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蓝桥杯踩坑题
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<h2 id="前言"><a href="#前言" class="headerlink" title="前言"></a>前言</h2><p>今天又是一年一度的蓝桥（<del>baoli</del>）杯大赛，小 J 同学经过了两年的参赛，对于前三题秉承着能 baoli 就不要思考的出发点来做题，结果他今天在第三题填空题用暴力法踩坑了，比赛中卡了一小时居然还没做出来。睡了一觉下午清醒之后，用三分钟想到了一个简单的做法，写下了这篇博文。</p>
<a id="more"></a>
<h2 id="题面"><a href="#题面" class="headerlink" title="题面"></a>题面</h2><p>想不起完整的题目了，下面大概复述一下。</p>
<p>小明是个富二代，家里仓库非常大，里面常常装满了他网购来的各种好东西。有一天，他刚学完 C++，想算一算自己家仓库的长宽高（都是整数）最多可以有多少种组合，但他想了好久也没想出来，你能帮帮他吗？</p>
<p>比如仓库的体积为 4，那么就有 1×4×1，1×1×4，4×1×1，1×2×2，2×1×2，2×2×1 这六种放法，下面要求的是当体积为 2021041820210418 的情况下，共有几种放法。</p>
<h2 id="心路历程"><a href="#心路历程" class="headerlink" title="心路历程"></a>心路历程</h2><p>这个非常直白也没有什么难度的题为什么卡了我那么久？ 大概还是因为自己一开始想的太过简单吧，因为太久没练算法题，当时一看到是个没有时间限制的填空题，就想着直接三重循环开干了。后来发现不太对，十六位数的三重循环算一天也未必算得完，后来断断续续又加了一些约束条件，比如：只用两个循环，最后一个数用除法来确定；第一层循环中的 $i$ 只遍历到 $\sqrt{N}$ ，第二层循环中的 $j$ 只遍历到 $\sqrt{N/i}$并且用 gcd 判断是否为互质的数，但，这些小伎俩在十六位数面前都是徒劳。十六位数面前只有使用 $O(\sqrt{N})$ 的算法才可以在短时间内得到解，当时在最后时间内明白了这一点后，再去想其他方法时间已经不太够了。</p>
<p>直到下午一觉睡醒，才突然醒悟，最常见的 $O(\sqrt{N})$ 算法不就是找因子数算法吗，大一就学过了，果然还是太久没有做题，并且在比赛的时候从一个错的方向进行思考，然后坑越踩越深，想回头也比较困难了。</p>
<p>虽然题解真的比较简单，但还是在下面写一下留个记录吧。</p>
<h2 id="题解"><a href="#题解" class="headerlink" title="题解"></a>题解</h2><figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br></pre></td><td class="code"><pre><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><iostream></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">include</span> <span class="meta-string"><cmath></span></span></span><br><span class="line"><span class="meta">#<span class="meta-keyword">define</span> ll long long</span></span><br><span class="line"></span><br><span class="line">ll factor[<span class="number">10005</span>];<span class="comment">//因子数组 </span></span><br><span class="line"></span><br><span class="line"><span class="keyword">using</span> <span class="keyword">namespace</span> <span class="built_in">std</span>;</span><br><span class="line"><span class="function"><span class="keyword">int</span> <span class="title">main</span><span class="params">()</span></span></span><br><span class="line"><span class="function"></span>{</span><br><span class="line"> <span class="comment">//cin\cout 输入输出加速 </span></span><br><span class="line"> ios::sync_with_stdio(<span class="literal">false</span>);</span><br><span class="line"> <span class="built_in">cin</span>.tie(<span class="number">0</span>);</span><br><span class="line"> <span class="built_in">cout</span>.tie(<span class="number">0</span>);</span><br><span class="line"> <span class="keyword">int</span> cnt=<span class="number">0</span>;</span><br><span class="line"> <span class="keyword">int</span> ans=<span class="number">0</span>;</span><br><span class="line"> ll N;</span><br><span class="line"> <span class="comment">//2021041820210418 仓库体积大小 </span></span><br><span class="line"> <span class="built_in">cin</span> >> N;</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">1</span>;i<=<span class="built_in">sqrt</span>(N);i++)</span><br><span class="line"> {</span><br><span class="line"> <span class="keyword">if</span>(N%i==<span class="number">0</span>)</span><br><span class="line"> {</span><br><span class="line"> factor[cnt++] = i;</span><br><span class="line"> <span class="keyword">if</span>(N/i != i)</span><br><span class="line"> {</span><br><span class="line"> factor[cnt++] = N/i;</span><br><span class="line"> } </span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="comment">//三重循环遍历所有因子组合（也挺暴力的其实） </span></span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> i=<span class="number">0</span>;i<cnt;i++)</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> j=<span class="number">0</span>;j<cnt;j++)</span><br><span class="line"> {</span><br><span class="line"> <span class="comment">// 在第二个循环里判断一下，不符合的因子可以跳过 </span></span><br><span class="line"> <span class="keyword">if</span>(factor[i]*factor[j]>N)</span><br><span class="line"> {</span><br><span class="line"> <span class="keyword">continue</span>;</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span>(<span class="keyword">int</span> k=<span class="number">0</span>;k<cnt;k++)</span><br><span class="line"> { </span><br><span class="line"> <span class="keyword">if</span>(factor[i] * factor[j] * factor[k] == N)</span><br><span class="line"> {</span><br><span class="line"> ans++;</span><br><span class="line"> }</span><br><span class="line"> } </span><br><span class="line"> }</span><br><span class="line"> <span class="built_in">cout</span> << ans << <span class="built_in">endl</span>;</span><br><span class="line"> <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<h2 id="总结"><a href="#总结" class="headerlink" title="总结"></a>总结</h2><p>今年蓝桥杯的题整体难度上升了不少，填空题已经开始考图论题目了，稍微简单一点的填空题也没有以前好做了，甚至在大题第二题还涉及了比较少见的博弈论内容，总体来说题目可以说是更有区分度了吧。</p>
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