Find Minimum Log Transportation Cost
<p>You are given integers <code>n</code>, <code>m</code>, and <code>k</code>.</p>

<p>There are two logs of lengths <code>n</code> and <code>m</code> units, which need to be transported in three trucks where each truck can carry one log with length <strong>at most</strong> <code>k</code> units.</p>

<p>You may cut the logs into smaller pieces, where the cost of cutting a log of length <code>x</code> into logs of length <code>len1</code> and <code>len2</code> is <code>cost = len1 * len2</code> such that <code>len1 + len2 = x</code>.</p>

<p>Return the <strong>minimum total cost</strong> to distribute the logs onto the trucks. If the logs don&#39;t need to be cut, the total cost is 0.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 6, m = 5, k = 5</span></p>

<p><strong>Output:</strong> <span class="example-io">5</span></p>

<p><strong>Explanation:</strong></p>

<p>Cut the log with length 6 into logs with length 1 and 5, at a cost equal to <code>1 * 5 == 5</code>. Now the three logs of length 1, 5, and 5 can fit in one truck each.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 4, m = 4, k = 6</span></p>

<p><strong>Output:</strong> <span class="example-io">0</span></p>

<p><strong>Explanation:</strong></p>

<p>The two logs can fit in the trucks already, hence we don&#39;t need to cut the logs.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= k &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= n, m &lt;= 2 * k</code></li>
	<li>The input is generated such that it is always possible to transport the logs.</li>
</ul>
Math
class Solution:
    def minCuttingCost(self, n: int, m: int, k: int) -> int:
        x = max(n, m)
        return 0 if x <= k else k * (x - k)
