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2023/07/05/trace-of-line-01/index.html

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 <meta property="og:description" content="不存在的历史，勾起了石月仙的注意，随着找寻真相的深入，一个隐瞒了全世界的阴谋浮出水面。而对于人类，阴谋何来，何从，何去？「第三次科技革命后，人类进入信息时代，信息化开始影响人类的历史进程……」「为什么初中高中上过的还要再上一遍啊？我都快烦死了。」石月仙在心中自言自语道，眼睛无神地看着教授在全息…">
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2023/07/05/zhihu-aac-old/index.html

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  <a href="/tags/%E5%8F%8D%E7%88%AC/" class="print-no-link">#反爬</a>
 
+ <a href="/tags/%E9%80%86%E5%90%91/" class="print-no-link">#逆向</a>
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+ <a href="/tags/%E7%88%AC%E8%99%AB/" class="print-no-link">#爬虫</a>
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2023/07/06/zerotier-planet-convert/index.html

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 <meta property="og:description" content="由于国内特殊的网络原因，Zerotier 官方提供的 Planet 用户体验不佳。为此，不少人选择自建私有 Moon，甚至私有 Planet 服务器。然而，正如官方文档所说，使用私有 Planet 服务器会使你的节点无法找到其他的标准节点。本文试图提出一种方案在使用私有 Planet 服务器的同时与标准节点通信。">
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2023/07/10/trace-of-line-02/index.html

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2023/08/04/permission-system-design-share/index.html

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2023/08/31/cell-structure/index.html

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2023/10/03/busuanzi-bug/index.html

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2024/07/11/json-format/index.html

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 <meta property="og:description" content="你是否曾经疑惑过，当我们提到 JSON 的时候，我们在说什么？JSON 以 JS 开头，那么其与 JS 的联系到底是什么？在这篇文章中，我们将尽可能通俗地了解 JSON 的起源、基本格式和意义，并解答一些常见的疑惑。">
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2024/09/06/htb-rsaiseasy/index.html

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 <meta property="og:site_name" content="创新者老王的博客">
 <meta property="og:description" content="原题地址：HackTheBox - RSAisEasy 之前做过这个题，是与另一位朋友一起分析的，但是当时没有写 writeup，于是再梳理一遍。 从标题可以看出，这道题与 RSA 有点相关，因此会用到一部分基于 RSA 公式的数学分析。（说实话，在拉着咱朋友一起来分析数学部分之前，是有那么一点不敢做这题的……） 0. 题目内容题目提供了一个 zip 文件，解压后有两个文件 123456# o">
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  <i class="iconfont icon-date-fill" aria-hidden="true"></i>
- <time datetime="2024-09-06 17:53" pubdate>
- 2024年9月6日 下午
+ <time datetime="2024-09-06 18:01" pubdate>
+ 2024年9月6日 晚上
  </time>
  </span>
 
@@ -270,7 +270,7 @@
  <span class="post-meta mr-2">
  <i class="iconfont icon-chart"></i>
 
- 6.9k 字
+ 7k 字
 
  </span>
 
@@ -366,7 +366,7 @@ <h2 id="5-n2"><a href="#5-n2" class="headerlink" title="5. n2"></a>5. n2</h2><p>
 <p>为方便展示过程，我们将 <code>(n1 * E) + n2</code> 记为 <code>nE</code>。</p>
 <p>$$<br>n2 \equiv nE \pmod {n1}<br>$$</p>
 <p>即</p>
-<p>$$<br>n2 &#x3D; nE\ %\ n1 + n1 \cdot k<br>$$</p>
+<p>$$<br>\def\mod{\text{ mod }}<br>n2 &#x3D; nE \mod n1 + n1 \cdot k<br>$$</p>
 <p>其中 <code>k</code> 是未知的，你可以直接猜测 <code>k = 0</code> 即 <code>n2 = nE % n1</code>，但我还是想证明它。</p>
 <p>注意到 <code>n2</code> 是两个 <code>getPrime(512)</code> 相乘，因此 <code>0 &lt; k &lt; 2^1024</code>。接下来用代码测试 <code>k</code> 可以的取值。</p>
 <figure class="highlight python"><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></pre></td><td class="code"><pre><code class="hljs python">n1 = <span class="hljs-number">101302608234750530215072272904674037076286246679691423280860345380727387460347553585319149306846617895151397345134725469568034944362725840889803514170441153452816738520513986621545456486260186057658467757935510362350710672577390455772286945685838373154626020209228183673388592030449624410459900543470481715269</span><br>ne = <span class="hljs-number">601613204734044874510382122719388369424704454445440856955212747733856646787417730534645761871794607755794569926160226856377491672497901427125762773794612714954548970049734347216746397532291215057264241745928752782099454036635249993278807842576939476615587990343335792606509594080976599605315657632227121700808996847129758656266941422227113386647519604149159248887809688029519252391934671647670787874483702292498358573950359909165677642135389614863992438265717898239252246163</span><br><br><span class="hljs-keyword">for</span> k <span class="hljs-keyword">in</span> (-<span class="hljs-number">1</span>, <span class="hljs-number">0</span>, <span class="hljs-number">1</span>): <br> n2 = ne % n1 + n1 * k<br> <span class="hljs-built_in">print</span>(k, <span class="hljs-number">0</span> &lt; n2 &lt; <span class="hljs-number">2</span>**<span class="hljs-number">1024</span>, n2)<br></code></pre></td></tr></table></figure>

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+ <a href="/tags/%E9%80%86%E5%90%91/">#逆向</a>
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