Submission #1095230

#	Time	Username	Problem	Language	Result	Execution time	Memory
1095230		LonlyR	Split the sequence (APIO14_sequence)	C++17	78 / 100	696 ms	96696 KiB

sequence

#include<bits/stdc++.h>
#define int long long
#define ld long double
#define sz(u) (int)u.size()
using namespace std;
const int maxn = 1e5 + 5;
int n, k;
int a[maxn], dp[2][205], pf[maxn];
signed pre[maxn][205];

struct iii{
    int a, b;
    ld x;
    int p;
};

inline ld ori(iii a, iii b)
{
    int u = a.a, v = a.b;
    int x = b.a, y = b.b;
    if (u == x) return (ld)1e18;
    return (ld)(y - v) / (ld)(u - x);
}

void trace(int n, int k)
{
    if (pre[n][k] == 0) return;
    trace(pre[n][k], k - 1);
    assert(pre[n][k] != n);
    cout << pre[n][k] << " ";
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
//    freopen("test.inp", "r", stdin);
//    freopen("test.out", "w", stdout);
    cin >> n >> k;
    assert(k > 0);
    for (int i = 1; i <= n; i++)
        cin >> a[i], pf[i] = pf[i - 1] + a[i];
    /// dp[i][j] = dp[k - 1][j - 1] + (pf[i] - pf[k - 1]) * (pf[n] - pf[i])
    /// dp[i][j] = pf[k - 1] * pf[i] + dp[k - 1][j - 1] + pf[n] * pf[i] - pf[i] ^ 2 - pf[k - 1] * pf[n]
    vector<iii> b[2];
    int be = 0, cu = 1;
    b[0].push_back({0, 0, -1e18, 0});
    k++;
    for (int j = 0; j < k; j++)
    {
        b[cu].clear();
        for (int i = 1 + j, pt = -1; i <= n; i++)
        {
            pt = min(pt, sz(b[be]) - 2);
            while (pt + 1 < sz(b[be]) && (ld)pf[i] >= b[be][pt + 1].x) pt++;
            dp[cu][j + 1] = b[be][pt].a * pf[i] + b[be][pt].b - pf[i] * pf[i] + pf[n] * pf[i];
            pre[i][j + 1] = b[be][pt].p;
            iii now;
            now.a = pf[i];
            now.b = dp[cu][j + 1] - pf[i] * pf[n];
            now.p = i;
            now.x = -1e18;
            while (sz(b[cu]) >= 2 && ori(b[cu][sz(b[cu]) - 2], now) <= ori(b[cu][sz(b[cu]) - 2], b[cu][sz(b[cu]) - 1]))
                b[cu].pop_back();
            if (b[cu].size()) now.x = ori(now, b[cu].back());
            b[cu].emplace_back(now);
        }
        swap(cu, be);
    }
    cout << dp[be][k] << "\n";
    trace(n, k);
}

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