Submission #23313

#	Time	Username	Problem	Language	Result	Execution time	Memory
23313		Nurlykhan	Split the sequence (APIO14_sequence)	C++14	39 / 100	2000 ms	25924 KiB

sequence

#include <bits/stdc++.h>

#define pii pair<int, int>
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define ll long long 
#define ld long double
#define sz(v) int(v.size())
#define all(v) v.begin(), v.end()

using namespace std;

const int N = (int) 1e4 + 7;
const int M = (int) 2e6 + 7;
const int K = 202;
const ll LINF = (ll) 1e18;
const int INF = (int) 1e9 + 7;
const double EPS = (double) 1e-9;
ll dp[N][K], a[N];
int p[N][K];
int n, k;

ll sum(int l, int r) {
    return a[r] - a[l - 1];
}

ld inter(int i, int x, int y) {
    return (dp[y][i] - dp[x][i]) * 1.0 / (a[y] - a[x]);
}

vector<int> st;

ll f(int i, int j, ll x) {
    return x *  1ll * a[i] + dp[i][j];
}

int get(int j, int x) {
    int id = 0;
    for (auto it : st) {
        if (f(it, j, x) > f(id, j, x)) 
            id = it;
    }
    return id;
}

int main() {
    #define fn "balls"
    #ifdef witch
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #else
//        freopen(fn".in", "r", stdin);
//        freopen(fn".out", "w", stdout);
    #endif
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        a[i] += a[i - 1];
    }
    for (int i = 1; i <= n; i++) {
        dp[i][1] = sum(1, i) * sum(i + 1, n);
    }
    for (int i = 2; i <= k; i++) {
        st.clear();
        for (int j = i - 1; j <= n; j++) {
            while (sz(st) > 1 && inter(i - 1, st[sz(st) - 1], j) - inter(i - 1, st[sz(st) - 2], st[sz(st) - 1]) > -EPS) {
                st.pop_back();
            }
            /*
            y[t] = (a[j] - a[t]) * (a[n] - a[j]) + dp[t][i - 1]
            a[j]*a[n]-a[j]*a[j]-a[t]*a[n]+a[t]*a[j]
            
            a[t]*(-a[n]+a[j])+dp[t][i-1]+a[j]*(a[n]-a[j])

            k=a[j]-a[t]
            */
            if (!st.empty()) {
                int t = get(i - 1, -sum(j + 1, n));
                dp[j][i] = dp[t][i - 1] + sum(j + 1, n) * sum(t + 1, j);
                p[j][i] = t;
            }
            st.pb(j);
        }
    }
    int j = 1;
    for (int i = 1; i <= n; i++) {
        if (dp[j][k] < dp[i][k]) j = i;
    }
    cout << dp[j][k] << endl;
    vector<int> ans;
    while (j > 0 && k > 0) {
        ans.pb(j);
        j = p[j][k];
        k--;
    }
    reverse(all(ans));
    for (auto it : ans) cout << it << " ";
    return 0;
}

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