Submission #30160

#	Time	Username	Problem	Language	Result	Execution time	Memory
30160		Andrei1998	Split the sequence (APIO14_sequence)	C++14	0 / 100	0 ms	1844 KiB

sequence

#include <bits/stdc++.h>

using namespace std;

const int NMAX = 100000 + 5;
const int KMAX = 200 + 5;
typedef long long int lint;
const lint INF = 2E18;

int N, K;
int sPart[NMAX];
lint dpBrute[KMAX][NMAX];
int father[KMAX][NMAX];

void brute() {
    dpBrute[1][0] = INF;
    for (int j = 1; j <= N; ++ j)
        dpBrute[1][j] = 1LL * sPart[j] * sPart[j];
    for (int i = 2; i <= K; ++ i) {
            dpBrute[i & 1][0] = INF;
            for (int j = 1; j <= N; ++ j) {
            pair <lint, int> sol = {1LL * (sPart[j] - sPart[j - 1]) * (sPart[j] - sPart[j - 1]) + dpBrute[(i - 1) & 1][j - 1], j - 1};
            for (int k = 1; k <= j; ++ k)
                sol = min(sol, {1LL * (sPart[j] - sPart[k - 1]) * (sPart[j] - sPart[k - 1]) + dpBrute[(i - 1) & 1][k - 1], k - 1});
            dpBrute[i & 1][j] = sol.first;
            father[i][j] = sol.second;
        }
    }
}

int main()
{
    //freopen("data.in", "r", stdin);

    cin >> N >> K;
    ++ K;
    for (int i = 1; i <= N; ++ i) {
        int val;
        cin >> val;
        sPart[i] = val + sPart[i - 1];
    }

    brute();
    cout << (1LL * sPart[N] * sPart[N] - dpBrute[K & 1][N]) / 2 << '\n';

    vector <int> sol;

    int where = N;
    int k = K;
    while (father[k][where]) {
        sol.push_back(father[k][where]);
        where = father[k --][where];
    }

    reverse(sol.begin(), sol.end());
    for (int i = 0; i < sol.size(); ++ i)
        cout << sol[i] << " \n"[i + 1 == sol.size()];
    return 0;
}

Compilation message (stderr)

sequence.cpp: In function 'int main()':
sequence.cpp:56:23: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
     for (int i = 0; i < sol.size(); ++ i)
                       ^
sequence.cpp:57:39: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
         cout << sol[i] << " \n"[i + 1 == sol.size()];
                                       ^

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