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链接直达~~（并非一定不可用）</del></a></td>~~	100% vibe coding 构建的个人网站。内置 AI 小模型， vibe coder 的好帮手。服务器日常关机。</del></td></tr> </tbody></table> Hello Zola! Hello oak.md!! 2026-04-19T19:23:00+08:00 我的博客上线了！这是一个测试页面。</p> 数学公式</h3> 这里的公式 $a^2 + b^2 = c^2$ 会在行内显示。 $$ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} $$</p> 图片</h3> </p> 代码块</h3> C++14 code:</p> using </span> ll R</span></span> ll tot </span> cin tot </span> Java17 code:</p> </span> public </span> mp</span> </span> </span> System</span> Rust code:</p> </span> pub phpsessid</span> id</span> phpsessid<&# id</span> </p> [f44] 防止 PipeWire 发现小米电视 2026-02-21T00:00:00+00:00 #1E1E2E;">#include</span> <bits/stdc++.h></span></span> #define</span> ll</span> long long</span></span> namespace</span> std</span>;</span></span> char</span> mat</span>[</span>110</span>][</span>110</span>];</span></span> style="color: #9399B2;">,</span> C</span>,</span> K</span>;</span></span> int</span> main</span>()</span></span> {</span></span> italic;"> scanf</span>(</span>"%lld %lld %lld"</span>,</span> &</span>R</span>,</span> &</span>C</span>,</span> &</span>K</span>);</span></span> style="color: #94E2D5;">=</span> 0</span>;</span></span> for</span> (</span>ll i </span>=</span> 0</span>;</span> i </span><</span> R</span>;</span> i</span>++</span>)</span></span> for</span> (</span>ll j </span>=</span> 0</span>;</span> j </span><</span> C</span>;</span> j</span>++</span>)</span></span> </span>>></span> mat</span>[</span>i</span>][</span>j</span>];</span></span> style="color: #94E2D5;">=</span> R </span></span> (</span>C </span>-</span> K </span>+</span> 1</span>)</span> +</span> (</span>R </span>-</span> K </span>+</span> 1</span>)</span> </span> C</span>;</span></span> italic;"> // subtask1</span></span> for</span> (</span>ll i </span>=</span> 0</span>;</span> i </span><</span> R</span>;</span> i</span>++</span>)</span></span> for</span> (</span>ll j </span>=</span> 0</span>;</span> j </span><=</span> C </span>-</span> K</span>;</span> j</span>++</span>)</span></span> for</span> (</span>ll pos </span>=</span> j</span>;</span> pos </span><</span> j </span>+</span> K</span>;</span> pos</span>++</span>)</span></span> if</span> (</span>mat</span>[</span>i</span>][</span>pos</span>]</span> !=</span> '.'</span>)</span></span> {</span></span> tot</span>--</span>;</span></span> break</span>;</span></span> }</span></span> italic;"> // subtask2</span></span> for</span> (</span>ll i </span>=</span> 0</span>;</span> i </span><=</span> R </span>-</span> K</span>;</span> i</span>++</span>)</span></span> for</span> (</span>ll j </span>=</span> 0</span>;</span> j </span><</span> C</span>;</span> j</span>++</span>)</span></span> for</span> (</span>ll pos </span>=</span> i</span>;</span> pos </span><</span> i </span>+</span> K</span>;</span> pos</span>++</span>)</span></span> if</span> (</span>mat</span>[</span>pos</span>][</span>j</span>]</span> !=</span> '.'</span>)</span></span> {</span></span> tot</span>--</span>;</span></span> break</span>;</span></span> }</span></span> if</span> (</span>K </span>==</span> 1</span>)</span></span> return</span> printf</span>(</span>"%lld</span> \n</span>"</span>,</span> tot </span>/</span> 2</span>),</span> 0</span>;</span></span> italic;"> printf</span>(</span>"%lld</span> \n</span>"</span>,</span> tot</span>);</span></span> return</span> 0</span>;</span></span> }</span></span></code></pre> #1E1E2E;">import</span> java</span>.</span>util</span>.</span>Scanner</span>;</span></span> class</span> Main</span> {</span></span> public static void</span> main</span>(</span>String</span>[]</span> args</span>) {</span></span> Scanner</span> input</span> =</span> new</span> Scanner</span>(</span>System</span>.</span>in</span>);</span></span> int</span> N</span> =</span> input</span>.</span>nextInt</span>();</span></span> int</span> size</span> =</span> N </span>+</span> 1</span>;</span></span> int</span>[][]</span> mp</span> =</span> new int</span>[</span>size</span>][</span>size</span>];</span></span> int</span>[][][]</span> dp</span> =</span> new int</span>[</span>size </span>+</span> N</span>][</span>size</span>][</span>size</span>];</span></span> while</span> (</span>true</span>) {</span></span> int</span> x</span> =</span> input</span>.</span>nextInt</span>(),</span> y </span>=</span> input</span>.</span>nextInt</span>(),</span> v </span>=</span> input</span>.</span>nextInt</span>();</span></span> if</span> (</span>x </span>==</span> 0</span> &&</span> y </span>==</span> 0</span> &&</span> v </span>==</span> 0</span>)</span> break</span>;</span></span> style="color: #9399B2;">[</span>x</span>][</span>y</span>]</span> =</span> v</span>;</span></span> }</span></span> for</span> (</span>int</span> i</span> =</span> 2</span>;</span> i </span><=</span> N </span></span> 2</span>;</span> i </span>++</span>) {</span></span> for</span> (</span>int</span> j</span> =</span> 1</span>;</span> j </span><=</span> N</span>;</span> j </span>++</span>) {</span></span> for</span> (</span>int</span> k</span> =</span> 1</span>;</span> k </span><=</span> N</span>;</span> k </span>++</span>) {</span></span> int</span> j1</span> =</span> i </span>-</span> j</span>,</span> j2 </span>=</span> i </span>-</span> k</span>;</span></span> if</span> (</span>j1 </span>>=</span> 1</span> &&</span> j1 </span><=</span> N </span>&&</span> j2 </span>>=</span> 1</span> &&</span> j2 </span><=</span> N</span>) {</span></span> int</span> t</span> =</span> mp</span>[</span>j</span>][</span>j1</span>];</span></span> if</span> (</span>j </span>!=</span> k</span>)</span> t </span>+=</span> mp</span>[</span>k</span>][</span>j2</span>];</span></span> dp</span>[</span>i</span>][</span>j</span>][</span>k</span>]</span> =</span> Math</span>.</span>max</span>(</span>dp</span>[</span>i </span>-</span> 1</span>][</span>j </span>-</span> 1</span>][</span>k </span>-</span> 1</span>]</span> +</span> t</span>,</span> Math</span>.</span>max</span>(</span>dp</span>[</span>i </span>-</span> 1</span>][</span>j </span>-</span> 1</span>][</span>k</span>]</span> +</span> t</span>,</span></span> Math</span>.</span>max</span>(</span>dp</span>[</span>i </span>-</span> 1</span>][</span>j</span>][</span>k </span>-</span> 1</span>]</span> +</span> t</span>,</span> dp</span>[</span>i </span>-</span> 1</span>][</span>j</span>][</span>k</span>]</span> +</span> t</span>)));</span></span> }</span></span> }</span></span> }</span></span> }</span></span> style="color: #9399B2;">.</span>out</span>.</span>println</span>(</span>dp</span>[</span>N </span>+</span> N</span>][</span>N</span>][</span>N</span>]);</span></span> }</span></span> }</span></span></code></pre> #1E1E2E;">use super</span>::</span>User</span>;</span></span> italic;">#</span>[</span>derive</span>(</span>Debug</span>,</span> Clone</span>)]</span></span> struct</span> AuthUser</span> {</span></span> style="color: #94E2D5;">:</span> String</span>,</span></span> style="color: #94E2D5;">:</span> String</span>,</span></span> }</span></span> impl</span> AuthUser</span> {</span></span> pub fn</span> new</span>(</span>phpsessid</span>: &</span>str</span>)</span> -></span> anyhow</span>::</span>Result</span><</span>Self</span>></span> {</span></span> italic;"> Ok</span>(</span>Self</span> {</span></span> x2F;span>:</span> phpsessid</span>.</span>to_owned</span>(),</span></span> style="color: #94E2D5;">:</span> if let</span> Some</span>((</span>id</span>,</span> _</span>))</span> =</span> phpsessid</span>.</span>split_once</span>('</span>_</span>') {</span></span> id</span>.</span>to_owned</span>()</span></span> }</span> else</span> {</span></span> italic;"> panic!</span>(</span>"invalid phpsessid"</span>);</span></span> },</span></span> })</span></span> }</span></span> pub fn</span> phpsessid</span>(</span>&</span>self</span>)</span> -> &</span>str</span> {</span></span> &</span>self</span>.</span>phpsessid</span></span> }</span></span> }</span></span> impl</span> User</span> for</span> AuthUser</span> {</span></span> fn</span> id</span>(</span>&</span>self</span>)</span> -> &</span>str</span> {</span></span> &</span>self</span>.</span>id</span></span> }</span></span> }</span></span></code></pre> 你不会喜欢片的声音自动传到客厅电视，对吧。</p> 为了永远不发生这样的事情，应该按下面的步骤，禁用网络发现！</p> 1. 创建配置文件目录</h3> mkdir</span> -p ~/.config/pipewire/pipewire.conf.d</span></span> cd</span> ~/.config/pipewire/pipewire.conf.d</span></span></code></pre>2. 写入配置</h3> 把下面的配置写到当前目录的 99-disable-network-sink.conf</code> 文件里，你应该创建这个文件。</p> context.properties = {</span></span> # 禁用 AirPlay 发现 (通常是小米电视出现的来源)</span></span> module.raop = false</span></span> </span></span> # 禁用 Zeroconf/Avahi 发现 (Chromecast/DNLA 等)</span></span> module.zeroconf.discover = false</span></span> }</span></span></code></pre>3. 重启音频服务</h3> systemctl</span> --user restart pipewire</span></span></code></pre> 然后音频设备列表里的小米电视就永远消失了，耶！</p> 4. 反安装</h3> 你喜欢片的声音自动传到客厅电视，对吧。</p> 按照下面的步骤，启用网络发现。 :-)</p> 删除碍事的配置 rm ~/.config/pipewire/pipewire.conf.d/99-disable-network-sink.conf</code></li> 重启音频服务 systemctl --user restart pipewire</code></li> </ol> 题解：P11232 [CSP-S 2024] 超速检测（民间数据） 2024-10-26T23:06:00+08:00 挺麻烦的一个题，赛时调了 $2h$ 。</p> 先写暴力</h3> $O(nm)$ 算出来第一个输出，然后 $O(2^mnm)$ 枚举删哪些测速点，可以骗到一点分。</p> 考虑优化掉 $O(nm)$</h3> 对 $p$ 从小到大排序，设 $v_{i, q}$ 是第 $i$ 辆车在 $q$ 点的速度，则对于第 $i$ 辆车 $A$ ：</p> 如果 $a_i<0$， $A$ 超速当且仅当 $v_{i, P}>V$ ，其中 $P$ 是最小的 $p_j$ 满足 $p_j\space{\geq}{\space}d_i$</p> </li> 如果 $a_i\space{\geq}\space0$， $A$ 超速当且仅当 $v_{i, p_m}>V$</p> </li> </ul> 对于 $a_i<0$ ，显然可以二分，整体时间复杂度 $O(n\log{m})$ 。</p> 考虑优化掉 $O(2^mnm)$</h3> $p$ 是有序的，可以把每辆车的测超速范围看成一个线段 $X$ ，具体来说，对于第 $i$ 辆车：</p> 如果 $a_i<0$， $X_i=[P,Q]$ ，其中 $P$ 是最小的 $j$ 满足 $v_{i, p_j}>V$ ， $Q$ 是最大的 $j$ 满足 $v_{i, p_j}>V$ 。</li> 如果 $a_i\space{\geq}\space0$， $X_i=[P,m]$ ，其中 $P$ 是最小的 $j$ 满足 $v_{i, p_j}>V$ 。</li> </ul> 因为 $v_{i, p_j}$ 在 $p_j\space{\geq}{\space}d_i$ 上具有单调性，所以二分求 $X$，时间复杂度 $O(n\log{m})$ 。</p> 此时问题转化成 P1803 线段覆盖，整体时间复杂度 $O(n\log{m})$ 。</p> 代码略（</p> 本文最初发表于 Luogu</a>，于 2026-04-19 同步至本站。</p> 题面： https://www.luogu.com.cn/problem/P11232</a></p> 题解：P11231 [CSP-S 2024] 决斗（民间数据） 2024-10-26T22:02:00+08:00 因为攻击顺序可以随意指定，那么 $r$ 的顺序就无所谓了。</p> 一看数据范围 ${10}^5$ ，不妨给 $r$ 排个序，发现 $r_i$ 能打败 $r_j$ 仅当 $j<i$ 。</p> 那么维护一个队列，把排序的 $r$ 依次放进去，每次让 $r_i$ 攻击队首。</p> 最后看看队里剩下几个数就是答案。</p> #include</span> <bits/stdc++.h></span></span> #define</span> i64</span> long long</span></span> </span> const</span> i64 MAXN </span>=</span> 1</span>e</span>5</span> +</span> 64</span>;</span></span> </span> std</span>::</span>queue</span><</span>i64</span>></span> q</span>;</span></span> </span> i64 a</span>[</span>MAXN</span>],</span> n</span>;</span></span> </span> void</span> solve</span>()</span></span> {</span></span> assert</span>(</span>scanf</span>(</span>"%lld"</span>,</span> &</span>n</span>));</span></span> for</span> (</span>i64 i </span>=</span> 1</span>;</span> i </span><=</span> n</span>;</span> ++</span>i</span>)</span> assert</span>(</span>scanf</span>(</span>"%lld"</span>,</span> &</span>a</span>[</span>i</span>]));</span></span> std</span>::</span>sort</span>(</span>a </span>+</span> 1</span>,</span> a </span>+</span> 1</span> +</span> n</span>,</span> std</span>::</span>less</span><</span>i64</span>>());</span></span> for</span> (</span>i64 i </span>=</span> 1</span>;</span> i </span><=</span> n</span>;</span> ++</span>i</span>)</span></span> {</span></span> if</span> (</span>!</span>q</span>.</span>empty</span>()</span> &&</span> q</span>.</span>front</span>()</span> <</span> a</span>[</span>i</span>])</span> q</span>.</span>pop</span>();</span></span> q</span>.</span>push</span>(</span>a</span>[</span>i</span>]);</span></span> }</span></span> printf</span>(</span>"%lld</span>\n</span>"</span>, (</span>i64</span>)</span>q</span>.</span>size</span>());</span></span> }</span></span> </span> signed</span> main</span>()</span></span> {</span></span> solve</span>();</span></span> return</span> 0</span>;</span></span> }</span></span> </span></code></pre> 很难想到这其实就是 $r$ 的众数在 $r$ 中出现的次数，所以有 $O(n)$ 做法。</p> 本文最初发表于 Luogu</a>，于 2026-04-19 同步至本站。</p> 题面： https://www.luogu.com.cn/problem/P11231</a></p> 题解：P11214 【MX-J8-T2】黑洞 2024-10-20T23:28:00+08:00 非常难的一道题，想了 $150$ 分钟。（怎么降黄了</p> 根据题意，对每一维同时增加或减少一个正整数 $C$，能得到被黑洞吞噬的点 $A$。</p> 把每一维对应成一根竹竿，表示能够增减的范围</strong>，以样例 #3</strong> 为例：</p> </p> 方便起见，调换这些竹竿的方向，把距离 $0$ 较长的一端放在上面，按另一端的长度排序。</p> 因为有竹竿 $[3, -1]$ ，所以 $C \leq 3$，那么对于所有竹竿，绝对值大于 $3$ 的部分是没用的。</p> 也就是说，对于所有输入，可以转化成 $n$ 个竹竿，其中第 $i$ 个竹竿上端长度为 $L$，下端长度为 $x_i$，保证 $x_{i-1} \le x_i$ 且 $x_i \le L$ 。</p> 考虑第 $i$ 个竹竿对 $A$ 取值数量的贡献 $p_i$ ，可以写成： $$ p_i = 2^{n - i + 1}(x_i-x_{i-1}) $$ 答案是 $1 + \Sigma {p} + (L-x_n)$ ，其中 $1$ 是输入的点本身，$L-x_n$ 是 $C > x_n$ 的情况。</p> 时间复杂度 $O(n \log n)$ ，瓶颈在排序，代码不难写：</p> #include</span> <bits/stdc++.h></span></span> #define</span> u64</span> unsigned long long</span></span> #define</span> inf</span> 0x</span>3f3f3f3f3f3f3f3f</span></span> #define</span> i64</span> long long</span></span> </span> const</span> i64 MAXN </span>=</span> 2</span>e</span>5</span> +</span> 64</span>;</span></span> </span> const</span> i64 MOD </span>=</span> 1000000007</span>;</span></span> #define</span> mod</span>(</span>x</span>) ((</span>x</span>)</span> %</span> MOD</span>)</span></span> </span> i64 m</span>[</span>MAXN</span>],</span> a</span>[</span>MAXN</span>],</span> n</span>;</span></span> </span> i64 x</span>[</span>MAXN</span>],</span> L </span>=</span> inf</span>,</span> sigp</span>;</span></span> </span> void</span> solve</span>()</span></span> {</span></span> assert</span>(</span>scanf</span>(</span>"%lld"</span>,</span> &</span>n</span>));</span></span> for</span> (</span>i64 i </span>=</span> 1</span>;</span> i </span><=</span> n</span>;</span> ++</span>i</span>)</span> assert</span>(</span>scanf</span>(</span>"%lld"</span>,</span> &</span>m</span>[</span>i</span>]));</span></span> for</span> (</span>i64 i </span>=</span> 1</span>;</span> i </span><=</span> n</span>;</span> ++</span>i</span>)</span> assert</span>(</span>scanf</span>(</span>"%lld"</span>,</span> &</span>a</span>[</span>i</span>]));</span></span> for</span> (</span>i64 i </span>=</span> 1</span>,</span> l</span>;</span> i </span><=</span> n</span>;</span> ++</span>i</span>)</span></span> {</span></span> l </span>=</span> m</span>[</span>i</span>]</span> -</span> a</span>[</span>i</span>],</span> x</span>[</span>i</span>]</span> =</span> a</span>[</span>i</span>]</span> -</span> 1</span>;</span></span> if</span> (</span>l </span><</span> x</span>[</span>i</span>])</span> std</span>::</span>swap</span>(</span>l</span>,</span> x</span>[</span>i</span>]);</span></span> L </span>=</span> std</span>::</span>min</span>(</span>L</span>,</span> l</span>);</span></span> }</span></span> for</span> (</span>i64 i </span>=</span> 1</span>;</span> i </span><=</span> n</span>;</span> ++</span>i</span>)</span> x</span>[</span>i</span>]</span> =</span> std</span>::</span>min</span>(</span>L</span>,</span> x</span>[</span>i</span>]);</span></span> std</span>::</span>sort</span>(</span>x </span>+</span> 1</span>,</span> x </span>+</span> 1</span> +</span> n</span>,</span> std</span>::</span>less</span><</span>i64</span>>());</span></span> for</span> (</span>i64 i </span>=</span> n</span>,</span> k </span>=</span> 2</span>;</span> i</span>;</span> --</span>i</span>,</span> k </span>=</span> mod</span>(</span>k </span><<</span> 1</span>))</span></span> {</span></span> sigp </span>=</span> mod</span>(</span>sigp </span>+</span> (</span>x</span>[</span>i</span>]</span> -</span> x</span>[</span>i </span>-</span> 1</span>])</span> </span> k</span>);</span></span> }</span></span> printf</span>(</span>"%lld</span>\n</span>"</span>,</span> mod</span>(</span>1</span> +</span> sigp </span>+</span> (</span>L </span>-</span> x</span>[</span>n</span>])));</span></span> }</span></span> </span> signed</span> main</span>()</span></span> {</span></span> solve</span>();</span></span> return</span> 0</span>;</span></span> }</span></span> </span></code></pre> 本文最初发表于 Luogu</a>，于 2026-04-19 同步至本站。</p> 题面： https://www.luogu.com.cn/problem/P11214</a></p> 题解：P10765 「CROI · R2」在相思树下 I 2024-07-17T16:09:00+08:00 当序列是 $ak-b,k>0$，</p> 如果删去奇数项，剩下 $a(2k)-b$，即 $2ak-b$。</p> </li> 如果删去偶数项，剩下 $a(2k-1)-b$，即 $2ak-(b+a)$。</p> </li> </ul> 我们发现无论如何操作，新的序列仍可以写成 $Ak-B$ 的形式。</p> 初始状态 $a=1, b=0$，每次操作维护 $a,b$，最后取 $k=1$ 即可。</p> #include</span> <bits/stdc++.h></span></span> #define</span> u64</span> unsigned long long</span></span> </span> long long</span> qread</span>()</span></span> {</span></span> long long</span> w </span>=</span> 1</span>,</span> c</span>,</span> ret</span>;</span></span> while</span> ((</span>c </span>=</span> getchar</span>())</span> ></span> '9'</span> \|\|</span> c </span><</span> '0'</span>)</span></span> w </span>=</span> (</span>c </span>==</span> '-'</span> ? -</span>1</span> :</span> 1</span>);</span></span> ret </span>=</span> c </span>-</span> '0'</span>;</span></span> while</span> ((</span>c </span>=</span> getchar</span>())</span> >=</span> '0'</span> &&</span> c </span><=</span> '9'</span>)</span></span> ret </span>=</span> ret </span></span> 10</span> +</span> c </span>-</span> '0'</span>;</span></span> return</span> ret </span></span> w</span>;</span></span> }</span></span> </span> u64 T</span>,</span> n</span>,</span> k</span>,</span> a</span>,</span> b</span>,</span> oper</span>;</span></span> </span> signed</span> main</span>()</span></span> {</span></span> T </span>=</span> qread</span>();</span></span> while</span> (</span>T</span>--</span>)</span></span> {</span></span> n </span>=</span> qread</span>(),</span> k </span>=</span> qread</span>(),</span> a </span>=</span> 1</span>,</span> b </span>=</span> 0</span>;</span></span> while</span> (</span>k</span>--</span>)</span></span> {</span></span> oper </span>=</span> qread</span>();</span></span> if</span> (</span>oper </span>==</span> 1</span>)</span></span> {</span></span> a </span><<=</span> 1</span>;</span></span> }</span></span> if</span> (</span>oper </span>==</span> 2</span>)</span></span> {</span></span> b </span>+=</span> a</span>;</span></span> a </span><<=</span> 1</span>;</span></span> }</span></span> }</span></span> std</span>::</span>cout </span><<</span> a </span>-</span> b </span><<</span> std</span>::</span>endl</span>;</span></span> }</span></span> return</span> 0</span>;</span></span> }</span></span> </span></code></pre> 本文最初发表于 Luogu</a>，于 2026-04-19 同步至本站。</p> 题面： https://www.luogu.com.cn/problem/P10765</a></p>

100% vibe coding 构建的个人网站。内置 AI 小模型， vibe coder 的好帮手。服务器日常关机。</del></td></tr> </tbody></table>

Hello Zola! Hello oak.md!!

2026-04-19T19:23:00+08:00

我的博客上线了！这是一个测试页面。
数学公式</h3>
这里的公式 $a^2 + b^2 = c^2$ 会在行内显示。 $$ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} $$
图片</h3>

代码块</h3>
C++14 code:
using ll R ll tot cin tot 
Java17 code:
 public mp System
Rust code:
 pub phpsessid id phpsessid<&# id


[f44] 防止 PipeWire 发现小米电视

2026-02-21T00:00:00+00:00 #1E1E2E;">#include <bits/stdc++.h> #define ll long long namespace std; char mat[110][110]; style="color: #9399B2;">, C, K; int main() { italic;"> scanf("%lld %lld %lld", &R, &C, &K); style="color: #94E2D5;">= 0; for (ll i = 0; i < R; i++) for (ll j = 0; j < C; j++) >> mat[i][j]; style="color: #94E2D5;">= R * (C - K + 1) + (R - K + 1) * C; italic;"> // subtask1 for (ll i = 0; i < R; i++) for (ll j = 0; j <= C - K; j++) for (ll pos = j; pos < j + K; pos++) if (mat[i][pos] != '.') { tot--; break; } italic;"> // subtask2 for (ll i = 0; i <= R - K; i++) for (ll j = 0; j < C; j++) for (ll pos = i; pos < i + K; pos++) if (mat[pos][j] != '.') { tot--; break; } if (K == 1) return printf("%lld \n", tot / 2), 0; italic;"> printf("%lld \n", tot); return 0; }</code></pre> #1E1E2E;">import java.util.Scanner; class Main { public static void main(String[] args) { Scanner input = new Scanner(System.in); int N = input.nextInt(); int size = N + 1; int[][] mp = new int[size][size]; int[][][] dp = new int[size + N][size][size]; while (true) { int x = input.nextInt(), y = input.nextInt(), v = input.nextInt(); if (x == 0 && y == 0 && v == 0) break; style="color: #9399B2;">[x][y] = v; } for (int i = 2; i <= N * 2; i ++) { for (int j = 1; j <= N; j ++) { for (int k = 1; k <= N; k ++) { int j1 = i - j, j2 = i - k; if (j1 >= 1 && j1 <= N && j2 >= 1 && j2 <= N) { int t = mp[j][j1]; if (j != k) t += mp[k][j2]; dp[i][j][k] = Math.max(dp[i - 1][j - 1][k - 1] + t, Math.max(dp[i - 1][j - 1][k] + t, Math.max(dp[i - 1][j][k - 1] + t, dp[i - 1][j][k] + t))); } } } } style="color: #9399B2;">.out.println(dp[N + N][N][N]); } }</code></pre> #1E1E2E;">use super::User; italic;">#[derive(Debug, Clone)] struct AuthUser { style="color: #94E2D5;">: String, style="color: #94E2D5;">: String, } impl AuthUser { pub fn new(phpsessid: &str) -> anyhow::Result<Self> { italic;"> Ok(Self { x2F;span>: phpsessid.to_owned(), style="color: #94E2D5;">: if let Some((id, _)) = phpsessid.split_once('_') { id.to_owned() } else { italic;"> panic!("invalid phpsessid"); }, }) } pub fn phpsessid(&self) -> &str { &self.phpsessid } } impl User for AuthUser { fn id(&self) -> &str { &self.id } }</code></pre>
你不会喜欢片的声音自动传到客厅电视，对吧。 为了永远不发生这样的事情，应该按下面的步骤，禁用网络发现！ 1. 创建配置文件目录</h3> mkdir -p ~/.config/pipewire/pipewire.conf.d cd ~/.config/pipewire/pipewire.conf.d</code></pre>2. 写入配置</h3> 把下面的配置写到当前目录的 99-disable-network-sink.conf</code> 文件里，你应该创建这个文件。 context.properties = { # 禁用 AirPlay 发现 (通常是小米电视出现的来源) module.raop = false # 禁用 Zeroconf/Avahi 发现 (Chromecast/DNLA 等) module.zeroconf.discover = false }</code></pre>3. 重启音频服务</h3> systemctl --user restart pipewire</code></pre> 然后音频设备列表里的小米电视就永远消失了，耶！ 4. 反安装</h3> 你喜欢片的声音自动传到客厅电视，对吧。 按照下面的步骤，启用网络发现。 :-) 删除碍事的配置 rm ~/.config/pipewire/pipewire.conf.d/99-disable-network-sink.conf</code></li> 重启音频服务 systemctl --user restart pipewire</code></li> </ol>
题解：P11232 [CSP-S 2024] 超速检测（民间数据） 2024-10-26T23:06:00+08:00 挺麻烦的一个题，赛时调了 $2h$ 。 先写暴力</h3> $O(nm)$ 算出来第一个输出，然后 $O(2^mnm)$ 枚举删哪些测速点，可以骗到一点分。 考虑优化掉 $O(nm)$</h3> 对 $p$ 从小到大排序，设 $v_{i, q}$ 是第 $i$ 辆车在 $q$ 点的速度，则对于第 $i$ 辆车 $A$ ： 如果 $a_i<0$， $A$ 超速当且仅当 $v_{i, P}>V$ ，其中 $P$ 是最小的 $p_j$ 满足 $p_j\space{\geq}{\space}d_i$ </li> 如果 $a_i\space{\geq}\space0$， $A$ 超速当且仅当 $v_{i, p_m}>V$ </li> </ul> 对于 $a_i<0$ ，显然可以二分，整体时间复杂度 $O(n\log{m})$ 。 考虑优化掉 $O(2^mnm)$</h3> $p$ 是有序的，可以把每辆车的测超速范围看成一个线段 $X$ ，具体来说，对于第 $i$ 辆车： 如果 $a_i<0$， $X_i=[P,Q]$ ，其中 $P$ 是最小的 $j$ 满足 $v_{i, p_j}>V$ ， $Q$ 是最大的 $j$ 满足 $v_{i, p_j}>V$ 。</li> 如果 $a_i\space{\geq}\space0$， $X_i=[P,m]$ ，其中 $P$ 是最小的 $j$ 满足 $v_{i, p_j}>V$ 。</li> </ul> 因为 $v_{i, p_j}$ 在 $p_j\space{\geq}{\space}d_i$ 上具有单调性，所以二分求 $X$，时间复杂度 $O(n\log{m})$ 。 此时问题转化成 P1803 线段覆盖，整体时间复杂度 $O(n\log{m})$ 。 代码略（ 本文最初发表于 Luogu</a>，于 2026-04-19 同步至本站。 题面： https://www.luogu.com.cn/problem/P11232</a> 题解：P11231 [CSP-S 2024] 决斗（民间数据） 2024-10-26T22:02:00+08:00 因为攻击顺序可以随意指定，那么 $r$ 的顺序就无所谓了。 一看数据范围 ${10}^5$ ，不妨给 $r$ 排个序，发现 $r_i$ 能打败 $r_j$ 仅当 $j<i$ 。 那么维护一个队列，把排序的 $r$ 依次放进去，每次让 $r_i$ 攻击队首。 最后看看队里剩下几个数就是答案。 #include <bits/stdc++.h> #define i64 long long const i64 MAXN = 1e5 + 64; std::queue<i64> q; i64 a[MAXN], n; void solve() { assert(scanf("%lld", &n)); for (i64 i = 1; i <= n; ++i) assert(scanf("%lld", &a[i])); std::sort(a + 1, a + 1 + n, std::less<i64>()); for (i64 i = 1; i <= n; ++i) { if (!q.empty() && q.front() < a[i]) q.pop(); q.push(a[i]); } printf("%lld\n", (i64)q.size()); } signed main() { solve(); return 0; } </code></pre> 很难想到这其实就是 $r$ 的众数在 $r$ 中出现的次数，所以有 $O(n)$ 做法。 本文最初发表于 Luogu</a>，于 2026-04-19 同步至本站。 题面： https://www.luogu.com.cn/problem/P11231</a> 题解：P11214 【MX-J8-T2】黑洞 2024-10-20T23:28:00+08:00 非常难的一道题，想了 $150$ 分钟。（怎么降黄了 根据题意，对每一维同时增加或减少一个正整数 $C$，能得到被黑洞吞噬的点 $A$。 把每一维对应成一根竹竿，表示能够增减的范围，以样例 #3 为例： 方便起见，调换这些竹竿的方向，把距离 $0$ 较长的一端放在上面，按另一端的长度排序。 因为有竹竿 $[3, -1]$ ，所以 $C \leq 3$，那么对于所有竹竿，绝对值大于 $3$ 的部分是没用的。 也就是说，对于所有输入，可以转化成 $n$ 个竹竿，其中第 $i$ 个竹竿上端长度为 $L$，下端长度为 $x_i$，保证 $x_{i-1} \le x_i$ 且 $x_i \le L$ 。 考虑第 $i$ 个竹竿对 $A$ 取值数量的贡献 $p_i$ ，可以写成： $$ p_i = 2^{n - i + 1}(x_i-x_{i-1}) $$ 答案是 $1 + \Sigma {p} + (L-x_n)$ ，其中 $1$ 是输入的点本身，$L-x_n$ 是 $C > x_n$ 的情况。 时间复杂度 $O(n \log n)$ ，瓶颈在排序，代码不难写： #include <bits/stdc++.h> #define u64 unsigned long long #define inf 0x3f3f3f3f3f3f3f3f #define i64 long long const i64 MAXN = 2e5 + 64; const i64 MOD = 1000000007; #define mod(x) ((x) % MOD) i64 m[MAXN], a[MAXN], n; i64 x[MAXN], L = inf, sigp; void solve() { assert(scanf("%lld", &n)); for (i64 i = 1; i <= n; ++i) assert(scanf("%lld", &m[i])); for (i64 i = 1; i <= n; ++i) assert(scanf("%lld", &a[i])); for (i64 i = 1, l; i <= n; ++i) { l = m[i] - a[i], x[i] = a[i] - 1; if (l < x[i]) std::swap(l, x[i]); L = std::min(L, l); } for (i64 i = 1; i <= n; ++i) x[i] = std::min(L, x[i]); std::sort(x + 1, x + 1 + n, std::less<i64>()); for (i64 i = n, k = 2; i; --i, k = mod(k << 1)) { sigp = mod(sigp + (x[i] - x[i - 1]) * k); } printf("%lld\n", mod(1 + sigp + (L - x[n]))); } signed main() { solve(); return 0; } </code></pre> 本文最初发表于 Luogu</a>，于 2026-04-19 同步至本站。 题面： https://www.luogu.com.cn/problem/P11214</a> 题解：P10765 「CROI · R2」在相思树下 I 2024-07-17T16:09:00+08:00 当序列是 $ak-b,k>0$， 如果删去奇数项，剩下 $a(2k)-b$，即 $2ak-b$。 </li> 如果删去偶数项，剩下 $a(2k-1)-b$，即 $2ak-(b+a)$。 </li> </ul> 我们发现无论如何操作，新的序列仍可以写成 $Ak-B$ 的形式。 初始状态 $a=1, b=0$，每次操作维护 $a,b$，最后取 $k=1$ 即可。 #include <bits/stdc++.h> #define u64 unsigned long long long long qread() { long long w = 1, c, ret; while ((c = getchar()) > '9' || c < '0') w = (c == '-' ? -1 : 1); ret = c - '0'; while ((c = getchar()) >= '0' && c <= '9') ret = ret * 10 + c - '0'; return ret * w; } u64 T, n, k, a, b, oper; signed main() { T = qread(); while (T--) { n = qread(), k = qread(), a = 1, b = 0; while (k--) { oper = qread(); if (oper == 1) { a <<= 1; } if (oper == 2) { b += a; a <<= 1; } } std::cout << a - b << std::endl; } return 0; } </code></pre> 本文最初发表于 Luogu</a>，于 2026-04-19 同步至本站。 题面： https://www.luogu.com.cn/problem/P10765</a>

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Hello Zola! Hello oak.md!!

数学公式</h3> 这里的公式 $a^2 + b^2 = c^2$ 会在行内显示。 $$ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} $$</p>

图片</h3> </p>

代码块</h3> C++14 code:</p> using </span> ll R</span></span> ll tot </span> cin tot </span> Java17 code:</p> </span> public </span> mp</span> </span> </span> System</span> Rust code:</p> </span> pub phpsessid</span> id</span> phpsessid<&# id</span> </p>

[f44] 防止 PipeWire 发现小米电视

题解：P11232 [CSP-S 2024] 超速检测（民间数据）

先写暴力</h3> $O(nm)$ 算出来第一个输出，然后 $O(2^mnm)$ 枚举删哪些测速点，可以骗到一点分。</p>

题解：P11231 [CSP-S 2024] 决斗（民间数据）

题解：P11214 【MX-J8-T2】黑洞

题解：P10765 「CROI · R2」在相思树下 I

数学公式</h3>
这里的公式 $a^2 + b^2 = c^2$ 会在行内显示。 $$ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} $$</p>

图片</h3>
</p>

代码块</h3>
C++14 code:</p>
using </span> ll R</span></span> ll tot </span> cin tot </span>
Java17 code:</p>
</span> public </span> mp</span> </span> </span> System</span>
Rust code:</p>
</span> pub phpsessid</span> id</span> phpsessid<&# id</span>
</p>

先写暴力</h3>
$O(nm)$ 算出来第一个输出，然后 $O(2^mnm)$ 枚举删哪些测速点，可以骗到一点分。</p>