Overview

This problem asks you to find the minimum total length from end to end when arranging \(N\) books and a bookend in a row, with at least a specified minimum gap of \(K\) between them.

Analysis

At first glance, it seems like you need to consider “in what order should the books be arranged,” but in fact, the order of the books does not affect the answer. Let’s break down the components that make up the total length.

The total length is the sum of the following 3 components:

1. Width of the bookend: \(S\) centimeters (fixed)
2. Total of the gaps (spaces) between items:
   • A gap of at least \(K\) centimeters is required between the bookend and the first book (1 location).
   • A gap of at least \(K\) centimeters is required between each pair of adjacent books (among \(N\) books) (\(N-1\) locations).
   • Therefore, a total of \(1 + (N-1) = N\) gaps are needed. To minimize the total length, we should make all of these gaps exactly \(K\) centimeters, the minimum value.
3. Total thickness of all books:
   • If the thickness of each book is \(D_1, D_2, \ldots, D_N\), then the total is \(\sum_{i=1}^{N} D_i\). No matter what order the books are arranged in, this sum remains the same.

Adding all of these together, the desired minimum value can be expressed by the following formula: $\(\text{Minimum distance} = S + (N \times K) + \sum_{i=1}^{N} D_i\)$

Algorithm

1. Read \(N, S, K\) and the thickness \(D_i\) of each book from the input.
2. Compute the total thickness \(D_{sum}\) of the \(N\) books.
3. Calculate the result using the formula \(S + (N \times K) + D_{sum}\).
4. Output the result.

Complexity

• Time complexity: \(O(N)\)
  • To compute the total thickness of the books, we need to scan through the \(N\) input values once.
• Space complexity: \(O(N)\)
  • If storing the thickness of all books in a list or similar structure, \(O(N)\) memory is used.

Implementation Notes

• Since the values of \(N\) and \(D_i\) can be large, the total may exceed the range of 32-bit integers. In Python, this is not an issue because integers are automatically handled as arbitrary-precision integers. However, if using other languages (such as C++), be careful to use 64-bit integer types (such as long long).

• Note that the number of gaps is not \(N-1\), but \(N\), because the gap between the bookend and the first book must also be included.

  Source Code

import sys

def main():
    # 標準入力からすべてのデータを読み込み、空白（スペース、改行など）で分割します
    input_data = sys.stdin.read().split()
    
    if not input_data:
        return
        
    # 各値を取得
    # N: 本の冊数
    # S: ブックエンドの幅
    # K: 必要な最小間隔
    n = int(input_data[0])
    s = int(input_data[1])
    k = int(input_data[2])
    
    # 各本の厚さ D_i の総和を計算
    # 本を並べる順番に関わらず、厚さの合計と隙間の合計は一定です
    d_sum = sum(map(int, input_data[3:3+n]))
    
    # 最小距離を計算
    # 距離 = ブックエンドの幅 + (N 個の隙間 × K) + 全ての本の厚さの合計
    # - ブックエンドと1冊目の本の間に隙間 K が必要
    # - 隣り合う本同士の間 (N-1 箇所) にそれぞれ隙間 K が必要
    # 合計で N 箇所の隙間が必要となります
    ans = s + (n * k) + d_sum
    
    # 結果を出力
    print(ans)

if __name__ == '__main__':
    main()

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