Submission #1133488

#	Time	Username	Problem	Language	Result	Execution time	Memory
1133488		totoro	Split the sequence (APIO14_sequence)	C++20	50 / 100	2096 ms	24132 KiB

sequence

#include <iostream>
#include <vector>

const long long NEGINFINITY = -1000000000000000ll;

int main() {
    int n, k;
    std::cin >> n >> k;
    std::vector<long long> v(n);
    for (int i = 0; i < n; ++i) std::cin >> v[i];
    std::vector<long long> p(n + 1);
    for (int i = 1; i <= n; ++i) p[i] = p[i - 1] + v[i - 1];
    /*
        dp[kk][i] is the best value using kk splits with last one 'at' i
        so if we know dp[kk-1] for all i, then we can calculate dp[kk][i] as:
        max from prev=0 to prev = i-1 of (dp[kk-1][prev] + (p[i] - p[prev]) * (p.back() - p[i]))
    */
    std::vector<std::vector<long long>> dp(k+2, std::vector<long long>(n+1));
    std::vector<std::vector<int>> bestfrom(k+2, std::vector<int>(n+1));
    /*
        when kk=0, it must be that we can get no score
    */
    for (int i = 0; i <= n; ++i) dp[0][i] = 0;
    /*
        when i = 0 but kk > 0 then we cannot even do anything, not even score 0
    */
    for (int kk = 1; kk <= k+1; ++kk) dp[kk][0] = NEGINFINITY;
    for (int kk = 1; kk <= k+1; ++kk) {
        for (int i = 1; i <= n; ++i) {
            dp[kk][i] = -1;
            for (int prev = 0; prev < i; ++prev) {
                long long candidate = dp[kk-1][prev] + (p[i] - p[prev])*(p.back() - p[i]);
                if (candidate > dp[kk][i]) {
                    dp[kk][i] = candidate;
                    bestfrom[kk][i] = prev;
                }
            }
        }
    }
    std::cout << dp[k+1][n] << '\n';
    int cur = bestfrom[k+1][n];
    while (k) {
        std::cout << cur << " \n"[k==1];
        cur = bestfrom[k][cur];
        --k;
    }
}

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