Submission #766607

#	Time	Username	Problem	Language	Result	Execution time	Memory
766607		Shreyan_Paliwal	Split the sequence (APIO14_sequence)	C++17	100 / 100	527 ms	82908 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 ll long long

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

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

struct F {
    ll a, b; 
    F() {}
    F(ll A, ll B) { if (B < 0) B *= -1, A *= -1; a = A, b = B; }
    bool operator<(F o) const { return a*o.b<b*o.a; } 
    bool operator<=(F o) const { return a*o.b<=b*o.a; }
};
struct L {
    ll m, b; int i;
    ll f(ll x) { return m*x+b; }
    F operator*(L o) { if (m==o.m) return F{o.b, 1}; return F{o.b-b,m-o.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(ll x) {
        while (d.size() >= 2 && d[1].f(x) >= d[0].f(x)) d.pop_front();
        return d[0];
    }
};

signed main() {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) { cin >> p[i]; p[i] += p[i-1]; }
    
    fill(dp[0] + (maxn+1), dp[0] + (maxn+1)*2, -INF);
    fill(dp[0], dp[0] + (maxn+1), 0);
    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] << '\n';
    int x = pre[n][k--];
    while (k >= 0) {
        cout << x << ' ';
        x = pre[x][k--];
    }

    return 0;
}

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