游戏王残局简化版

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User & Date: 顽雨沉风 on 2023-09-25 02:01:06
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2023-09-25
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Modified 解法参考/MH-20210116.html from [ab4e6c2e45] to [ed311cb628].

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  <!--[if lt IE 9]>
    <script src="//cdnjs.cloudflare.com/ajax/libs/html5shiv/3.7.3/html5shiv-printshiv.min.js"></script>
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</head>
<body>
<header id="title-block-header">
<h1 class="title">MH-20210116</h1>
</header>
<nav id="TOC" role="doc-toc">
<ul>
<li><a href="#原解" id="toc-原解"><span class="toc-section-number">1</span> 原解</a></li>
<li><a href="#扩展" id="toc-扩展"><span class="toc-section-number">2</span> 扩展</a>
<ul>
<li><a href="#问" id="toc-问"><span class="toc-section-number">2.1</span> 问</a>
<ul>
<li><a href="#答" id="toc-答"><span class="toc-section-number">2.1.1</span> 答</a></li>
<li><a href="#解" id="toc-解"><span class="toc-section-number">2.1.2</span> 解</a></li>
</ul></li>
</ul></li>
</ul>
</nav>
<h1 data-number="1" id="原解"><span class="header-section-number">1</span> 原解</h1>
<pre><code>1 * 5 = 5

2 * 4 = 8

3 * 3 = 9</code></pre>
<h1 data-number="2" id="扩展"><span class="header-section-number">2</span> 扩展</h1>
<h2 data-number="2.1" id="问"><span class="header-section-number">2.1</span> 问</h2>
<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>
<h3 data-number="2.1.1" id="答"><span class="header-section-number">2.1.1</span> 答</h3>
<p>非 3 即 2</p>
<h3 data-number="2.1.2" id="解"><span class="header-section-number">2.1.2</span> 解</h3>
<p>可能的值有 1 ~ 正无穷</p>
<p>由指数爆炸性可知，只需要考虑 2 3 4</p>
<p>由 4 是 2 的倍数可知，只需要考虑 2 3</p>
<p>由某些正整数不能被 2 整除可知，需要 3</p>
<p>由某些正整数不能被 3 整除可知，需要 2</p>
<p>故，非 3 即 2</p>
<p>取 2 3 的最小公倍数 6</p>
<pre><code>2 ^ 3 = 8

3 ^ 3 = 9</code></pre>
<p>故，优先选 3</p>
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  <!--[if lt IE 9]>
    <script src="//cdnjs.cloudflare.com/ajax/libs/html5shiv/3.7.3/html5shiv-printshiv.min.js"></script>
  <![endif]-->
</head>
<body>
<header id="title-block-header">
<h1 class="title">MH-20210116</h1>
</header>
<nav id="TOC" role="doc-toc">
<ul>
<li><a href="#原解" id="toc-原解"><span class="toc-section-number">1</span> 原解</a></li>
<li><a href="#扩展" id="toc-扩展"><span class="toc-section-number">2</span> 扩展</a></li>







</ul>
</nav>
<h1 data-number="1" id="原解"><span class="header-section-number">1</span> 原解</h1>
<pre><code>1 * 5 = 5

2 * 4 = 8

3 * 3 = 9</code></pre>
<h1 data-number="2" id="扩展"><span class="header-section-number">2</span> 扩展</h1>
<pre><code>6 = 3 + 3 = 3 + 3 + 1 - 1 = 3 + 1 + 3 - 1 = (3 + 1) + (3 - 1)

(3 + 1) * (3 - 1) = 3 * 3 + 3 * (-1) + 1 * 3 + 1 * (-1) = 3 * 3 + 1 * (-1) = 3 * 3 - 1 ^ 2 = 3 ^ 2 - 1 ^ 2











3 ^ 2 &gt; 3 ^ 2 - 1 ^ 2</code></pre>

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︙︙

Modified 解法参考/MH-20210116.md from [afa3ff62be] to [df95e3fc0c].

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2 * 4 = 8

3 * 3 = 9
~~~

# 扩展

## 问

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

### 答

非 3 即 2

### 解

可能的值有 1 ~ 正无穷

由指数爆炸性可知，只需要考虑 2 3 4

由 4 是 2 的倍数可知，只需要考虑 2 3

由某些正整数不能被 2 整除可知，需要 3

由某些正整数不能被 3 整除可知，需要 2

故，非 3 即 2

取 2 3 的最小公倍数 6

~~~
2 ^ 3 = 8

3 ^ 3 = 9
~~~

故，优先选 3









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2 * 4 = 8

3 * 3 = 9
~~~

# 扩展


~~~

6 = 3 + 3 = 3 + 3 + 1 - 1 = 3 + 1 + 3 - 1 = (3 + 1) + (3 - 1)



(3 + 1) * (3 - 1) = 3 * 3 + 3 * (-1) + 1 * 3 + 1 * (-1) = 3 * 3 + 1 * (-1) = 3 * 3 - 1 ^ 2 = 3 ^ 2 - 1 ^ 2



3 ^ 2 > 3 ^ 2 - 1 ^ 2












~~~