Overview

This problem asks for the minimum distance from the left end to the right end when arranging a bookend and \(N\) books on a bookshelf. The key insight is that the answer is constant regardless of the order in which the books are arranged.

Analysis

First, let’s illustrate the arrangement on the bookshelf from left to right.

|<-- S -->|<- K ->|<- D_a ->|<- K ->|<- D_b ->|<- K ->| ... |<- K ->|<- D_? ->|
bookend     gap     book1     gap     book2     gap           gap     bookN

Key insight: No matter what order the books are arranged in, the total distance remains the same.

Let’s think about why this is the case. The total distance can be decomposed as follows:

1. Width of the bookend: \(S\) (fixed)
2. Total gaps: \(K\) between the bookend and the first book, then \(K\) for each of the \(N-1\) gaps between consecutive books. Total: \(N \times K\)
3. Total thickness of the books: \(D_1 + D_2 + \cdots + D_N\) (the sum is the same regardless of the order)

The gaps must be “at least \(K\)”, but since we want to minimize the total distance, it is optimal to set each gap to exactly \(K\) (the minimum value).

Also, changing the order of the books only changes the order of addition, which does not affect the total sum. For example, with \(N = 3\), \(S = 2\), \(K = 1\), \(D = [3, 5, 7]\):

\[\text{Total} = S + N \times K + \sum_{i=1}^{N} D_i = 2 + 3 \times 1 + (3 + 5 + 7) = 2 + 3 + 15 = 20\]

The answer is \(20\) regardless of the order in which the books are arranged.

Algorithm

From the analysis above, the answer can be obtained with the following simple formula:

\[\text{Answer} = S + N \times K + \sum_{i=1}^{N} D_i\]

Steps: 1. Read \(N\), \(S\), \(K\), and the thickness \(D_i\) of each book 2. Compute and output the above formula

Complexity

• Time complexity: \(O(N)\) (to compute the sum of \(D_i\))
• Space complexity: \(O(N)\) (to store the array \(D\))

Implementation Notes

• Since \(N\), \(S\), \(K\), and \(D_i\) can each be up to \(10^9\), the answer can be approximately \(10^9 + 10^5 \times 10^9 + 10^5 \times 10^9 \approx 2 \times 10^{14}\). In C++ and similar languages, you need to use long long, but in Python there is no need to worry about integer overflow.

• The key point is to interpret the condition “at least \(K\) gap” as exactly \(K\), since this is a minimization problem.

  Source Code

N, S, K = map(int, input().split())
D = list(map(int, input().split()))

# The layout from left to right:
# [Bookend: width S] [gap >= K] [Book1] [gap >= K] [Book2] ... [gap >= K] [BookN]
# Total distance = S + N * K + sum(D)
# The order of books doesn't matter since all gaps are exactly K and all books must be placed.

total = S + N * K + sum(D)
print(total)

This editorial was generated by claude4.6opus-thinking.