Overview

When delivering packages to \(N\) destinations, given that the daily delivery capacity can be increased from \(K\) up to a maximum of \(K+M\), the problem asks to minimize the total number of delivery days \(\sum_{i=1}^{N} \lceil A_i / (K+x) \rceil\).

Analysis

Key Insight: Increasing delivery capacity reduces (never increases) the number of days

The core of this problem lies in the property that the ceiling function \(\lceil A_i / c \rceil\) is monotonically non-increasing with respect to \(c\) (delivery capacity).

Let’s verify with a concrete example. When \(A_i = 10\):

Delivery capacity \(c\)	\(\lceil 10 / c \rceil\)
3	4
4	3
5	2
10	1
11	1

As delivery capacity \(c\) increases, more packages can be delivered per visit, so the required number of visits decreases (or stays the same).

If it holds for each destination, it holds for the sum too

If each \(\lceil A_i / c \rceil\) is monotonically non-increasing with respect to \(c\), then the sum \(\sum_{i=1}^{N} \lceil A_i / c \rceil\) is also monotonically non-increasing with respect to \(c\).

Conclusion

Since the total number of days is smaller (or unchanged) as the delivery capacity \(c = K + x\) increases, setting \(x\) to its maximum value \(M\) is always optimal.

In other words, no complex search or binary search is needed at all — the answer can be obtained simply by computing the total number of days with \(c = K + M\).

Algorithm

1. Set the delivery capacity to \(c = K + M\).
2. For each destination \(i\), compute the required number of days \(\lceil A_i / c \rceil\).
   • Ceiling division can be computed as integer division (floor) of \((A_i + c - 1) / c\).
3. Sum up the number of days for all destinations and output the result.

Supplementary Note on Ceiling Division

\(\lceil a / b \rceil\) (where \(a, b\) are positive integers) can be computed with the following formula:

\[\lceil a / b \rceil = \lfloor (a + b - 1) / b \rfloor\]

For example, when \(a = 10, b = 3\): \((10 + 3 - 1) / 3 = 12 / 3 = 4 = \lceil 10/3 \rceil\), which gives the correct result.

Complexity

• Time complexity: \(O(N)\) — only one ceiling division per destination
• Space complexity: \(O(N)\) — for storing the array \(A\)

Implementation Notes

• Since \(K, M, A_i\) can each be up to \(10^9\), \(K + M\) can be up to \(2 \times 10^9\). In Python there is no concern about integer overflow, but in C++ and similar languages, long long must be used.

• Computing ceiling division with math.ceil(A_i / c) may introduce floating-point errors, so it is safer to use integer arithmetic: (A_i + c - 1) // c.

• When reading the problem, one might think “Don’t we need to search for the optimal \(x\)?”, but once you notice the monotonicity, it becomes immediately clear that \(x = M\) is optimal.

  Source Code

import sys

def compute_days(A, cap):
    total = 0
    for a in A:
        total += (a + cap - 1) // cap
    return total

def solve():
    input_data = sys.stdin.buffer.read().split()
    N = int(input_data[0])
    K = int(input_data[1])
    M = int(input_data[2])
    A = [int(input_data[3 + i]) for i in range(N)]
    
    # We want to minimize sum of ceil(A_i / (K+x)) for x in [0, M]
    # As x increases (capacity increases), total days decrease or stay same.
    # So the minimum is achieved at x = M.
    # Wait, that's not necessarily optimal if the function is monotonically non-increasing...
    # Actually, ceil(A_i / c) is non-increasing in c, so the sum is non-increasing in c.
    # Therefore x = M always gives the minimum.
    
    # But wait, let me reconsider. The problem says "find the minimum days when x is chosen optimally".
    # Since increasing capacity can only decrease or maintain the number of days for each destination,
    # the optimal is simply x = M.
    
    cap = K + M
    print(compute_days(A, cap))

solve()

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