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chapter_greedy/max_product_cutting_problem/index.html
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@@ -3143,22 +3143,22 @@
<ul class="md-nav__list" data-md-component="toc" data-md-scrollfix>
<li class="md-nav__item">
<a href="#_1" class="md-nav__link">
贪心策略确定
<a href="#1" class="md-nav__link">
1. &nbsp; 贪心策略确定
</a>
</li>
<li class="md-nav__item">
<a href="#_2" class="md-nav__link">
代码实现
<a href="#2" class="md-nav__link">
2. &nbsp; 代码实现
</a>
</li>
<li class="md-nav__item">
<a href="#_3" class="md-nav__link">
正确性证明
<a href="#3" class="md-nav__link">
3. &nbsp; 正确性证明
</a>
</li>
@@ -3390,22 +3390,22 @@
<ul class="md-nav__list" data-md-component="toc" data-md-scrollfix>
<li class="md-nav__item">
<a href="#_1" class="md-nav__link">
贪心策略确定
<a href="#1" class="md-nav__link">
1. &nbsp; 贪心策略确定
</a>
</li>
<li class="md-nav__item">
<a href="#_2" class="md-nav__link">
代码实现
<a href="#2" class="md-nav__link">
2. &nbsp; 代码实现
</a>
</li>
<li class="md-nav__item">
<a href="#_3" class="md-nav__link">
正确性证明
<a href="#3" class="md-nav__link">
3. &nbsp; 正确性证明
</a>
</li>
@@ -3450,7 +3450,7 @@ n = \sum_{i=1}^{m}n_i
\max(\prod_{i=1}^{m}n_i)
\]</div>
<p>我们需要思考的是：切分数量 <span class="arithmatex">\(m\)</span> 应该多大，每个 <span class="arithmatex">\(n_i\)</span> 应该是多少？</p>
<h3 id="_1">贪心策略确定<a class="headerlink" href="#_1" title="Permanent link">&para;</a></h3>
<h3 id="1">1. &nbsp; 贪心策略确定<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
<p>根据经验，两个整数的乘积往往比它们的加和更大。假设从 <span class="arithmatex">\(n\)</span> 中分出一个因子 <span class="arithmatex">\(2\)</span> ，则它们的乘积为 <span class="arithmatex">\(2(n-2)\)</span> 。我们将该乘积与 <span class="arithmatex">\(n\)</span> 作比较：</p>
<div class="arithmatex">\[
\begin{aligned}
@@ -3477,7 +3477,7 @@ n &amp; \geq 4
<li>当余数为 <span class="arithmatex">\(2\)</span> 时，不继续划分，保留之。</li>
<li>当余数为 <span class="arithmatex">\(1\)</span> 时，由于 <span class="arithmatex">\(2 \times 2 &gt; 1 \times 3\)</span> ，因此应将最后一个 <span class="arithmatex">\(3\)</span> 替换为 <span class="arithmatex">\(2\)</span> 。</li>
</ol>
<h3 id="_2">代码实现<a class="headerlink" href="#_2" title="Permanent link">&para;</a></h3>
<h3 id="2">2. &nbsp; 代码实现<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
<p>在代码中，我们无需通过循环来切分整数，而可以利用向下整除运算得到 <span class="arithmatex">\(3\)</span> 的个数 <span class="arithmatex">\(a\)</span> ，用取模运算得到余数 <span class="arithmatex">\(b\)</span> ，此时有：</p>
<div class="arithmatex">\[
n = 3 a + b
@@ -3672,7 +3672,7 @@ n = 3 a + b
<li>函数 <code>math.pow()</code> 内部调用 C 语言库的 <code>pow()</code> 函数，其执行浮点取幂，时间复杂度为 <span class="arithmatex">\(O(1)\)</span> 。</li>
</ul>
<p>变量 <span class="arithmatex">\(a\)</span> , <span class="arithmatex">\(b\)</span> 使用常数大小的额外空间，<strong>因此空间复杂度为 <span class="arithmatex">\(O(1)\)</span></strong> 。</p>
<h3 id="_3">正确性证明<a class="headerlink" href="#_3" title="Permanent link">&para;</a></h3>
<h3 id="3">3. &nbsp; 正确性证明<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
<p>使用反证法，只分析 <span class="arithmatex">\(n \geq 3\)</span> 的情况。</p>
<ol>
<li><strong>所有因子 <span class="arithmatex">\(\leq 3\)</span></strong> :假设最优切分方案中存在 <span class="arithmatex">\(\geq 4\)</span> 的因子 <span class="arithmatex">\(x\)</span> ，那么一定可以将其继续划分为 <span class="arithmatex">\(2(x-2)\)</span> ，从而获得更大的乘积。这与假设矛盾。</li>