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音频系统必然要使用舒适，就不能简单，最起码的功能：
音乐播放单独一个音频通道
让手机声音进入宿主进行统一管理
要有一个备用提供给其他软件的音频通道
人声通道 和 效果通道
音乐通道环绕声
个人喜欢整点花活 所以必须有 变声
前置准备
PC1
PC2
主力机
辅机
192.168.3.2
192.168.3.1
还有声卡设备 ， 我选用的是 midiplus studio m 和一个手机声卡
一根双绞线
主力机声卡驱动 midiplus
主力机声音设置录制
播放
主力机宿主跳线输入
输出
创建音轨后
这一套下来音乐伴奏就已经独立通道了，它的音乐环绕效果靠插件实现 ，DearVR pro 插件
让插件读取我画在包络线上的路径去控制 插件中的 X 和 Z 轴 并且 将包络线路径循环读取运作就实现了音乐环绕效果
手机声音处理将手机输出 到 手机声卡 ， 后将手机声卡与主力机连接 ，此时Voicemeeter Banana 会识别到设备，将设备Inout 选中手机声卡
将Out 处的 A1 输出到 5 ...</div></div></div><div class="recent-post-item"><div class="post_cover right"><a href="/2024/09/01/2024.9.1-leetcode-Delete%20the%20Nth%20bottom%20node%20of%20the%20linked%20list/" title="LeetCode题解 19.删除链表的倒数第 N 个节点"><img class="post-bg" src="/Src/LeetCode/LeetCode-logo1.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="LeetCode题解 19.删除链表的倒数第 N 个节点"></a></div><div class="recent-post-info"><a class="article-title" href="/2024/09/01/2024.9.1-leetcode-Delete%20the%20Nth%20bottom%20node%20of%20the%20linked%20list/" title="LeetCode题解 19.删除链表的倒数第 N 个节点">LeetCode题解 19.删除链表的倒数第 N 个节点</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time class="post-meta-date-created" datetime="2024-09-01T07:27:00.000Z" title="Created 2024-09-01 15:27:00">2024-09-01</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">Updated</span><time class="post-meta-date-updated" datetime="2024-09-01T07:30:34.171Z" title="Updated 2024-09-01 15:30:34">2024-09-01</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E7%BC%96%E7%A8%8B/">编程</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E7%BC%96%E7%A8%8B/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/">数据结构与算法</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E7%BC%96%E7%A8%8B/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/c-plus-plus/">C++</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/">数据结构与算法</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E7%BC%96%E7%A8%8B/">编程</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/C/">C++</a></span></div><div class="content">19.删除链表的倒数第 N 个节点
解题思路：
1.首先拿到链表需要做预处理，如果为空或 <=0 就返回原链表
2.然后开始用 current指针遍历 ，并计数有多少节点
3.处理 counter 和 position 相同就是删除头节点
4.遍历结束就可以 将 current 重置 指向头节点， counter（链表节点总数） - position 就得到正向索引值
5.用for 指向目标节点的前一个节点，将目标节点的 前后节点相连,目标节点自然就会脱离链表，最后delete 将其释放
123456789101112131415161718192021222324252627282930313233343536373839// struct ListNode {// int val;// ListNode *next;// ListNode() : val(0), next(nullptr) {}// ListNode(int x) : val(x), next(n ...</div></div></div><div class="recent-post-item"><div class="post_cover left"><a href="/2024/08/19/2024.8.19-REVERSE%20ENGINEERING%20-%20PLUG-IN%20PACIFY%20PACIFICATION/" title="逆向工程 - 外挂 PACIFY安抚"><img class="post-bg" src="https://s2.loli.net/2024/09/01/lDnIu7CJfU8HLQm.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="逆向工程 - 外挂 PACIFY安抚"></a></div><div class="recent-post-info"><a class="article-title" href="/2024/08/19/2024.8.19-REVERSE%20ENGINEERING%20-%20PLUG-IN%20PACIFY%20PACIFICATION/" title="逆向工程 - 外挂 PACIFY安抚">逆向工程 - 外挂 PACIFY安抚</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time class="post-meta-date-created" datetime="2024-08-19T04:00:00.000Z" title="Created 2024-08-19 12:00:00">2024-08-19</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">Updated</span><time class="post-meta-date-updated" datetime="2024-09-01T12:05:33.226Z" title="Updated 2024-09-01 20:05:33">2024-09-01</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%8A%80%E6%9C%AF/">技术</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%8A%80%E6%9C%AF/%E9%80%86%E5%90%91%E5%B7%A5%E7%A8%8B/">逆向工程</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%8A%80%E6%9C%AF/">技术</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E9%80%86%E5%90%91%E5%B7%A5%E7%A8%8B/">逆向工程</a></span></div><div class="content">逆向工程 - 外挂 PACIFY安抚
人在焦虑中总是要逃避一小段时间 ， 比如玩游戏 ，可是多巴胺消退不再给我刺激时 ，焦虑还是会萦绕于心 ，不面对就永远无法摆脱。 ——jelly
兴趣于学业结合我也会想短短的逃避一下，但又不能抛弃学业，陷入其中
我会在玩游戏的时候 ，去对一个游戏逆向工程，这就是兴趣于学业结合
我做了什么
首先就是内存地址
我采用 Github开源的 ImGui框架 编写
我获取到了游戏的 关键的 模块地址，基址，内存偏移,我把他们写在如下
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124 ...</div></div></div><div class="recent-post-item"><div class="post_cover right"><a href="/2024/08/06/2024.8.6-Open%20node%20resources/" title="开放节点资源"><img class="post-bg" src="https://s2.loli.net/2024/08/06/qQ7E5zRvW9irIXx.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="开放节点资源"></a></div><div class="recent-post-info"><a class="article-title" href="/2024/08/06/2024.8.6-Open%20node%20resources/" title="开放节点资源">开放节点资源</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time class="post-meta-date-created" datetime="2024-08-05T18:33:22.000Z" title="Created 2024-08-06 02:33:22">2024-08-06</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">Updated</span><time class="post-meta-date-updated" datetime="2024-08-06T06:13:47.183Z" title="Updated 2024-08-06 14:13:47">2024-08-06</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E8%B5%84%E6%BA%90/">资源</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E8%B5%84%E6%BA%90/">资源</a></span></div><div class="content">开放分享的节点，点击节点页面可获取Linked：加入群组：Butterfly
节点页面：Butterfly
V2rayN: Butterfly
V2rayNG: Butterfly
Xray-core: Butterfly
</div></div></div><div class="recent-post-item"><div class="post_cover left"><a href="/2024/08/03/2024.8.3-Cheat%20Engine/" title="Cheat Engine编译"><img class="post-bg" src="https://s2.loli.net/2024/08/04/5UP8Fc2m3rtT1iA.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="Cheat Engine编译"></a></div><div class="recent-post-info"><a class="article-title" href="/2024/08/03/2024.8.3-Cheat%20Engine/" title="Cheat Engine编译">Cheat Engine编译</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time class="post-meta-date-created" datetime="2024-08-02T18:33:22.000Z" title="Created 2024-08-03 02:33:22">2024-08-03</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">Updated</span><time class="post-meta-date-updated" datetime="2024-08-03T17:21:35.378Z" title="Updated 2024-08-04 01:21:35">2024-08-04</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%8A%80%E6%9C%AF/">技术</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%8A%80%E6%9C%AF/">技术</a></span></div><div class="content">事情起因我想自己重新编译一个最新的Cheat Engine在配置文件和项目设置里改了一通。
点击编译，结果发现 ？？？？
Cheat Engine Release 64-Bit O4 AVX2 ,Error: Identifier not found “ExtractFileNameWithoutExt”
然后我上 stackoverflow 也没找到解决办法 最后在一则 Issues 找到了原因https://github.com/cheat-engine/cheat-engine/issues/2642
原来是环境问题 新版本的lazarus编译不起作用 ，我尝试了2.2.4版本的lazarus 编译正常通过
从Lazarus 2.2开始，FileUtil中的某些功能在较新的版本中已更改为LazFileUtils ，ExtractFileNameWithoutExt这个函数在 LazFileUtils 中，而不是在较新的 Lazaruses 中的 FileUtil 中
还是太粗心大意了，这就是不看文档的后果，踩到不该踩的坑
解决采用旧版本lazarus ，2.2.4 或2. ...</div></div></div><div class="recent-post-item"><div class="post_cover right"><a href="/2024/06/24/2024.7.29-C++%20linked%20list%20from%200%20to%201%20in%20practice/" title="C++ 链接表从 0 到 1 的实践应用"><img class="post-bg" src="/Src/C++-logo.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="C++ 链接表从 0 到 1 的实践应用"></a></div><div class="recent-post-info"><a class="article-title" href="/2024/06/24/2024.7.29-C++%20linked%20list%20from%200%20to%201%20in%20practice/" title="C++ 链接表从 0 到 1 的实践应用">C++ 链接表从 0 到 1 的实践应用</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time class="post-meta-date-created" datetime="2024-06-23T18:03:26.000Z" title="Created 2024-06-24 02:03:26">2024-06-24</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">Updated</span><time class="post-meta-date-updated" datetime="2024-08-01T10:47:03.800Z" title="Updated 2024-08-01 18:47:03">2024-08-01</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E7%BC%96%E7%A8%8B/">编程</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E7%BC%96%E7%A8%8B/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/">数据结构与算法</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E7%BC%96%E7%A8%8B/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/c-plus-plus/">C++</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/">数据结构与算法</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E7%BC%96%E7%A8%8B/">编程</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/C/">C++</a></span></div><div class="content">C++ 链接表从 0 到 1 的实践应用
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195 ...</div></div></div><div class="recent-post-item"><div class="post_cover left"><a href="/2024/06/05/2024.6.5-GPT%20Questions%20and%20Answers-1/" title="与GPT的题目问答1"><img class="post-bg" src="/Src/GPT.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="与GPT的题目问答1"></a></div><div class="recent-post-info"><a class="article-title" href="/2024/06/05/2024.6.5-GPT%20Questions%20and%20Answers-1/" title="与GPT的题目问答1">与GPT的题目问答1</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time class="post-meta-date-created" datetime="2024-06-04T16:00:00.000Z" title="Created 2024-06-05 00:00:00">2024-06-05</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">Updated</span><time class="post-meta-date-updated" datetime="2024-08-01T10:52:34.378Z" title="Updated 2024-08-01 18:52:34">2024-08-01</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%8A%80%E6%9C%AF/">技术</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%8A%80%E6%9C%AF/%E9%97%AE%E7%AD%94/">问答</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%8A%80%E6%9C%AF/">技术</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E9%97%AE%E7%AD%94/">问答</a></span></div><div class="content">容器的选择
GPT: 在什么情况下，你会选择使用std::vector而不是std::list？请解释你的理由。
答：在存储大量且连续的数据会使用 vector，因为vector的容量的扩充是 以幂次扩充的它特别适合存储大量且连续的数据，相对于list 它还有迭代器的随机访问能力，时间复杂度是O1,这是list做不到的,std::vector在需要频繁随机访问和插入/删除操作主要集中在末尾时性能更优。相比之下，std::list在需要频繁在中间插入或删除元素时更合适，因为它的插入和删除操作不涉及大量元素的移动。
GPT: 描述std::map和std::unordered_map的主要区别，以及它们各自的使用场景。
答：他们两个的区别仅仅在于 ，std::map是基于红黑树实现的，键值对有序且键不允许重复。std::unordered_map是基于哈希表实现的，键值对无序但键同样不允许重复。如果需要保持元素的排序顺序，应该使用std::map；如果不需要排序但更关注查找效率，可以使用std::unordered_map，它提供平均O(1)的查找复杂度
迭代器的类型
GPT: 请解 ...</div></div></div><div class="recent-post-item"><div class="post_cover right"><a href="/2024/06/02/2024.6.2-.leetcode-Roman%20numerals%20to%20integers/" title="LeetCode题解 13.罗马数字转整数"><img class="post-bg" src="/Src/LeetCode/LeetCode-logo2.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="LeetCode题解 13.罗马数字转整数"></a></div><div class="recent-post-info"><a class="article-title" href="/2024/06/02/2024.6.2-.leetcode-Roman%20numerals%20to%20integers/" title="LeetCode题解 13.罗马数字转整数">LeetCode题解 13.罗马数字转整数</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time class="post-meta-date-created" datetime="2024-06-01T18:33:22.000Z" title="Created 2024-06-02 02:33:22">2024-06-02</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">Updated</span><time class="post-meta-date-updated" datetime="2024-08-01T10:47:44.624Z" title="Updated 2024-08-01 18:47:44">2024-08-01</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/">数据结构与算法</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/LeetCode/">LeetCode</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/">数据结构与算法</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/LeetCode/">LeetCode</a></span></div><div class="content">13.罗马数字转整数
解题思路：
1.首先 既然是罗马转阿拉伯数字 那就是接受参数无外乎是字符串的 ‘I’,’V’,’X’,’L’,’C’,’D’,’M’
2.然后将接受的字符串 转换成 字符数组
3.接下来很自然的用到 switch() ,将罗马数字按照依次判断，然后将对应的阿拉伯数组 放在对应整形数组中
通常情况下，罗马数字中小的数字在大的数字的右边。但也存在特例，例如 4 不写做 IIII，而是 IV。数字 1 在数字 5 的左边，所表示的数等于大数 5 减小数 1 得到的数值 4 。同样地，数字 9 表示为 IX。
4.IV是表示4 ， 相当于 V 减去 前面的 I，这样就可以转换阿拉伯数字后，按照这样运算就能求出正确结果，只需要判断 if(num[n] < num[n+1]) 如果成立，则需要减去 num[n] 的数，如果不成立则需要把 num[n] 加进sum
5.最后再累加整形数组中的最后一位元素,返回sum
1234567891011121314151617181920212223242526272829303132333435class S ...</div></div></div><div class="recent-post-item"><div class="post_cover left"><a href="/2024/06/01/2024.6.1-leetcode-Search%20Insertion%20Location/" title="LeetCode题解 35.搜索插入位置"><img class="post-bg" src="/Src/LeetCode/LeetCode-logo1.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="LeetCode题解 35.搜索插入位置"></a></div><div class="recent-post-info"><a class="article-title" href="/2024/06/01/2024.6.1-leetcode-Search%20Insertion%20Location/" title="LeetCode题解 35.搜索插入位置">LeetCode题解 35.搜索插入位置</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time class="post-meta-date-created" datetime="2024-05-31T18:33:22.000Z" title="Created 2024-06-01 02:33:22">2024-06-01</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">Updated</span><time class="post-meta-date-updated" datetime="2024-08-01T10:47:59.179Z" title="Updated 2024-08-01 18:47:59">2024-08-01</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/">数据结构与算法</a><i class="fas fa-angle-right article-meta-link"></i><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/LeetCode/">LeetCode</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E6%95%B0%E6%8D%AE%E7%BB%93%E6%9E%84%E4%B8%8E%E7%AE%97%E6%B3%95/">数据结构与算法</a><span class="article-meta-link">•</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/LeetCode/">LeetCode</a></span></div><div class="content">35. 搜索插入位置
解题思路：
1.函数传入 2个参数 ，一个是vector容器以引用传递 ， 一个是target目标值
2.用for循环遍历容器 去判断当前循环的nums[i] 是否 == target ,成立则返回其索引，反之则判断 target < nums[i] 如成立则 插入到 nums.begin{} + i 处的索引，并返回索引数值
3.循环外是末尾添加的情况 ，在nums数组末尾添加target，并返回最后数组的下标
123456789101112131415161718class Solution {public: int searchInsert(vector<int> &nums, int target) { for (int i = 0 ; i < nums.size() ; ++i) { if (nums[i] == target) { ...</div></div></div><div class="recent-post-item"><div class="post_cover right"><a href="/2024/03/05/2024.7.29-Self-Regulatory%20Convention/" title="自我约束"><img class="post-bg" src="/Src/2024.7.29-Self-Regulatory%20Convention.jpg" onerror="this.onerror=null;this.src='/img/404.jpg'" alt="自我约束"></a></div><div class="recent-post-info"><a class="article-title" href="/2024/03/05/2024.7.29-Self-Regulatory%20Convention/" title="自我约束">自我约束</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time class="post-meta-date-created" datetime="2024-03-04T18:23:07.000Z" title="Created 2024-03-05 02:23:07">2024-03-05</time><span class="article-meta-separator">|</span><i class="fas fa-history"></i><span class="article-meta-label">Updated</span><time class="post-meta-date-updated" datetime="2024-07-28T19:23:13.211Z" title="Updated 2024-07-29 03:23:13">2024-07-29</time></span><span class="article-meta"><span class="article-meta-separator">|</span><i class="fas fa-inbox"></i><a class="article-meta__categories" href="/categories/%E5%A3%B0%E6%98%8E/">声明</a></span><span class="article-meta tags"><span class="article-meta-separator">|</span><i class="fas fa-tag"></i><a class="article-meta__tags" href="/tags/%E5%A3%B0%E6%98%8E/">声明</a></span></div><div class="content">公约
每日任务：
每日一题：每天至少完成一道数据结构或算法题。可以从LeetCode、HackerRank、CodeSignal等平台选择题目。
学习笔记：完成题目后，写一篇简短的学习笔记，记录解题思路、遇到的困难及解决方法。
长期任务：
每月总结：每月末总结本月完成的题目，分析进步和需要改进的地方。
专项突破：每月选择一个数据结构或算法专题（如排序算法、图算法、动态规划等）进行深入学习，并至少完成相关的10道题目。
惩罚机制
每日惩罚：若未完成每日任务，逐次减少1小时娱乐时间，如达到2次以上失去所有娱乐时间，将任务完成即恢复。
每周惩罚：若未完成一周目标，周末取消一次休闲活动
Habitica
需要搭配 Habitica使用 ，设定自定义的任务
工具
地址
Habitica
https://habitica.com/
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