Overview

This is a problem where you calculate the total number of work sessions needed for each area and determine how much it exceeds Takahashi’s team’s work limit \(M\).

Analysis

The key points for solving this problem are “how to calculate the number of work sessions needed for each area” and “how to efficiently process a large amount of data”.

1. Calculating the Number of Work Sessions (Ceiling Division)

The \(i\)-th area has \(A_i\) chairs, and at most \(K\) chairs can be carried per work session. The number of work sessions needed is the ceiling of \(A_i \div K\), namely \(\lceil A_i / K \rceil\). In programming, the ceiling division \(\lceil a / b \rceil\) for integers \(a, b\) is commonly computed using the following formula: $\(\text{ceil}(a, b) = (a + b - 1) // b\)$ (Here // denotes integer division)

For example, if \(A_i = 10, K = 3\), then \(\lceil 10 / 3 \rceil = 4\) sessions are needed. Applying the formula gives \((10 + 3 - 1) // 3 = 12 // 3 = 4\), which is correct.

2. Number of Backup Requests

Let the total number of work sessions across all areas be \(S = \sum_{i=1}^{N} \lceil A_i / K \rceil\). - If \(S\) is at most \(M\), Takahashi’s team can finish on their own, so the answer is 0 times. - If \(S\) is greater than \(M\), they need to ask Aoki for backup exactly \(S - M\) times to cover the shortfall.

3. Handling the Constraints

Since the number of areas \(N\) can be as large as \(10^6\), calling input() one by one in Python may not finish within the time limit (TL). Therefore, techniques such as reading all input at once are necessary.

Algorithm

1. Read the input \(N, M, K\) and the array \(A\).
2. For each \(A_i\), compute the required number of work sessions \(c_i = (A_i + K - 1) // K\).
3. Calculate the total number of work sessions across all areas \(S = \sum c_i\).
4. Compute \(S - M\). If this value is negative, output \(0\); if positive, output that value as the answer.

Complexity

• Time complexity: \(O(N)\)
  • We scan through \(N\) elements once to compute the total, so the processing time is proportional to the number of areas.
• Space complexity: \(O(N)\)
  • If all input values \(A_i\) are stored in memory, the memory usage is proportional to \(N\).

Implementation Notes

• Fast input: By using sys.stdin.read().split(), a large number of values can be read at once, reducing execution time.

• Memory efficiency: When computing the total, using a generator expression sum( (a + K - 1) // K for a in A ) instead of creating a new list can reduce memory consumption.

• Handling huge numbers: Python natively supports arbitrary-precision integers (integers of any number of digits), so even if \(M\) or \(A_i\) are extremely large, calculations can be performed without worrying about overflow.

  Source Code

import sys

def solve():
    # 競技プログラミングにおいて、大量の入力を高速に読み込むため、
    # sys.stdin.read().split() を使用して全ての入力を一度に読み込み、空白で分割します。
    input_data = sys.stdin.read().split()
    
    # 入力が空の場合、何もせず終了します。
    if not input_data:
        return
    
    # N: エリアの数
    # M: 高橋君のチームがこなせる作業回数の上限
    # K: 1回の作業で運べる椅子の最大脚数
    N = int(input_data[0])
    M = int(input_data[1])
    K = int(input_data[2])
    
    # 各エリア A_i の椅子の数は、input_data のインデックス 3 以降に格納されています。
    # 必要な作業回数の合計を計算します。
    # i 番目のエリアに必要な作業回数は ceil(A_i / K) です。
    # Python の整数除算を用いて、(A_i + K - 1) // K と計算することで
    # 切り上げ値（ceil）を効率的に求めることができます。
    
    # map(int, input_data[3:3+N]) を使用して文字列を整数に変換し、
    # ジェネレータ式と sum() を組み合わせることで、メモリ消費を抑えつつ高速に合計を求めます。
    total_tasks_needed = sum((a + K - 1) // K for a in map(int, input_data[3:3+N]))
    
    # 青木君に依頼する最小回数は、(必要な作業回数の合計 - 高橋君のチームの上限 M) です。
    # ただし、高橋君のチームだけで完了できる場合は 0 回となります。
    ans = total_tasks_needed - M
    if ans < 0:
        ans = 0
        
    # 結果を出力します。
    print(ans)

if __name__ == '__main__':
    solve()

This editorial was generated by gemini-3-flash-thinking.