Match’s model#3
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Match’s model (Editorial)

The problem is fundamentally about arranging cubes in 3D space to maximize edge sharing. When two cubes are placed adjacent to each other, they share a face, and the 4 edges around that face become internal to the structure rather than requiring separate matches.

The optimal arrangement is a rectangular prism with dimensions as close to each other as possible. Think of it like packing cubes into an a \times b \times c box. A cube-shaped arrangement (like 3 \times 3 \times 3) has minimal surface area, which means fewer exposed edges and thus fewer matches needed.

We can count matches by thinking of the arrangement as a 3D grid. If we have a \times b \times c cubes, the grid has (a+1) \times (b+1) \times (c+1) vertices. The matches are edges connecting these vertices in three directions: horizontal, vertical, and depth.

Solution Approach

We start by finding k \approx \sqrt[3]{n}, approximately the cube root of . This gives us a baseline for dimensions that are roughly equal. Since int(n**(1/3)) can be off by one due to floating-point precision, we adjust upward if (k+1)^3 \leq n, or downward if k^3 > n, until we find the correct integer cube root.

Next, we try to fit cubes into one of three near-cubic arrangements: k \times k \times k, k \times k \times (k+1), or k \times (k+1) \times (k+1). We pick the largest arrangement that doesn't exceed cubes. This becomes our base block.

For the base block, we calculate the number of matches using the edge-counting formula:

\text{matches} = a(b+1)(c+1) + b(a+1)(c+1) + c(a+1)(b+1)

In each direction, we count how many edges span that direction. For example, in the -direction, we have edges across, repeated (b+1) \times (c+1) times for each row and layer.

If there are leftover cubes after filling the base block, we arrange them optimally on top. We try to form a rectangle (approximately \sqrt{r} \times \sqrt{r} where is the remainder) to maintain compactness. Any remaining cubes that don't fit the rectangle are added in a line, which costs matches per cube plus to attach to the structure.

Implementation

import math

n = int(input())

if n == 0:
    print(0)
else:
    # Find k ≈ cube root of n, adjusting for floating point precision
    k = int(n**(1/3))
    while (k+1)**3 <= n:
        k += 1
    while k**3 > n:
        k -= 1

    # Choose base block dimensions
    # Try k×k×k, k×k×(k+1), or k×(k+1)×(k+1)
    if n <= k * k * (k + 1):
        a = k
        b = k
        c = k
        rem = n - k**3
    elif n <= k * (k+1) * (k+1):
        a = k
        b = k
        c = k + 1
        rem = n - k * k * (k + 1)
    else:
        a = k
        b = k + 1
        c = k + 1
        rem = n - k * (k+1) * (k+1)

    # Count matches for base block: edges in x, y, and z directions
    result = a * (b+1) * (c+1) + b * (a+1) * (c+1) + c * (a+1) * (b+1)

    # Add matches for remainder cubes if any
    if rem > 0:
        # Arrange remainder as an s×t rectangle
        s = int(math.isqrt(rem))

        if s > 0:
            t = rem // s
        else:
            t = 0

        left = rem - s * t

        # Add edges for the rectangle on top of base block
        result += (s+1) * (t+1) + s * (t+1) + t * (s+1)

        # Add any leftover cubes in a line
        if left > 0:
            result += 3 * left + 2

    print(result)
Python

Complexity Analysis: