Cyclists – Codeforces Round 1086 (Div. 2) Writeup

Problem Summary

Bob is playing a mobile card game with n cards arranged in a queue (deck). He can only pick cards from the first k positions of the deck at any moment. Each card costs energy to play, and one special card (the win-condition) is at position p. After playing a card, it goes to the end of the deck.

Bob wants to play the win-condition card as many times as possible without exceeding a total energy limit of m.

We are given multiple test cases. For each, we must determine the maximum times the win-condition card can be played.

Constraints:

• $1 \le t \le 5000$ test cases
• $1 \le k, p \le n \le 5000$, $1 \le m \le 5000$
• $1 \le a_i \le m$ (cost of each card)
• Sum of n over all test cases ≤ 5000

Output: For each test case, print one integer, the maximum times the win-condition card can be played.

Observations

1. The naive idea of always picking the win-condition card may fail if it is not in the first k cards.
2. If the win-condition is beyond the first k, we need to remove cheaper cards in front of it to bring it within reach.
3. If multiple cards have the same cost as the win-condition, we must track the exact index of the win-condition card, not just its value, because card costs can repeat.

Key insight: simulate the game while tracking the win-condition by its position. Whenever a card is removed before it, the index shifts left; if the card is played, it moves to the end.

Step-by-Step Solution

1. Read t test cases.

2. For each test case:

   • Track the win-condition index (win_pos = p-1).

   • While we have energy (money >= 0):

     1. If the win-condition is in the first k cards:

        • Pay its cost, increment the counter, move it to the end (win_pos = n-1).

     2. Otherwise:

        • Pick the cheapest card among the first k, pay its cost, move it to the end.
        • Adjust win_pos if a card before it was removed.

3. Print the total times the win-condition card was played.

Example Walkthrough

Input:

1
4
2
2 1 2 42
3
42 1
4
3 3 2 6
5
2 1 2
6
3 2 2 6
7
2 1 2
8
8 4 7 10
9
3 4 4 2 1 1 4 2

Output:

1
0
2
6
3
2
4
1

• Test case 1: Only the first card can be played. Win-condition is second → can’t play → 0.
• Test case 2: Win-condition is already in first 3; energy allows 6 plays → 6.
• Test case 3: Must remove first card once, then win-condition is playable → 2.
• Test case 4: Only one energy-efficient path → play win-condition once → 1.

Code Implementation (Python)

1
t = int(input())
2
for _ in range(t):
3
    n, k, p, m = map(int, input().split())
4
    deck = list(map(int, input().split()))
5

6
    win_pos = p - 1      # track the winning card by index
7
    money = m
8
    count = 0
9

10
    while True:
11
        if win_pos < k:
12
            cost = deck[win_pos]
13
            if money < cost:
14
                break
15
            money -= cost
16
            count += 1
17
            deck.pop(win_pos)
18
            deck.append(cost)
19
            win_pos = n - 1
20
        else:
21
            pickable = deck[:k]
22
            min_cost = min(pickable)
23
            if money < min_cost:
24
                break
25
            idx = pickable.index(min_cost)
26
            money -= min_cost
27
            deck.pop(idx)
28
            deck.append(min_cost)
29
            if idx < win_pos:
30
                win_pos -= 1
31

32
    print(count)

Explanation

• Track the win-condition by index, not value.
• Adjust win_pos whenever a card before it is removed.
• Pick cheapest card if the win-condition is not within first k.
• Count how many times the win-condition card is played before energy runs out.

Complexity Analysis

• Time complexity: $O(t \cdot n \cdot k)$ in worst case, since we check the first k cards repeatedly.
• Space complexity: $O(n)$ per test case.

Given constraints, this simulation is efficient enough.

Takeaways

• When items have duplicate values, always track them by position to avoid ambiguity.
• Greedy simulations often require careful index adjustments.
• A small, correct simulation beats a complex formula in tricky games like this.