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ebdeddb639 2023-09-15 31: <h1 class="title">MH-20210116</h1>
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ebdeddb639 2023-09-15 34: <ul>
ebdeddb639 2023-09-15 35: <li><a href="#原解" id="toc-原解"><span class="toc-section-number">1</span> 原解</a></li>
ebdeddb639 2023-09-15 36: <li><a href="#衍生" id="toc-衍生"><span class="toc-section-number">2</span> 衍生</a></li>
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ebdeddb639 2023-09-15 39: <h1 data-number="1" id="原解"><span class="header-section-number">1</span> 原解</h1>
ebdeddb639 2023-09-15 40: <pre><code>1 * 5 = 5
ebdeddb639 2023-09-15 41:
ebdeddb639 2023-09-15 42: 2 * 4 = 8
ebdeddb639 2023-09-15 43:
ebdeddb639 2023-09-15 44: 3 * 3 = 9</code></pre>
ebdeddb639 2023-09-15 45: <h1 data-number="2" id="衍生"><span class="header-section-number">2</span> 衍生</h1>
ebdeddb639 2023-09-15 46: <p>给定一个由数个小正整数累加起来的大正整数，问这些小正整数的值为多少才能让这些小正整数的累乘值最大</p>
ebdeddb639 2023-09-15 47: <p>非 3 即 2</p>
ebdeddb639 2023-09-15 48: <p>解法：</p>
ebdeddb639 2023-09-15 49: <p>可能的值有 1 ~ 正无穷</p>
ebdeddb639 2023-09-15 50: <p>由指数爆炸性可知，只需要考虑 2 3 4</p>
ebdeddb639 2023-09-15 51: <p>取 2 3 4 的最小公倍数 12</p>
ebdeddb639 2023-09-15 52: <p>拆分 12 可得</p>
ebdeddb639 2023-09-15 53: <pre><code>2 ^ 6 = 64
ebdeddb639 2023-09-15 54:
ebdeddb639 2023-09-15 55: 3 ^ 4 = 81
ebdeddb639 2023-09-15 56:
ebdeddb639 2023-09-15 57: 4 ^ 3 = 48</code></pre>
ebdeddb639 2023-09-15 58: <p>至此，4 被淘汰</p>
ebdeddb639 2023-09-15 59: <p>最优解为 3</p>
ebdeddb639 2023-09-15 60: <p>但考虑到有些整数不能被 3 整数，故也需要 2</p>
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