DMOPC '20 Contest 1 P2 - Victor's Moral Dilemma

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Points: 5
Time limit: 0.6s
Memory limit: 128M

Problem type

Is killing an innocent person strictly wrong? ~ Victor, 2019

Victor has become obsessed with the Trolley Problem! Victor found the original trolley problem too boring, so he has devised his own version. In Victor's trolley problem, there is initially an array of N trolleys, D days, and the kth trolley contains A_k people. On the ith day, Victor lines up all the remaining trolleys, and picks a number n_i. He will then partition his array into two subarrays, F = [A_1,A_2,\dots,A_{n_i}] and S = [A_{n_i+1},A_{n_i+2},\dots,A_{m_i}] (where m_i is the total number of trolleys on day i). If \mathrm{sum}(F) \ge \mathrm{sum}(S), then Victor will snap all the trolleys in F out of existence and set A equal to S. Otherwise, he will snap all the trolleys in S out of existence and set A equal to F.

Calculate the number of people Victor snaps on each day!


  • 1 \le N \le 10^6
  • 1 \le D \le N
  • 1 \le a_k \le 10^3
  • m_i \ge 1 for all i.
  • The order of the trolleys will always remain the same.
  • 1 \le n_i \le m_i for all i.

Input Specification

The first line will contain two space-separated integers N and D, denoting the initial number of trolleys and the number of days respectively.

The next line will contain N space-separated integers a_1, a_2, \dots, a_N, denoting the number of people in each trolley.

The next D lines will each contain a single integer n_i.

Output Specification

For each day, output the number of people that Victor will snap out of existence on a new line.

Sample Input

8 3
6 1 3 2 9 10 2 4

Sample Output


Explanation of Sample Output

On the first day, F=[6,1,3,2] and S=[9,10,2,4]. Then, \mathrm{sum}(F)=12 and \mathrm{sum}(S)=25. Since 12<25, Victor will snap trolleys 5 to 8 out of existence, leaving [6,1,3,2] as our array of trolleys.

On the second day, F=[6] and S=[1,3,2]. Then, \mathrm{sum}(F)=6 and \mathrm{sum}(S)=6. Since 6 \ge 6, Victor will snap the first trolley and leave [1,3,2] as our array.

On the third and last day, F=[1] and S=[3,2]. Then, \mathrm{sum}(F)=1 and \mathrm{sum}(S)=5. Since 1 < 5, Victor will snap the last two trolleys and leave [1] as our array.


  • -4
    ross_cleary  commented on Oct. 13, 2020, 8:53 a.m.

    It's too bad that brute force O(ND) solutions passed during the contest, this was a nice prefix sum array problem.

    • 15
      Kirito  commented on Oct. 13, 2020, 9:01 a.m. edited

      You can prove that the brute force solution only takes \mathcal O(N \log N) time, and was in fact intended.

      • 0
        ross_cleary  commented on Oct. 13, 2020, 2:12 p.m.

        Interesting, if there was no upper bound on a_k, would O(ND) be worst case.

        • 5
          Tzak  commented on Oct. 13, 2020, 3:16 p.m.

          Water is wet