Update code highlight

authors/weily/index.html

@@ -460,7 +460,7 @@ <h1 class="gblog-post__title">
 
 <span class="flex align-center no-wrap gblog-post__meta--readtime">
   <svg class="gblog-icon gblog_timer"><use xlink:href="#gblog_timer"></use></svg>
-  <span class="gblog-post__tag">1 min read</span>
+  <span class="gblog-post__tag">2 min read</span>
 </span>
 
 

index.html

@@ -746,7 +746,7 @@ <h1 class="gblog-post__title">
 
 <span class="flex align-center no-wrap gblog-post__meta--readtime">
   <svg class="gblog-icon gblog_timer"><use xlink:href="#gblog_timer"></use></svg>
-  <span class="gblog-post__tag">1 min read</span>
+  <span class="gblog-post__tag">2 min read</span>
 </span>
 
 
@@ -921,7 +921,7 @@ <h1 class="gblog-post__title">
 
 <span class="flex align-center no-wrap gblog-post__meta--readtime">
   <svg class="gblog-icon gblog_timer"><use xlink:href="#gblog_timer"></use></svg>
-  <span class="gblog-post__tag">1 min read</span>
+  <span class="gblog-post__tag">2 min read</span>
 </span>
 
 

post/bezouts/index.html

@@ -67,7 +67,7 @@
       "url" : "https://weilycoder.github.io/post/bezouts/",
       "headline": "辗转相除法和裴蜀定理",
       "description": "数论基础第二弹！",
-      "wordCount" : "314",
+      "wordCount" : "324",
       "license": "CC BY-SA 4.0",
       "inLanguage": "en",
       "isFamilyFriendly": "true",
@@ -465,8 +465,15 @@ <h1 class="gblog-post__title">辗转相除法和裴蜀定理</h1>
 <p>既然公约数相同，那么最大公约数也相同。</p>
 </blockquote>
 <p>容易写出递归版：</p>
-<div class="highlight"><pre tabindex="0" class="chroma"><code class="language-cpp" data-lang="cpp"><span class="line"><span class="cl"><span class="kt">int</span> <span class="nf">gcd</span><span class="p">(</span><span class="kt">int</span> <span class="n">a</span><span class="p">,</span> <span class="kt">int</span> <span class="n">b</span><span class="p">)</span> <span class="p">{</span> <span class="k">return</span> <span class="n">b</span><span class="o">?</span> <span class="n">gcd</span><span class="p">(</span><span class="n">b</span><span class="p">,</span> <span class="n">a</span> <span class="o">%</span> <span class="n">b</span><span class="p">)</span><span class="o">:</span> <span class="n">a</span><span class="p">;</span> <span class="p">}</span>
-</span></span></code></pre></div><div class="flex align-center gblog-post__anchorwrap">
+<div class="highlight"><div class="chroma">
+<table class="lntable"><tr><td class="lntd">
+<pre tabindex="0" class="chroma"><code><span class="lnt">1
+</span></code></pre></td>
+<td class="lntd">
+<pre tabindex="0" class="chroma"><code class="language-Cpp" data-lang="Cpp"><span class="line"><span class="cl"><span class="kt">int</span> <span class="nf">gcd</span><span class="p">(</span><span class="kt">int</span> <span class="n">a</span><span class="p">,</span> <span class="kt">int</span> <span class="n">b</span><span class="p">)</span> <span class="p">{</span> <span class="k">return</span> <span class="n">b</span><span class="o">?</span> <span class="n">gcd</span><span class="p">(</span><span class="n">b</span><span class="p">,</span> <span class="n">a</span> <span class="o">%</span> <span class="n">b</span><span class="p">)</span><span class="o">:</span> <span class="n">a</span><span class="p">;</span> <span class="p">}</span></span></span></code></pre></td></tr></table>
+</div>
+</div>
+<div class="flex align-center gblog-post__anchorwrap">
     <h3 id="时间复杂度"
     >
         时间复杂度
@@ -527,16 +534,31 @@ <h1 class="gblog-post__title">辗转相除法和裴蜀定理</h1>
 <p>故解为 $(x_0,y_0-\left\lfloor\dfrac{a}{b}\right\rfloor x_0)$，由数学归纳法，对任意 $\gcd(a,b)=1$，均能构造 $(x,y)$ 使 $ax+by=1$，进而原命题得证。</p>
 <p>裴蜀定理的逆定理也成立。</p>
 <p>容易写出代码：</p>
-<div class="highlight"><pre tabindex="0" class="chroma"><code class="language-cpp" data-lang="cpp"><span class="line"><span class="cl"><span class="kt">int</span> <span class="nf">exgcd</span><span class="p">(</span><span class="kt">int</span> <span class="n">a</span><span class="p">,</span> <span class="kt">int</span> <span class="n">b</span><span class="p">,</span> <span class="kt">int</span> <span class="o">&amp;</span><span class="n">x</span><span class="p">,</span> <span class="kt">int</span> <span class="o">&amp;</span><span class="n">y</span><span class="p">)</span> <span class="p">{</span>
+<div class="highlight"><div class="chroma">
+<table class="lntable"><tr><td class="lntd">
+<pre tabindex="0" class="chroma"><code><span class="lnt">1
+</span><span class="lnt">2
+</span><span class="lnt">3
+</span><span class="lnt">4
+</span><span class="lnt">5
+</span><span class="lnt">6
+</span><span class="lnt">7
+</span><span class="lnt">8
+</span><span class="lnt">9
+</span></code></pre></td>
+<td class="lntd">
+<pre tabindex="0" class="chroma"><code class="language-Cpp" data-lang="Cpp"><span class="line"><span class="cl"><span class="kt">int</span> <span class="nf">exgcd</span><span class="p">(</span><span class="kt">int</span> <span class="n">a</span><span class="p">,</span> <span class="kt">int</span> <span class="n">b</span><span class="p">,</span> <span class="kt">int</span> <span class="o">&amp;</span><span class="n">x</span><span class="p">,</span> <span class="kt">int</span> <span class="o">&amp;</span><span class="n">y</span><span class="p">)</span> <span class="p">{</span>
 </span></span><span class="line"><span class="cl">  <span class="k">if</span> <span class="p">(</span><span class="o">!</span><span class="n">b</span><span class="p">)</span> <span class="p">{</span>
 </span></span><span class="line"><span class="cl">    <span class="n">x</span> <span class="o">=</span> <span class="mi">1</span><span class="p">,</span> <span class="n">y</span> <span class="o">=</span> <span class="mi">0</span><span class="p">;</span>
 </span></span><span class="line"><span class="cl">    <span class="k">return</span> <span class="n">a</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="kt">int</span> <span class="n">d</span> <span class="o">=</span> <span class="n">exgcd</span><span class="p">(</span><span class="n">b</span><span class="p">,</span> <span class="n">a</span> <span class="o">%</span> <span class="n">b</span><span class="p">,</span> <span class="n">y</span><span class="p">,</span> <span class="n">x</span><span class="p">);</span>
 </span></span><span class="line"><span class="cl">  <span class="n">y</span> <span class="o">-=</span> <span class="n">a</span> <span class="o">/</span> <span class="n">b</span> <span class="o">*</span> <span class="n">x</span><span class="p">;</span>
 </span></span><span class="line"><span class="cl">  <span class="k">return</span> <span class="n">d</span><span class="p">;</span>
-</span></span><span class="line"><span class="cl"><span class="p">}</span>
-</span></span></code></pre></div><p>由于函数在求最大公因数的过程中计算可行解，这种算法被称作扩展欧几里得算法。</p>
+</span></span><span class="line"><span class="cl"><span class="p">}</span></span></span></code></pre></td></tr></table>
+</div>
+</div>
+<p>由于函数在求最大公因数的过程中计算可行解，这种算法被称作扩展欧几里得算法。</p>
 <p>有时我们担心，扩展欧几里得算法给出的可行解过大，使得变量发生溢出；幸运的是，可以证明，上述算法给出的解满足 $|x|&lt;b,|y|&lt;a.$</p>
 <p>归纳法易证。</p>
 <div class="flex align-center gblog-post__anchorwrap">

post/bin-inversion/index.html

@@ -535,6 +535,24 @@ <h1 class="gblog-post__title">二项式反演学习笔记</h1>
 
 
 
+
+
+
+
+
+        <span class="gblog-footer__item gblog-footer__item--row">
+
+            <svg class="gblog-icon gblog_link"><use xlink:href="#gblog_link"></use></svg>
+
+          <a
+            href="https://github.com/thegeeklab/hugo-geekblog"
+            class="gblog-footer__link"
+          >
+            Theme
+          </a>
+        </span>
+
+