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@@ -4787,7 +4787,7 @@ n & \geq 4
<li>输入整数 <span class="arithmatex">\(n\)</span> ，从其不断地切分出因子 <span class="arithmatex">\(3\)</span> ，直至余数为 <span class="arithmatex">\(0\)</span>、<span class="arithmatex">\(1\)</span>、<span class="arithmatex">\(2\)</span> 。</li>
<li>当余数为 <span class="arithmatex">\(0\)</span> 时，代表 <span class="arithmatex">\(n\)</span> 是 <span class="arithmatex">\(3\)</span> 的倍数，因此不做任何处理。</li>
<li>当余数为 <span class="arithmatex">\(2\)</span> 时，不继续划分，保留。</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>
<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> 和余数 <span class="arithmatex">\(1\)</span> 替换为两个 <span class="arithmatex">\(2\)</span> 。</li>
</ol>
<h3 id="2">2. &nbsp; 代码实现<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
<p>如图 15-16 所示，我们无须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>