0] + (maxn+1), dp[0] + (maxn+1)*2, -INF);
    dp[0][0] = -INF;

    for (int i = 1; i <= k; i++) {
        CHT h;
        dp[0][i&1] = -INF;
        for (int j = 1; j <= n; j++) {
            h.add(L{p[j-1], dp[j-1][!(i&1)] - p[j-1]*p[j-1], j-1});
            L q = h.qry(p[j]);
            dp[j][i&1] = max(q.f(p[j]), -INF);
            pre[j][i] = q.i;
        }
    }

    cout << dp[n][k & 1] << endl;
    int x = pre[n][k--];
    while (k >= 0) {
        cout << x << ' ';
        x = pre[x][k--];
    }
}

signed main() {
    // freopen("main.in", "r", stdin);
    int t; 
    // cin >> t;
    t=1;
    while(t--) solve();
    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