CCC '26 S4 - Minecarts

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Points: 15 (partial)
Time limit: 2.0s
Memory limit: 1000M

Problem types
Canadian Computing Competition: 2026 Stage 1, Senior #4

Steve has N

Steve wants the minecarts to be arranged in a row such that the number of gems in each minecart is non-decreasing from left to right. To do this, he plans to build a side track to the right of all of the minecarts that splits off from the main track. Steve can move minecarts that are on the left of the side track into the side track, allowing other minecarts on its left to move past it on the main track. Minecarts moved into the side track can be moved back into the main track one at a time: a minecart on the side track will move back to the left of the side track and to the right of all other minecarts which are on the left of the side track. The last minecart moved into the side track must be the first minecart to be moved back out: that is, the side track follows a last-in, first-out method. The side track can be used any number of times. Finally, once a minecart is moved to the right of the side track, it can no longer be moved to the left. Below is an example sequence of moves that can be made with 5

Steve has K

Input Specification

The first line of input will consist of two space-separated integers N

The next line contains N

The following table shows how the available 15 marks are distributed:

Marks Awarded Bounds on N Bounds on K
2 marks 1\leq N \leq 5\,000 K = 0
2 marks 1\leq N \leq 5\,000 K = 10^{12}
2 marks 1\leq N \leq 5\,000 0\leq K \leq 10^{12}
3 marks 1\leq N \leq 3\times10^5 K = 0
3 marks 1\leq N \leq 3\times10^5 K = 10^{12}
3 marks 1\leq N \leq 3\times10^5 0\leq K \leq 10^{12}

Output Specification

Output a single integer, the minimum capacity of the side track that needs to be built given that Steve distributes up to K

Sample Input 1

4 14 
5 0 4 0

Sample Output 1

1

Explanation for Sample Output 1

One optimal distribution is to put 6

We can then build a side track of capacity 1

  1. Move minecart 4

  2. Move minecart 3

  3. Move minecart 2

  4. Move minecart 1

  5. Move minecart 3

  6. Move minecart 3

Sample Input 2

4 8
5 0 4 0

Sample Output 2

2

Explanation for Sample Output 2

One optimal distribution is to put all 8

Then we can build a side track of capacity 2

  1. Move minecart 4

  2. Move minecart 3

  3. Move minecart 2

  4. Move minecart 1

  5. Move minecart 2

  6. Move minecart 3

  7. Move minecart 3

  8. Move minecart 2

Sample Input 3

4 123456789 
40 30 20 10

Sample Output 3

3

Explanation for Sample Output 3

Since there are no empty minecarts, there is only one possible distribution of spare gems: no spare gems are used at all.

Then we can build a side track of capacity 3

  1. Move minecart 4

  2. Move minecart 3

  3. Move minecart 2

  4. Move minecart 1

  5. Move minecart 2

  6. Move minecart 3

  7. Move minecart 4


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