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    <title>题解 on CrystalCore 的博客</title>
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    <item>
      <title>CF2225D</title>
      <link>https://crystalcoreqwq.github.io/posts/cf2225d/</link>
      <pubDate>Fri, 24 Apr 2026 12:20:54 &#43;0800</pubDate>
      <author>niurousaman@outlook.com (CrystalCore)</author>
      <guid>https://crystalcoreqwq.github.io/posts/cf2225d/</guid>
      <description>
        <![CDATA[<h1>CF2225D</h1><p>作者：CrystalCore（niurousaman@outlook.com）</p>
        
          <h2 id="做法">
<a class="header-anchor" href="#%e5%81%9a%e6%b3%95"></a>
做法？
</h2><p>开题时没思路，想着打表看看能不能找规律：哪些区间异或起来能得到 $0$？于是有了以下代码：</p>
<p>:::info[打表代码]</p>
<div class="highlight"><pre tabindex="0" class="chroma"><code class="language-cpp" data-lang="cpp"><span class="line"><span class="cl"><span class="cp">#include</span> <span class="cpf">&lt;bits/stdc++.h&gt;</span><span class="cp">
</span></span></span><span class="line"><span class="cl"><span class="cp">#define ll long long
</span></span></span><span class="line"><span class="cl"><span class="cp">#define str string
</span></span></span><span class="line"><span class="cl"><span class="cp">#define db double
</span></span></span><span class="line"><span class="cl"><span class="k">using</span> <span class="k">namespace</span> <span class="n">std</span><span class="p">;</span>
</span></span><span class="line"><span class="cl"><span class="kt">int</span> <span class="nf">main</span><span class="p">()</span> <span class="p">{</span>
</span></span><span class="line"><span class="cl">	<span class="n">ios</span><span class="o">::</span><span class="n">sync_with_stdio</span><span class="p">(</span><span class="nb">false</span><span class="p">);</span>
</span></span><span class="line"><span class="cl">	<span class="n">cin</span><span class="p">.</span><span class="n">tie</span><span class="p">(</span><span class="k">nullptr</span><span class="p">),</span> <span class="n">cout</span><span class="p">.</span><span class="n">tie</span><span class="p">(</span><span class="k">nullptr</span><span class="p">);</span>
</span></span><span class="line"><span class="cl">	<span class="c1">// 枚举右端点
</span></span></span><span class="line"><span class="cl">	<span class="k">for</span> <span class="p">(</span><span class="n">ll</span> <span class="n">i</span> <span class="o">=</span> <span class="mi">1</span><span class="p">;</span> <span class="n">i</span> <span class="o">&lt;=</span> <span class="mi">30</span><span class="p">;</span> <span class="o">++</span><span class="n">i</span><span class="p">)</span> <span class="p">{</span>
</span></span><span class="line"><span class="cl">		<span class="n">ll</span> <span class="n">cnt</span> <span class="o">=</span> <span class="n">i</span><span class="p">;</span>
</span></span><span class="line"><span class="cl">		<span class="n">cout</span> <span class="o">&lt;&lt;</span> <span class="n">i</span> <span class="o">&lt;&lt;</span> <span class="s">&#34;:   &#34;</span><span class="p">;</span>
</span></span><span class="line"><span class="cl">		<span class="c1">// 枚举左端点
</span></span></span><span class="line"><span class="cl">		<span class="k">for</span> <span class="p">(</span><span class="n">ll</span> <span class="n">j</span> <span class="o">=</span> <span class="n">i</span> <span class="o">-</span> <span class="mi">1</span><span class="p">;</span> <span class="n">j</span> <span class="o">&gt;=</span> <span class="mi">1</span><span class="p">;</span> <span class="n">j</span><span class="o">--</span><span class="p">)</span> <span class="p">{</span>
</span></span><span class="line"><span class="cl">			<span class="n">cnt</span> <span class="o">^=</span> <span class="n">j</span><span class="p">;</span>
</span></span><span class="line"><span class="cl">			<span class="k">if</span><span class="p">(</span><span class="n">cnt</span> <span class="o">==</span> <span class="mi">0</span><span class="p">)</span> <span class="n">cout</span> <span class="o">&lt;&lt;</span> <span class="n">j</span> <span class="o">&lt;&lt;</span> <span class="sc">&#39; &#39;</span><span class="p">;</span>
</span></span><span class="line"><span class="cl">		<span class="p">}</span>
</span></span><span class="line"><span class="cl">		<span class="n">cout</span> <span class="o">&lt;&lt;</span> <span class="sc">&#39;\n&#39;</span><span class="p">;</span>
</span></span><span class="line"><span class="cl">	<span class="p">}</span>
</span></span><span class="line"><span class="cl">	<span class="k">return</span> <span class="mi">0</span><span class="p">;</span>
</span></span><span class="line"><span class="cl"><span class="p">}</span>
</span></span></code></pre></div><p>:::</p>
        
        <hr><p>本文2026-04-24首发于<a href='https://crystalcoreqwq.github.io/'>CrystalCore 的博客</a>，最后修改于2026-04-24</p>]]>
      </description>
      
        <category>算法</category>
      
    </item>
    
    

    <item>
      <title>CF2225C</title>
      <link>https://crystalcoreqwq.github.io/posts/cf2225c/</link>
      <pubDate>Fri, 24 Apr 2026 12:20:54 &#43;0800</pubDate>
      <author>niurousaman@outlook.com (CrystalCore)</author>
      <guid>https://crystalcoreqwq.github.io/posts/cf2225c/</guid>
      <description>
        <![CDATA[<h1>CF2225C</h1><p>作者：CrystalCore（niurousaman@outlook.com）</p>
        
          <h2 id="做法">
<a class="header-anchor" href="#%e5%81%9a%e6%b3%95"></a>
做法
</h2><p>简单的 DP。</p>
<p>很容易想到，对于一个 $2 \times 2$ 的部分只有两种填法：</p>
<p><img src="https://cdn.luogu.com.cn/upload/image_hosting/5eo8ozcu.png" alt=""></p>
<p>很显然可以 DP 吧？</p>
<p>设 $dp[i]$ 表示前 $i$ 列满足要求的最小代价。</p>
<p>然后位置 $i$ 可以由 $i-2$ 通过加两个横着的矩形转移（填充方法 $1$），也可以由 $i-1$ 用过加一个竖着的矩形转移（填充方法 $2$ 取一半）。</p>
        
        <hr><p>本文2026-04-24首发于<a href='https://crystalcoreqwq.github.io/'>CrystalCore 的博客</a>，最后修改于2026-04-24</p>]]>
      </description>
      
        <category>算法</category>
      
    </item>
    
    

    <item>
      <title>SP22268</title>
      <link>https://crystalcoreqwq.github.io/posts/sp22268/</link>
      <pubDate>Fri, 09 Jan 2026 12:20:54 &#43;0800</pubDate>
      <author>niurousaman@outlook.com (CrystalCore)</author>
      <guid>https://crystalcoreqwq.github.io/posts/sp22268/</guid>
      <description>
        <![CDATA[<h1>SP22268</h1><p>作者：CrystalCore（niurousaman@outlook.com）</p>
        
          <h1 id="题目传送门">
<a class="header-anchor" href="#%e9%a2%98%e7%9b%ae%e4%bc%a0%e9%80%81%e9%97%a8"></a>
<a href="https://www.luogu.com.cn/problem/SP22268">题目传送门</a>
</h1><h1 id="sp22268-etfs---欧拉函数筛法-题解">
<a class="header-anchor" href="#sp22268-etfs---%e6%ac%a7%e6%8b%89%e5%87%bd%e6%95%b0%e7%ad%9b%e6%b3%95-%e9%a2%98%e8%a7%a3"></a>
SP22268 ETFS - 欧拉函数筛法 题解
</h1><h2 id="题目大意">
<a class="header-anchor" href="#%e9%a2%98%e7%9b%ae%e5%a4%a7%e6%84%8f"></a>
题目大意
</h2><p>给定区间 $[a, b]$，求区间内每个数的欧拉函数值 $\varphi(n)$。</p>
<h2 id="算法思路">
<a class="header-anchor" href="#%e7%ae%97%e6%b3%95%e6%80%9d%e8%b7%af"></a>
算法思路
</h2><p>很显然，$10^{12}$ 的数据是不可能线性做的，但是 $b - a \leq 10^6$，那么我们就可以只求出区间内的 $\varphi(i)$。</p>
        
        <hr><p>本文2026-01-09首发于<a href='https://crystalcoreqwq.github.io/'>CrystalCore 的博客</a>，最后修改于2026-01-09</p>]]>
      </description>
      
        <category>算法</category>
      
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