Submission #107914

#	Time	Username	Problem	Language	Result	Execution time	Memory
107914		SamAnd	Split the sequence (APIO14_sequence)	C++17	71 / 100	1053 ms	132096 KiB

sequence

#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
const long long INF = 1000000009000000009;

int n, m;
int a[N];
long long p[N];

long long dp[N][202];
int pp[N][202];

vector<int> v[202];
vector<long double> x[202];

long double hat(long long b1, long long k1, long long b2, long long k2)
{
    return 1.0 * (b2 - b1 + 0.0) / (k1 - k2 + 0.0);
}

void ave(int k, int j)
{
    while (!v[k].empty() && p[v[k].back()] == p[j])
    {
        v[k].pop_back();
        if (!x[k].empty())
            x[k].pop_back();
    }
    while (v[k].size() >= 2)
    {
        long double x1 = hat(dp[v[k].back()][k] - p[v[k].back()] * p[v[k].back()], p[v[k].back()], dp[j][k] - p[j] * p[j], p[j]);
        if (x1 > x[k].back())
            break;
        v[k].pop_back();
        x[k].pop_back();
    }
    if (v[k].empty())
    {
        v[k].push_back(j);
        return;
    }
    long double x1 = hat(dp[v[k].back()][k] - p[v[k].back()] * p[v[k].back()], p[v[k].back()], dp[j][k] - p[j] * p[j], p[j]);
    x[k].push_back(x1);
    v[k].push_back(j);
}

int byn(int i, int k)
{
    int l = 0, r = x[k - 1].size() - 1;
    while ((r - l) > 3)
    {
        int m = (l + r) / 2;
        if (p[i] >= x[k - 1][m])
            l = m;
        else
            r = m;
    }
    for (int m = r; m >= l; --m)
    {
        if (p[i] >= x[k - 1][m])
            return v[k - 1][m + 1];
    }
    return v[k - 1][0];
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; ++i)
        cin >> a[i];
    for (int i = 1; i <= n; ++i)
        p[i] = p[i - 1] + a[i];
    for (int i = 1; i <= n; ++i)
    {
        for (int k = 1; k <= min(m, i - 1); ++k)
        {
            int j = byn(i, k);
            dp[i][k] = dp[j][k - 1] - p[j] * p[j] + p[j] * p[i];
            pp[i][k] = j;
        }
        for (int k = 0; k <= min(m, i - 1); ++k)
        {
            ave(k, i);
        }
    }
    cout << dp[n][m] << endl;
    int i = n;
    for (int k = m; k > 0; --k)
    {
        cout << pp[i][k] << ' ';
        i = pp[i][k];
    }
    cout << endl;
    return 0;
}

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