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<rss xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:atom="http://www.w3.org/2005/Atom" xmlns:media="http://search.yahoo.com/mrss/" version="2.0"><channel><title>作者：lbr - Maochong</title><link>https://lbr.ink/authors/admin</link><atom:link href="https://lbr.ink/authors/admin/feed/authors/admin.xml" rel="self" type="application/rss+xml"/><description>Maochong</description><generator>Halo v2.25.0</generator><language>zh-cn</language><image><url>https://lbr.ink/upload/icon-Huqp.png</url><title>作者：lbr - Maochong</title><link>https://lbr.ink/authors/admin</link></image><lastBuildDate>Mon, 15 Jun 2026 11:45:11 GMT</lastBuildDate><item><title><![CDATA[Words Lists]]></title><link>https://lbr.ink/archives/wei-ming-ming-wen-zhang-Jgv70fHd</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=Words%20Lists&amp;url=/archives/wei-ming-ming-wen-zhang-Jgv70fHd" width="1" height="1" alt="" style="opacity:0;">]]></description><guid isPermaLink="false">/archives/wei-ming-ming-wen-zhang-Jgv70fHd</guid><dc:creator>lbr</dc:creator><pubDate>Sun, 10 May 2026 12:02:04 GMT</pubDate></item><item><title><![CDATA[Plan Holidays]]></title><link>https://lbr.ink/archives/wei-ming-ming-wen-zhang-xrEexWZm</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=Plan%20Holidays&amp;url=/archives/wei-ming-ming-wen-zhang-xrEexWZm" width="1" height="1" alt="" style="opacity:0;">]]></description><guid isPermaLink="false">/archives/wei-ming-ming-wen-zhang-xrEexWZm</guid><dc:creator>lbr</dc:creator><pubDate>Fri, 8 May 2026 05:38:10 GMT</pubDate></item><item><title><![CDATA[Endless holiday]]></title><link>https://lbr.ink/archives/wei-ming-ming-wen-zhang-28IHP9zp</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=Endless%20holiday&amp;url=/archives/wei-ming-ming-wen-zhang-28IHP9zp" width="1" height="1" alt="" style="opacity:0;">]]></description><guid isPermaLink="false">/archives/wei-ming-ming-wen-zhang-28IHP9zp</guid><dc:creator>lbr</dc:creator><pubDate>Wed, 6 May 2026 15:07:01 GMT</pubDate></item><item><title><![CDATA[操作系统笔记]]></title><link>https://lbr.ink/archives/wei-ming-ming-wen-zhang-9BDw47tq</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=%E6%93%8D%E4%BD%9C%E7%B3%BB%E7%BB%9F%E7%AC%94%E8%AE%B0&amp;url=/archives/wei-ming-ming-wen-zhang-9BDw47tq" width="1" height="1" alt="" style="opacity:0;">]]></description><guid isPermaLink="false">/archives/wei-ming-ming-wen-zhang-9BDw47tq</guid><dc:creator>lbr</dc:creator><category>笔记</category><pubDate>Mon, 13 Apr 2026 15:23:42 GMT</pubDate></item><item><title><![CDATA[C 属性大爆发]]></title><link>https://lbr.ink/archives/c-shu-xing-da-bao-fa</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=C%20%E5%B1%9E%E6%80%A7%E5%A4%A7%E7%88%86%E5%8F%91&amp;url=/archives/c-shu-xing-da-bao-fa" width="1" height="1" alt="" style="opacity:0;">null]]></description><guid isPermaLink="false">/archives/c-shu-xing-da-bao-fa</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Fri, 10 Apr 2026 05:15:46 GMT</pubDate></item><item><title><![CDATA[编译原理笔记]]></title><link>https://lbr.ink/archives/bian-yi-yuan-li-bi-ji</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=%E7%BC%96%E8%AF%91%E5%8E%9F%E7%90%86%E7%AC%94%E8%AE%B0&amp;url=/archives/bian-yi-yuan-li-bi-ji" width="1" height="1" alt="" style="opacity:0;">语言处理过程 源程序 --&gt; 编译器 --&gt; 目标汇编程序 --&gt; 汇编器 --&gt; 可重定位机器代码 --&gt; 连接器 --&gt; 机器码 翻译程序 把某一种语言程序等价地转换成另一种语言程序（目标语言程序）的程序 编译程序 把一种高级语言程序等价地转换成另一种低级语言程序的程序 解释程序 把源语言的源程]]></description><guid isPermaLink="false">/archives/bian-yi-yuan-li-bi-ji</guid><dc:creator>lbr</dc:creator><category>笔记</category><pubDate>Fri, 19 Dec 2025 00:00:00 GMT</pubDate></item><item><title><![CDATA[Double Sum 2]]></title><link>https://lbr.ink/archives/double-sum-2</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=Double%20Sum%202&amp;url=/archives/double-sum-2" width="1" height="1" alt="" style="opacity:0;">Double Sum 2 根据lowbit不同进行分组, 不同组间直接算, 同一组间用 0-1 trie #include &lt;bits/stdc++.h&gt; using namespace std; using ll = long long; #define int long long #define]]></description><guid isPermaLink="false">/archives/double-sum-2</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Wed, 17 Sep 2025 00:00:00 GMT</pubDate></item><item><title><![CDATA[MEX Queries]]></title><link>https://lbr.ink/archives/mex-queries</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=MEX%20Queries&amp;url=/archives/mex-queries" width="1" height="1" alt="" style="opacity:0;">MEX Queries You are given a set of integer numbers, initially it is empty. You should perform n queries. There are three different types of queries: 1]]></description><guid isPermaLink="false">/archives/mex-queries</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Wed, 13 Aug 2025 00:00:00 GMT</pubDate></item><item><title><![CDATA[Prefix LIS Query]]></title><link>https://lbr.ink/archives/prefix-lis-query</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=Prefix%20LIS%20Query&amp;url=/archives/prefix-lis-query" width="1" height="1" alt="" style="opacity:0;">Prefix LIS Query 把询问离线之后就变成了一个二分求最长上升子序列的问题，参考导弹拦截 求上升子序列 用一个数组f[i]记录 目前长度为i的结尾最小值 dp[i]记录以当前第i位结尾的最长上升子序列 f[i]随着i的增加单调递增，可以二分找位置 #include &lt;bits/stdc+]]></description><guid isPermaLink="false">/archives/prefix-lis-query</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Mon, 28 Jul 2025 00:00:00 GMT</pubDate></item><item><title><![CDATA[Development]]></title><link>https://lbr.ink/archives/development</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=Development&amp;url=/archives/development" width="1" height="1" alt="" style="opacity:0;">Development 题目大意是给一个无向图，只有500个点代表城市，一些双向边代表路，边权代表通行时间。 还有一些城市有飞机场，给定飞机场之间的通行时间都一样为t。 给q次询问，三种op,op=1,加边;op=2,加机场,op=3,查询所有城市两两通行的时间和，无法通行记为0 q=1000 数据]]></description><guid isPermaLink="false">/archives/development</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Sun, 27 Jul 2025 00:00:00 GMT</pubDate></item><item><title><![CDATA[Jump Distance Sum]]></title><link>https://lbr.ink/archives/jump-distance-sum</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=Jump%20Distance%20Sum&amp;url=/archives/jump-distance-sum" width="1" height="1" alt="" style="opacity:0;">Jump Distance Sum 转化后可以发现对于 x + y 为偶数的，只能跳到 x + y 为偶数的点上，因为每次跳跃改变 x + y 的值只能为 +2、-2 或 0。 那对于 x + y 分奇偶求。 观察发现这个最短距离等价于切比雪夫距离： \max]]></description><guid isPermaLink="false">/archives/jump-distance-sum</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Sun, 27 Jul 2025 00:00:00 GMT</pubDate></item><item><title><![CDATA[苹果树]]></title><link>https://lbr.ink/archives/ping-guo-shu</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=%E8%8B%B9%E6%9E%9C%E6%A0%91&amp;url=/archives/ping-guo-shu" width="1" height="1" alt="" style="opacity:0;">显然树链剖分，但是操作2暴力维护显然会T，发现一个性质，对于一条重链上，任意一个非链顶节点，其父亲的 重儿子必为该节点。 这启发我们，每次修改就只需要对父亲 Fa𝑥 与重儿子 son𝑥 进行单点修改。所有非链顶节点都可以正确更新，唯独链顶节点 不能正确更新。 不难发现，单点查询可以轻松 𝑂(1]]></description><guid isPermaLink="false">/archives/ping-guo-shu</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Mon, 21 Jul 2025 00:00:00 GMT</pubDate></item><item><title><![CDATA[中位数]]></title><link>https://lbr.ink/archives/zhong-wei-shu</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=%E4%B8%AD%E4%BD%8D%E6%95%B0&amp;url=/archives/zhong-wei-shu" width="1" height="1" alt="" style="opacity:0;">2025CC福建邀请 中位数的一个trick 大于等于某个数设置为1，小于设置为0 这里需要二分选择某个数作为可能的最大值，然后尽可能减少0的数量，最终比较0和1的数量 发现只有000 和 00100这样的情况有可能减少0和1的数量差值，其他情况操作0和1的数量差值都不会改变 用栈维护 有一个地方]]></description><guid isPermaLink="false">/archives/zhong-wei-shu</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Sun, 20 Jul 2025 00:00:00 GMT</pubDate></item><item><title><![CDATA[整除分块]]></title><link>https://lbr.ink/archives/zheng-chu-fen-kuai</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=%E6%95%B4%E9%99%A4%E5%88%86%E5%9D%97&amp;url=/archives/zheng-chu-fen-kuai" width="1" height="1" alt="" style="opacity:0;">ABC414E Count A%B=C 题目描述 请计算满足以下条件的整数三元组 $(a, b, c)$ 的个数，并将结果对 $998244353$ 取模。 $1 \leq a, b, c \leq N$。 $a, b, c$ 互不相同。 $a$ 除以 $b$ 的余数等于 $c$。 输入格式 输入从]]></description><guid isPermaLink="false">/archives/zheng-chu-fen-kuai</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Sun, 20 Jul 2025 00:00:00 GMT</pubDate></item><item><title><![CDATA[二维数点]]></title><link>https://lbr.ink/archives/er-wei-shu-dian</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=%E4%BA%8C%E7%BB%B4%E6%95%B0%E7%82%B9&amp;url=/archives/er-wei-shu-dian" width="1" height="1" alt="" style="opacity:0;">前置知识--二维数点 P10814 【模板】离线二维数点 前缀和配合树状数组 #include &lt;bits/stdc++.h&gt; using namespace std; using ll = long long; #define _ cout &lt;&lt; "----------" &lt;&lt; endl #def]]></description><guid isPermaLink="false">/archives/er-wei-shu-dian</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Thu, 10 Jul 2025 00:00:00 GMT</pubDate></item><item><title><![CDATA[F - Balanced Rectangles]]></title><link>https://lbr.ink/archives/f---balanced-rectangles</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=F%20-%20Balanced%20Rectangles&amp;url=/archives/f---balanced-rectangles" width="1" height="1" alt="" style="opacity:0;">Balanced Rectangles Problem Statement You are given an $H \times W$ grid, where each cell contains # or .. The information about the symbols written i]]></description><guid isPermaLink="false">/archives/f---balanced-rectangles</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Tue, 24 Jun 2025 00:00:00 GMT</pubDate></item><item><title><![CDATA[Popcount Sum 3]]></title><link>https://lbr.ink/archives/popcount-sum-3</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=Popcount%20Sum%203&amp;url=/archives/popcount-sum-3" width="1" height="1" alt="" style="opacity:0;">问题陈述 E - Popcount Sum 3 给你正整数 $N$ 和 $K$ 。 求所有不超过 $N$ 且满足以下条件的正整数 $x$ 的和，取模 $998244353$ ： $x$ 的 popcount 恰好是 $K$ 。 给你 $T$ 个测试用例，请逐个求解。 什么是 popcount？对于正]]></description><guid isPermaLink="false">/archives/popcount-sum-3</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Thu, 12 Jun 2025 12:07:00 GMT</pubDate></item><item><title><![CDATA[Sums of Sliding Window Maximum]]></title><link>https://lbr.ink/archives/sums-of-sliding-window-maximum</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=Sums%20of%20Sliding%20Window%20Maximum&amp;url=/archives/sums-of-sliding-window-maximum" width="1" height="1" alt="" style="opacity:0;">https://atcoder.jp/contests/abc407/tasks/abc407_f 对于每个a[i]点算 对答案区间的贡献，发现是一个两个等差数列区间， 最后跑两遍前缀和即可 #include &lt;bits/stdc++.h&gt; using namespace std; using ll]]></description><guid isPermaLink="false">/archives/sums-of-sliding-window-maximum</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Wed, 11 Jun 2025 16:00:00 GMT</pubDate></item><item><title><![CDATA[Incessant Rain]]></title><link>https://lbr.ink/archives/incessant-rain</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=Incessant%20Rain&amp;url=/archives/incessant-rain" width="1" height="1" alt="" style="opacity:0;">H. Incessant Rain 线段树离线 一个经典的问题 然后是线段树区间合并维护区间最大子段和 记录从前缀连续最大，后缀连续最大，区间连续最大 然后合并的时候维护即可 对于某个数x的k-majority,可以把原数组转化为 等于x转化为1，否则-1，然后最大k-majority就是这一段的最]]></description><guid isPermaLink="false">/archives/incessant-rain</guid><dc:creator>lbr</dc:creator><pubDate>Tue, 10 Jun 2025 00:00:00 GMT</pubDate></item><item><title><![CDATA[宝可梦图鉴]]></title><link>https://lbr.ink/archives/bao-ke-meng-tu-jian</link><description><![CDATA[<img src="https://lbr.ink/plugins/feed/assets/telemetry.gif?title=%E5%AE%9D%E5%8F%AF%E6%A2%A6%E5%9B%BE%E9%89%B4&amp;url=/archives/bao-ke-meng-tu-jian" width="1" height="1" alt="" style="opacity:0;">宝可梦图鉴 珂朵莉树 + 权值线段树 赛时想到了珂朵莉树的做法但是不知道这个板子 可惜了 一般的珂朵莉树，查询也是暴力遍历每个区间，因此有可能会被卡掉 这个做法结合了珂朵莉树的优点，由于查询是在权值线段树上的，所以不会被卡掉 #include &lt;bits/stdc++.h&gt; using namesp]]></description><guid isPermaLink="false">/archives/bao-ke-meng-tu-jian</guid><dc:creator>lbr</dc:creator><category>题解</category><pubDate>Sun, 8 Jun 2025 00:00:00 GMT</pubDate></item></channel></rss>