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<a href="/2022/07/28/Wireguard-NAT-to-NAT/" class="post-title-link" itemprop="url">使用 Wireguard 并利用服务器中转实现无公网 IP 异地组网</a>
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<h2 id="情景"><a href="#情景" class="headerlink" title="情景"></a>情景</h2><p>设备 A 与设备 B 均无公网 IP。需要 VPN 构建隧道。</p>
<h2 id="实现方法"><a href="#实现方法" class="headerlink" title="实现方法"></a>实现方法</h2><h3 id="服务器端"><a href="#服务器端" class="headerlink" title="服务器端"></a>服务器端</h3><p>服务器 Ubuntu 22.04</p>
<p>安装 wireguard</p>
<figure class="highlight bash"><table><tr><td class="gutter"><pre><span class="line">1</span><br></pre></td><td class="code"><pre><span class="line">sudo apt install wireguard</span><br></pre></td></tr></table></figure>
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<h1 id="前言"><a href="#前言" class="headerlink" title="前言"></a>前言</h1><p>彩虹六号（Tom Clancy’s Rainbow Six Siege）使用 Vivox 提供的语音聊天服务。由于部分机场使用 Xray/V2Ray 时，需要禁止 BT 或网络审计，在 inbound 开启了 sniffing，导致语音聊天服务器无法连接。<br>通过在 <a target="_blank" rel="noopener" href="https://www.virustotal.com/gui/domain/vivox.com/relations">virustotal</a> 的查询，以及在 <a target="_blank" rel="noopener" href="https://www.abuseipdb.com/whois/74.201.106.174">AbuseIPDB</a> 的查询，并凭借已知的彩虹六号语音服务器域名 rbspsxp.<a target="_blank" rel="noopener" href="http://www.vivox.com/">www.vivox.com</a> （来源：<a target="_blank" rel="noopener" href="https://github.com/XTLS/Xray-core/issues/250#issuecomment-778590896">XTLS/Xray-core#250</a>）以及 rbswp.<a target="_blank" rel="noopener" href="http://www.vivox.com/">www.vivox.com</a> （来源：<a target="_blank" rel="noopener" href="https://steamcommunity.com/app/359550/discussions/1/2524779067037299243/?l=english">steamcommunity: Voice chat not working?</a>），我们可以推测出彩虹六号的语音服务器域名应该以“rbs”开头，可能意为“rainbow six”，并得到以下域名/IP对：</p>
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<h1 id="Introduction"><a href="#Introduction" class="headerlink" title="Introduction"></a>Introduction</h1><p>Many people confuse proxy and Virtual Private Network (VPN). This article explains the technical principles and application scenarios of proxy and VPN.</p>
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<h1 id="Introduction"><a href="#Introduction" class="headerlink" title="Introduction"></a>Introduction</h1><p>WaveCloud提供了Emby影视库，拥有丰富的媒体资源。但多数人并不会使用Emby解锁，且普拉斯影业的使用教程并不全。本教程将介绍全平台的Emby解锁方法。</p>
<h1 id="Method"><a href="#Method" class="headerlink" title="Method"></a>Method</h1><h2 id="Windows-macOS-and-Linux"><a href="#Windows-macOS-and-Linux" class="headerlink" title="Windows, macOS and Linux"></a>Windows, macOS and Linux</h2><h3 id="Windows-and-Linux-Method-1"><a href="#Windows-and-Linux-Method-1" class="headerlink" title="Windows and Linux: Method 1"></a>Windows and Linux: Method 1</h3><p>参考<a target="_blank" rel="noopener" href="https://neko.re/archives/225.html">这篇文章</a></p>
<h3 id="Windows-macOS-and-Linux-Method-2"><a href="#Windows-macOS-and-Linux-Method-2" class="headerlink" title="Windows, macOS and Linux: Method 2"></a>Windows, macOS and Linux: Method 2</h3><h4 id="Step-1-Install-and-Trust-Certificate"><a href="#Step-1-Install-and-Trust-Certificate" class="headerlink" title="Step 1: Install and Trust Certificate"></a>Step 1: Install and Trust Certificate</h4><p>下载<a target="_blank" rel="noopener" href="http://app.wavecloud.us:2333/Emby/embyUnlockProxy.7z">这个文件</a>，并解压。</p>
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<p><em>转载声明：本文转载自微信公众号：不能使用该名称（微信号: ForbiddenSpeech），原文链接为：<a target="_blank" rel="noopener" href="https://mp.weixin.qq.com/s/cndzW_oXClkSdwOtam0qUw">https://mp.weixin.qq.com/s/cndzW_oXClkSdwOtam0qUw</a></em></p>
<p>本文旨在分析“翻墙”行为的法律风险，并基于现行规范性法律文件和相关案例进行学术讨论。在分析相关法条时可能需要对部分计算机专业术语进行释义。<font color=Red><strong>但本文不涉及有关“翻墙”的任何技术指导或方法的具体介绍。</strong></font><br><u><font color=Red>另，本文讨论的一切“XXX合法与违法”问题，分析的主体都是“单纯访问境外网站”，不包括“访问、发布、传播违法有害信息”，后者当然属于违法犯罪行为，但这和翻墙行为没有任何本质联系，因为在境内网站也可访问、发布、传播违法有害信息。</font></u> </p>
<p>在本文大致框架形成以前，我已查阅国内几乎所有关于“翻墙”的论文和文章，大部分探讨都存在着严重的问题，有些是法学常识性问题（或不讨论法律层面的问题），有些是计算机技术方面的常识性问题（常出现在法学类文章中）。<br>这是一个很有趣的现象——“翻墙”是一个同时牵涉法学和计算机技术的两个领域的跨学科问题。很多法学专家不了解技术；很多技术人员，也没有意愿探讨技术涉及的法律问题。这个疑难问题今天终于被我这个“既没学好法，又没学好计算机”的奇葩捡了漏。望两个学科的大佬们对于本文可能出现的纰漏予以指正！ </p>
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<h1 id="题目"><a href="#题目" class="headerlink" title="题目"></a>题目</h1><p>某公司为了丰富员工的业余文化生活，召开了一次趣味运动会. 甲、乙两人参加“射击气球”比赛，他们依次轮流射击气球一次，每人射击$n$次，射击气球只有两种结果：“中”或“不中”. 规则：甲先射击，若“中”得2分，否则1分；乙再射，若“中”，得第一次甲的得分加1，否则得1分；接着甲再射，若“中”，得第一次乙的得分加1，否则得1分；乙再射，若“中”，得第二次甲的得分加1，否则得1分；接着甲再射，若“中”，得第二次乙的得分加1，否则得1分. 按此规则，直至比赛结束. 已知射中概率均为$\dfrac{2}{3}$. 记$X_i$，$Y_i$分别为甲、乙第$i$次射击得分. 记$a_1=EX_1$，$a_2=EY_1$，$a_3=EX_2$，$\cdots$.</p>
<p>证明：$\lbrace a_n-3\rbrace$为等比数列.</p>
<h1 id="题解"><a href="#题解" class="headerlink" title="题解"></a>题解</h1><h2 id="全期望公式"><a href="#全期望公式" class="headerlink" title="全期望公式"></a>全期望公式</h2>
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<h1 id="原因（发现问题）"><a href="#原因（发现问题）" class="headerlink" title="原因（发现问题）"></a>原因（发现问题）</h1><p>买了乐学高考的化学课，只能在移动端观看。为了方便记笔记，以及更好的观看体验。决定解决这个问题。</p>
<h1 id="分析问题"><a href="#分析问题" class="headerlink" title="分析问题"></a>分析问题</h1><p>可能解决方法：</p>
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<li>安卓模拟器</li>
<li>手机投屏</li>
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<h1 id="解决问题"><a href="#解决问题" class="headerlink" title="解决问题"></a>解决问题</h1><h2 id="第一天"><a href="#第一天" class="headerlink" title="第一天"></a>第一天</h2><h3 id="第0xFF次尝试"><a href="#第0xFF次尝试" class="headerlink" title="第0xFF次尝试"></a>第0xFF次尝试</h3><p>想到了Android模拟器，但发现模拟器内的乐学高考不能正常播放。后来又想到了PhoenixOS，这也是个Android系统（基于x86架构），但是也发现乐学高考不能在这个系统内播放。直接Pass。</p>
<h3 id="第0x00次尝试"><a href="#第0x00次尝试" class="headerlink" title="第0x00次尝试"></a>第0x00次尝试</h3>
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<h2 id="条件最值例题"><a href="#条件最值例题" class="headerlink" title="条件最值例题"></a>条件最值例题</h2><p>已知$a+b=1(a>0,b>0)$，求$\dfrac{1}{a}+\dfrac{4}{b}$的最小值<br>解：<br>$$\begin{eqnarray}<br>原式&=&\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{4}{b}\right)\\<br>&=&\dfrac{b}{a}+\dfrac{4a}{b}+5\\<br>&\geq&9<br>\end{eqnarray}$$</p>
<p>那么这就是本题的常规解法，下面我来介绍另一种方法：拉格朗日乘子法(Lagrange multiplier)</p>
<h3 id="Lagrange-multiplier"><a href="#Lagrange-multiplier" class="headerlink" title="Lagrange multiplier"></a>Lagrange multiplier</h3><p>基本的拉格朗日乘子法就是求函数$f(x_1,x_2,…)$在约束条件$g(x_1,x_2,…)=0$下的极值的方法。<br>其主要思想是将约束条件函数与原函数联立，从而求出使原函数取得极值的各个变量的解。</p>
<h4 id="计算过程"><a href="#计算过程" class="headerlink" title="计算过程"></a>计算过程</h4>
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<h2 id="A-B-Problem描述"><a href="#A-B-Problem描述" class="headerlink" title="A+B Problem描述"></a>A+B Problem描述</h2><p>输入两个整数a,b，输出它们的和(|a|,|b|<=10^9)。</p>
<p>注意<br>1.pascal使用integer会爆<br>2.有负数<br>3.c/c++的main函数必须是int类型，而且最后要return 0</p>
<p>好吧，同志们，我们就从这一题开始，向着大牛的路进发。</p>