Overview
Comment: | 0.0.2548 |
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Downloads: | Tarball | ZIP archive | SQL archive |
Timelines: | family | ancestors | descendants | both | trunk |
Files: | files | file ages | folders |
SHA3-256: |
d975a7f729ce1e2855243106b872f48b |
User & Date: | 顽雨沉风 on 2023-09-15 02:44:15 |
Other Links: | manifest | tags |
Context
2023-09-15
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02:57 | 0.0.2549 check-in: 6a4a8a79e5 user: 顽雨沉风 tags: trunk | |
02:44 | 0.0.2548 check-in: d975a7f729 user: 顽雨沉风 tags: trunk | |
02:41 | 0.0.2547 check-in: ebdeddb639 user: 顽雨沉风 tags: trunk | |
Changes
Modified 解法参考/MH-20210116.html from [dc5d3d84e9] to [23211c273c].
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29 30 31 32 33 34 35 | <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> | | > > > > > > > > > < > | 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 | <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="#衍生" id="toc-衍生"><span class="toc-section-number">2</span> 衍生</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="衍生"><span class="header-section-number">2</span> 衍生</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>取 2 3 4 的最小公倍数 12</p> <p>拆分 12 可得</p> <pre><code>2 ^ 6 = 64 3 ^ 4 = 81 |
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Modified 解法参考/MH-20210116.md from [48a1adf6a3] to [1c8c412380].
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 | % MH-20210116 # 原解 ~~~ 1 * 5 = 5 2 * 4 = 8 3 * 3 = 9 ~~~ # 衍生 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 非 3 即 2 | > > > > | | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 | % MH-20210116 # 原解 ~~~ 1 * 5 = 5 2 * 4 = 8 3 * 3 = 9 ~~~ # 衍生 ## 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 取 2 3 4 的最小公倍数 12 |
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