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<?xml version="1.0" encoding="utf-8"?>
<feed xmlns="http://www.w3.org/2005/Atom">
<id>https://blog.ak-ioi.cf</id>
<title>rui_er 的博客</title>
<updated>2021-02-17T09:10:12.066Z</updated>
<generator>https://github.com/jpmonette/feed</generator>
<link rel="alternate" href="https://blog.ak-ioi.cf"/>
<link rel="self" href="https://blog.ak-ioi.cf/atom.xml"/>
<subtitle>三思而后码</subtitle>
<logo>https://blog.ak-ioi.cf/images/avatar.png</logo>
<icon>https://blog.ak-ioi.cf/favicon.ico</icon>
<rights>All rights reserved 2021, rui_er 的博客</rights>
<entry>
<title type="html"><![CDATA[序列分块入门教程]]></title>
<id>https://blog.ak-ioi.cf/post/sqrt-yyds-awa/</id>
<link href="https://blog.ak-ioi.cf/post/sqrt-yyds-awa/">
</link>
<updated>2021-02-17T07:02:34.000Z</updated>
<content type="html"><![CDATA[<p>学妹让我教她分块，翻了翻以前的一篇博客，发现写的太难、太简略了，所以写了这篇博客 QAQ。</p>
<p>真的是分块最基础的教程，大佬可以跳过了。</p>
<p>如果本文有可以改进的地方，可以在评论区提出。</p>
<h1 id="前言-序列分块是什么">前言 - 序列分块是什么</h1>
<p>顾名思义，序列分块就是把数列分成若干小块进行处理和维护，我们会把每一块当成整体，维护这些整体的有关信息。</p>
<p>最基础的分块当取块长为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msqrt><mi>n</mi></msqrt></mrow><annotation encoding="application/x-tex">\sqrt{n}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.04em;vertical-align:-0.23972em;"></span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span></span></span></span> 时的初始化复杂度是 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><mi>n</mi><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord mathdefault">n</span><span class="mclose">)</span></span></span></span>，单次操作复杂度是 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><msqrt><mi>n</mi></msqrt><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(\sqrt{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.05028em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mclose">)</span></span></span></span>，当然一些较为复杂的题需要对照题目具体分析。</p>
<h1 id="例题一-p3372-模板线段树-1">例题一 - <a href="https://www.luogu.com.cn/problem/P3372">P3372 【模板】线段树 1</a></h1>
<h2 id="简要分析">简要分析</h2>
<p>区间加区间和，参考题目名称可以发现，出题人貌似想让我们写线段树，于是我们考虑如何不写线段树。我们采用分块。</p>
<h2 id="变量定义">变量定义</h2>
<p>以下是代码中的一些数组或变量名称及其含义：</p>
<ul>
<li><span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>n</mi><mo separator="true">,</mo><mi>m</mi></mrow><annotation encoding="application/x-tex">n,m</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.625em;vertical-align:-0.19444em;"></span><span class="mord mathdefault">n</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault">m</span></span></span></span>：数列长度和操作次数。</li>
<li><span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>a</mi><mi>i</mi></msub></mrow><annotation encoding="application/x-tex">a_i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.58056em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span>：数列第 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi></mrow><annotation encoding="application/x-tex">i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.65952em;vertical-align:0em;"></span><span class="mord mathdefault">i</span></span></span></span> 项。</li>
<li><span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>s</mi><mi>i</mi></msub></mrow><annotation encoding="application/x-tex">s_i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.58056em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault">s</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span>：第 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi></mrow><annotation encoding="application/x-tex">i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.65952em;vertical-align:0em;"></span><span class="mord mathdefault">i</span></span></span></span> 块所有数的和。</li>
<li><span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>t</mi><mi>a</mi><msub><mi>g</mi><mi>i</mi></msub></mrow><annotation encoding="application/x-tex">tag_i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.80952em;vertical-align:-0.19444em;"></span><span class="mord mathdefault">t</span><span class="mord mathdefault">a</span><span class="mord"><span class="mord mathdefault" style="margin-right:0.03588em;">g</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:-0.03588em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span>：第 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi></mrow><annotation encoding="application/x-tex">i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.65952em;vertical-align:0em;"></span><span class="mord mathdefault">i</span></span></span></span> 块整体加法标记，详见下文。</li>
<li><span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>p</mi><mi>o</mi><msub><mi>s</mi><mi>i</mi></msub></mrow><annotation encoding="application/x-tex">pos_i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.625em;vertical-align:-0.19444em;"></span><span class="mord mathdefault">p</span><span class="mord mathdefault">o</span><span class="mord"><span class="mord mathdefault">s</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span>：数列第 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi></mrow><annotation encoding="application/x-tex">i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.65952em;vertical-align:0em;"></span><span class="mord mathdefault">i</span></span></span></span> 项所在块编号。</li>
<li><span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>L</mi><mi>i</mi></msub></mrow><annotation encoding="application/x-tex">L_i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.83333em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault">L</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span>：第 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi></mrow><annotation encoding="application/x-tex">i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.65952em;vertical-align:0em;"></span><span class="mord mathdefault">i</span></span></span></span> 块左端点编号。</li>
<li><span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>R</mi><mi>i</mi></msub></mrow><annotation encoding="application/x-tex">R_i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.83333em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault" style="margin-right:0.00773em;">R</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:-0.00773em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span>：第 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi></mrow><annotation encoding="application/x-tex">i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.65952em;vertical-align:0em;"></span><span class="mord mathdefault">i</span></span></span></span> 块右端点编号。</li>
<li><span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>s</mi><mi>z</mi></mrow><annotation encoding="application/x-tex">sz</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">s</span><span class="mord mathdefault" style="margin-right:0.04398em;">z</span></span></span></span>：块长。</li>
<li><span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>t</mi><mi>o</mi><mi>t</mi></mrow><annotation encoding="application/x-tex">tot</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.61508em;vertical-align:0em;"></span><span class="mord mathdefault">t</span><span class="mord mathdefault">o</span><span class="mord mathdefault">t</span></span></span></span>：总块数。</li>
</ul>
<h2 id="块状数组初始化">块状数组初始化</h2>
<p>首先取块长为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msqrt><mi>n</mi></msqrt></mrow><annotation encoding="application/x-tex">\sqrt{n}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.04em;vertical-align:-0.23972em;"></span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span></span></span></span>，然后枚举每一块，可以知道第 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi></mrow><annotation encoding="application/x-tex">i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.65952em;vertical-align:0em;"></span><span class="mord mathdefault">i</span></span></span></span> 块的左端点是第 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi><mo>−</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">i-1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.74285em;vertical-align:-0.08333em;"></span><span class="mord mathdefault">i</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">−</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span> 块的右端点的下一个元素，第 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi></mrow><annotation encoding="application/x-tex">i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.65952em;vertical-align:0em;"></span><span class="mord mathdefault">i</span></span></span></span> 块的右端点是 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>min</mi><mo></mo><mo>(</mo><mi>i</mi><mo>×</mo><mi>s</mi><mi>z</mi><mo separator="true">,</mo><mi>n</mi><mo>)</mo></mrow><annotation encoding="application/x-tex">\min(i\times sz,n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mop">min</span><span class="mopen">(</span><span class="mord mathdefault">i</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathdefault">s</span><span class="mord mathdefault" style="margin-right:0.04398em;">z</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault">n</span><span class="mclose">)</span></span></span></span>，我们通过这个式子对 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>L</mi><mo separator="true">,</mo><mi>R</mi></mrow><annotation encoding="application/x-tex">L,R</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8777699999999999em;vertical-align:-0.19444em;"></span><span class="mord mathdefault">L</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault" style="margin-right:0.00773em;">R</span></span></span></span> 进行初始化。</p>
<p>随后枚举 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>L</mi><mi>i</mi></msub><mo>∼</mo><msub><mi>R</mi><mi>i</mi></msub></mrow><annotation encoding="application/x-tex">L_i\sim R_i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.83333em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault">L</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">∼</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.83333em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault" style="margin-right:0.00773em;">R</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:-0.00773em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span> 的所有元素，用 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>s</mi><mi>i</mi></msub></mrow><annotation encoding="application/x-tex">s_i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.58056em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault">s</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span> 记录他们的和，同时把这些元素的 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>p</mi><mi>o</mi><mi>s</mi></mrow><annotation encoding="application/x-tex">pos</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.625em;vertical-align:-0.19444em;"></span><span class="mord mathdefault">p</span><span class="mord mathdefault">o</span><span class="mord mathdefault">s</span></span></span></span> 标记为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi></mrow><annotation encoding="application/x-tex">i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.65952em;vertical-align:0em;"></span><span class="mord mathdefault">i</span></span></span></span>。</p>
<p>至此，我们便完成了初始化，因为每一块的区间不重叠且没有遗漏，<span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>a</mi></mrow><annotation encoding="application/x-tex">a</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">a</span></span></span></span> 数组的每个元素下标恰好访问一遍，因此复杂度是 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><mi>n</mi><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord mathdefault">n</span><span class="mclose">)</span></span></span></span> 的。</p>
<p>此部分代码：</p>
<pre><code class="language-cpp">void initBlock() {
sz = sqrt(n);
while(++tot) {
L[tot] = R[tot-1] + 1;
R[tot] = min(sz*tot, n);
rep(i, L[tot], R[tot]) pos[i] = tot, s[tot] += a[i];
if(R[tot] == n) break;
}
}
</code></pre>
<h2 id="区间加">区间加</h2>
<p>此部分操作要求将 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>a</mi><mi>x</mi></msub><mo>∼</mo><msub><mi>a</mi><mi>y</mi></msub></mrow><annotation encoding="application/x-tex">a_x\sim a_y</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.58056em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.151392em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">x</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">∼</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.716668em;vertical-align:-0.286108em;"></span><span class="mord"><span class="mord mathdefault">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.15139200000000003em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight" style="margin-right:0.03588em;">y</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.286108em;"><span></span></span></span></span></span></span></span></span></span> 的所有数加上 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="normal">Δ</mi></mrow><annotation encoding="application/x-tex">\Delta</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.68333em;vertical-align:0em;"></span><span class="mord">Δ</span></span></span></span>，暴力单次加是 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><mi>n</mi><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord mathdefault">n</span><span class="mclose">)</span></span></span></span> 的，显然需要优化。</p>
<p>对于 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>x</mi></mrow><annotation encoding="application/x-tex">x</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">x</span></span></span></span> 与 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>y</mi></mrow><annotation encoding="application/x-tex">y</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.625em;vertical-align:-0.19444em;"></span><span class="mord mathdefault" style="margin-right:0.03588em;">y</span></span></span></span> 在同一块的情况，我们暴力从 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>x</mi></mrow><annotation encoding="application/x-tex">x</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">x</span></span></span></span> 枚举到 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>y</mi></mrow><annotation encoding="application/x-tex">y</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.625em;vertical-align:-0.19444em;"></span><span class="mord mathdefault" style="margin-right:0.03588em;">y</span></span></span></span>，将 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>a</mi><mi>i</mi></msub></mrow><annotation encoding="application/x-tex">a_i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.58056em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span> 和 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>s</mi><mrow><mi>p</mi><mi>o</mi><msub><mi>s</mi><mi>i</mi></msub></mrow></msub></mrow><annotation encoding="application/x-tex">s_{pos_{i}}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.716668em;vertical-align:-0.286108em;"></span><span class="mord"><span class="mord mathdefault">s</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.15139200000000003em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mathdefault mtight">p</span><span class="mord mathdefault mtight">o</span><span class="mord mtight"><span class="mord mathdefault mtight">s</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3280857142857143em;"><span style="top:-2.357em;margin-left:0em;margin-right:0.07142857142857144em;"><span class="pstrut" style="height:2.5em;"></span><span class="sizing reset-size3 size1 mtight"><span class="mord mtight"><span class="mord mathdefault mtight">i</span></span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.143em;"><span></span></span></span></span></span></span></span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.286108em;"><span></span></span></span></span></span></span></span></span></span> 加上 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="normal">Δ</mi></mrow><annotation encoding="application/x-tex">\Delta</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.68333em;vertical-align:0em;"></span><span class="mord">Δ</span></span></span></span>。由于块长为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msqrt><mi>n</mi></msqrt></mrow><annotation encoding="application/x-tex">\sqrt{n}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.04em;vertical-align:-0.23972em;"></span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span></span></span></span>，因此暴力枚举这一部分的复杂度为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><msqrt><mi>n</mi></msqrt><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(\sqrt{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.05028em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mclose">)</span></span></span></span>。</p>
<p>对于不在同一块的情况，对两侧的零散块（不完整块，即 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>x</mi><mo separator="true">,</mo><mi>y</mi></mrow><annotation encoding="application/x-tex">x,y</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.625em;vertical-align:-0.19444em;"></span><span class="mord mathdefault">x</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault" style="margin-right:0.03588em;">y</span></span></span></span> 所在块）按照同上的方法暴力处理，对中间的每个完整块记录整体加法标记 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>t</mi><mi>a</mi><mi>g</mi></mrow><annotation encoding="application/x-tex">tag</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.80952em;vertical-align:-0.19444em;"></span><span class="mord mathdefault">t</span><span class="mord mathdefault">a</span><span class="mord mathdefault" style="margin-right:0.03588em;">g</span></span></span></span>，并把块中所有数和同步记录。由于块长为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msqrt><mi>n</mi></msqrt></mrow><annotation encoding="application/x-tex">\sqrt{n}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.04em;vertical-align:-0.23972em;"></span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span></span></span></span>，不完整块最多操作次数为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>2</mn><mo>×</mo><msqrt><mi>n</mi></msqrt></mrow><annotation encoding="application/x-tex">2\times\sqrt{n}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">2</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:1.04em;vertical-align:-0.23972em;"></span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span></span></span></span>，完整块不超过 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msqrt><mi>n</mi></msqrt></mrow><annotation encoding="application/x-tex">\sqrt{n}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.04em;vertical-align:-0.23972em;"></span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span></span></span></span> 个，因此这一部分的复杂度为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><msqrt><mi>n</mi></msqrt><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(\sqrt{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.05028em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mclose">)</span></span></span></span>。</p>
<p>综上，块状数组区间加法的复杂度为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><msqrt><mi>n</mi></msqrt><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(\sqrt{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.05028em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mclose">)</span></span></span></span>。</p>
<p>此部分代码：</p>
<pre><code class="language-cpp">void modify(ll x, ll y, ll delta) {
ll l = pos[x], r = pos[y];
if(l == r) rep(i, x, y) a[i] += delta, s[l] += delta;
else {
rep(i, x, R[l]) a[i] += delta, s[l] += delta;
rep(i, l+1, r-1) tag[i] += delta, s[i] += delta * (R[i] - L[i] + 1);
rep(i, L[r], y) a[i] += delta, s[r] += delta;
}
}
</code></pre>
<h2 id="区间和">区间和</h2>
<p>查询区间 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>a</mi><mi>x</mi></msub><mo>∼</mo><msub><mi>a</mi><mi>y</mi></msub></mrow><annotation encoding="application/x-tex">a_x\sim a_y</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.58056em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.151392em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">x</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">∼</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.716668em;vertical-align:-0.286108em;"></span><span class="mord"><span class="mord mathdefault">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.15139200000000003em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight" style="margin-right:0.03588em;">y</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.286108em;"><span></span></span></span></span></span></span></span></span></span> 所有数的和，同样不能暴力加，考虑按照同样的方法优化：</p>
<p>对于同一块的情况，暴力枚举并累加 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>a</mi><mi>i</mi></msub><mo>+</mo><mi>t</mi><mi>a</mi><msub><mi>g</mi><mrow><mi>p</mi><mi>o</mi><msub><mi>s</mi><mi>i</mi></msub></mrow></msub></mrow><annotation encoding="application/x-tex">a_i+tag_{pos_i}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.73333em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.9011879999999999em;vertical-align:-0.286108em;"></span><span class="mord mathdefault">t</span><span class="mord mathdefault">a</span><span class="mord"><span class="mord mathdefault" style="margin-right:0.03588em;">g</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.15139200000000003em;"><span style="top:-2.5500000000000003em;margin-left:-0.03588em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mathdefault mtight">p</span><span class="mord mathdefault mtight">o</span><span class="mord mtight"><span class="mord mathdefault mtight">s</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3280857142857143em;"><span style="top:-2.357em;margin-left:0em;margin-right:0.07142857142857144em;"><span class="pstrut" style="height:2.5em;"></span><span class="sizing reset-size3 size1 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.143em;"><span></span></span></span></span></span></span></span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.286108em;"><span></span></span></span></span></span></span></span></span></span>，<strong>注意不要漏掉当前块的整体加法标记！</strong></p>
<p>对于不在同一块的情况，两侧的零散块同样方法处理，中间的整块直接累加块和即可。</p>
<p>同样的，查询区间和的复杂度为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><msqrt><mi>n</mi></msqrt><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(\sqrt{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.05028em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mclose">)</span></span></span></span>。</p>
<p>此部分代码：</p>
<pre><code class="language-cpp">ll query(ll x, ll y) {
ll l = pos[x], r = pos[y], ans = 0;
if(l == r) rep(i, x, y) ans += a[i] + tag[l];
else {
rep(i, x, R[l]) ans += a[i] + tag[l];
rep(i, l+1, r-1) ans += s[i];
rep(i, L[r], y) ans += a[i] + tag[r];
}
return ans;
}
</code></pre>
<h2 id="ac-本题">AC 本题</h2>
<p>将以上代码结合起来即可 AC 本题，总复杂度为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><mi>n</mi><mo>+</mo><mi>m</mi><msqrt><mi>n</mi></msqrt><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(n+m\sqrt{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord mathdefault">n</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:1.05028em;vertical-align:-0.25em;"></span><span class="mord mathdefault">m</span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mclose">)</span></span></span></span>，<a href="https://www.luogu.com.cn/paste/d0eyurmp">完整代码戳我</a>。</p>
<h1 id="习题一-p2357-守墓人">习题一 - <a href="https://www.luogu.com.cn/problem/P2357">P2357 守墓人</a></h1>
<p>基本上是双倍经验，可以稍加修改通过本题。</p>
<h1 id="例题二-p2801-教主的魔法">例题二 - <a href="https://www.luogu.com.cn/problem/P2801">P2801 教主的魔法</a></h1>
<h2 id="简要分析-2">简要分析</h2>
<p>简要题意：区间加，查询区间内多少个数大于等于 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>c</mi></mrow><annotation encoding="application/x-tex">c</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">c</span></span></span></span>。</p>
<p>我们可以采用分块，对每一块进行重构排序，然后询问时直接使用二分。</p>
<h2 id="块内排序">块内排序</h2>
<p>块内排序在代码中用的比较多，因此我专门写了个函数重构块。</p>
<pre><code class="language-cpp">void reconstruct(int x) {
rep(i, L[x], R[x]) b[i] = a[i];
sort(b+L[x], b+1+R[x]);
}
</code></pre>
<h2 id="初始化">初始化</h2>
<p>初始化部分主要有一个更改：对于每一块，需要将该块内元素拷贝一遍并排序。</p>
<p>新初始化复杂度 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><mi>n</mi><mi>log</mi><mo></mo><msqrt><mi>n</mi></msqrt><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(n\log\sqrt{n})</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.05028em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord mathdefault">n</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mclose">)</span></span></span></span>，使用对数换底公式可以知道这个复杂度与 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><mi>n</mi><mi>log</mi><mo></mo><mi>n</mi><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(n\log n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord mathdefault">n</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault">n</span><span class="mclose">)</span></span></span></span> 相差为常数，可以视为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><mi>n</mi><mi>log</mi><mo></mo><mi>n</mi><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(n\log n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord mathdefault">n</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault">n</span><span class="mclose">)</span></span></span></span>。</p>
<h2 id="区间加-2">区间加</h2>
<p>区间加对零散块直接暴力加，然后进行重构；因为对整块的加法不改变块内元素顺序，打上整体加法标记即可。</p>
<p>复杂度 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><msqrt><mi>n</mi></msqrt><mi>log</mi><mo></mo><msqrt><mi>n</mi></msqrt><mo>)</mo><mo>=</mo><mi mathvariant="script">O</mi><mo>(</mo><msqrt><mi>n</mi></msqrt><mi>log</mi><mo></mo><mi>n</mi><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(\sqrt{n}\log\sqrt{n})=\mathcal{O}(\sqrt{n}\log n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.05028em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mclose">)</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:1.05028em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault">n</span><span class="mclose">)</span></span></span></span>。</p>
<h2 id="区间大于等于-c-数个数">区间大于等于 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>c</mi></mrow><annotation encoding="application/x-tex">c</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">c</span></span></span></span> 数个数</h2>
<p>零散块直接暴力查询，<strong>记得加上整体加法标记。</strong></p>
<p>对于完整块，在排好序的块内二分找 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>c</mi><mo>−</mo><mi>t</mi><mi>a</mi><msub><mi>g</mi><mi>i</mi></msub></mrow><annotation encoding="application/x-tex">c-tag_i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.66666em;vertical-align:-0.08333em;"></span><span class="mord mathdefault">c</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">−</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.80952em;vertical-align:-0.19444em;"></span><span class="mord mathdefault">t</span><span class="mord mathdefault">a</span><span class="mord"><span class="mord mathdefault" style="margin-right:0.03588em;">g</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:-0.03588em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span></span></span></span>，因为是排好序的，直接根据下标算即可。</p>
<p>复杂度 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><msqrt><mi>n</mi></msqrt><mi>log</mi><mo></mo><msqrt><mi>n</mi></msqrt><mo>)</mo><mo>=</mo><mi mathvariant="script">O</mi><mo>(</mo><msqrt><mi>n</mi></msqrt><mi>log</mi><mo></mo><mi>n</mi><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(\sqrt{n}\log\sqrt{n})=\mathcal{O}(\sqrt{n}\log n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.05028em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mclose">)</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:1.05028em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault">n</span><span class="mclose">)</span></span></span></span>。</p>
<h2 id="ac-本题-2">AC 本题</h2>
<p>结合一下以上部分即可 AC 本题，总复杂度为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><mi>n</mi><mi>log</mi><mo></mo><mi>n</mi><mo>+</mo><mi>Q</mi><msqrt><mi>n</mi></msqrt><mi>log</mi><mo></mo><mi>n</mi><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(n\log n+Q\sqrt{n}\log n)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord mathdefault">n</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault">n</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:1.05028em;vertical-align:-0.25em;"></span><span class="mord mathdefault">Q</span><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8002800000000001em;"><span class="svg-align" style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span class="mord" style="padding-left:0.833em;"><span class="mord mathdefault">n</span></span></span><span style="top:-2.76028em;"><span class="pstrut" style="height:3em;"></span><span class="hide-tail" style="min-width:0.853em;height:1.08em;"><svg width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,
-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z'/></svg></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.23972em;"><span></span></span></span></span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mop">lo<span style="margin-right:0.01389em;">g</span></span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault">n</span><span class="mclose">)</span></span></span></span>。看到数据范围 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>Q</mi></mrow><annotation encoding="application/x-tex">Q</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8777699999999999em;vertical-align:-0.19444em;"></span><span class="mord mathdefault">Q</span></span></span></span> 是比较小的，可以通过。</p>
<p><a href="https://www.luogu.com.cn/paste/ixq9b1lp">完整代码接着戳我</a>。</p>
<h1 id="习题二-u147831-分块问题-3">习题二 - <a href="https://www.luogu.com.cn/problem/U147831">U147831 分块问题 3</a></h1>
<p>区间加，求区间内 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>c</mi></mrow><annotation encoding="application/x-tex">c</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">c</span></span></span></span> 的前驱。</p>
<p>依然是块内排序、查询时二分的思想。</p>
<h1 id="习题三-u147840-分块问题-6">习题三 - <a href="https://www.luogu.com.cn/problem/U147840">U147840 分块问题 6</a></h1>
<p>单点插入，单点查询。</p>
<p>对于每一块维护 vector，插入时遍历每一块找插入点，查询时遍历每一块用 vector 查值即可。</p>
<p>这一方法可以通过本题，但是因为插入的数随机生成，每一块的插入次数大致均匀。可以通过构造数据的方式卡掉这一做法。</p>
<p>我选择当一块的 vector 大小超过 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>8</mn><mo>×</mo><mi>s</mi><mi>z</mi></mrow><annotation encoding="application/x-tex">8\times sz</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">8</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">s</span><span class="mord mathdefault" style="margin-right:0.04398em;">z</span></span></span></span> 的时候将所有 vector 清空并重构，因为重构的次数比较少可以通过本题。</p>
<h1 id="习题四-p4145-上帝造题的七分钟-2-花神游历各国">习题四 - <a href="https://www.luogu.com.cn/problem/P4145">P4145 上帝造题的七分钟 2 / 花神游历各国</a></h1>
<p>区间开平方根，区间和。</p>
<p>与线段树做法想法类似，维护每一块的最小值。</p>
<h1 id="习题五-u147877-分块问题-8">习题五 - <a href="https://www.luogu.com.cn/problem/U147877">U147877 分块问题 8</a></h1>
<p>查询区间内多少个数为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>c</mi></mrow><annotation encoding="application/x-tex">c</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">c</span></span></span></span>，并把区间改为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>c</mi></mrow><annotation encoding="application/x-tex">c</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">c</span></span></span></span>。</p>
<p>经过我和几个初二学长的探讨，本题珂朵莉树解法的复杂度没有问题。</p>
<p>但是毕竟是练习分块，我还是老老实实地写了分块。</p>
<p>大致想法就是，对每一块记录区间赋值标记，优先级高于每个数的值，初始区间赋值标记设置一个特殊值表示不存在。其他的就比较基础了。</p>
<h1 id="习题六-p4168-violet蒲公英"><s>习题六 - <a href="https://www.luogu.com.cn/problem/P4168">P4168 [Violet]蒲公英</a></s></h1>
<p>这题貌似被我咕了好久，到现在都没做。。。</p>
<p>先放在这里吧，有时间做完再补上（（（</p>
<h1 id="挑战-学分块怎么能不做-ynoi-呢"><s>挑战 - 学分块怎么能不做 Ynoi 呢</s></h1>
<p>写了这么多字，这部分先咕着吧，以后有时间再写（</p>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[CodeForces 思维题集锦]]></title>
<id>https://blog.ak-ioi.cf/post/codeforces-si-wei-ti-ji-jin/</id>
<link href="https://blog.ak-ioi.cf/post/codeforces-si-wei-ti-ji-jin/">
</link>
<updated>2021-02-08T11:25:27.000Z</updated>
<content type="html"><![CDATA[<p><a href="https://www.luogu.com.cn/blog/ak-ioi/codeforces-0">Link</a></p>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[NOI WC2021 游记]]></title>
<id>https://blog.ak-ioi.cf/post/i-ak-noi-wc2021/</id>
<link href="https://blog.ak-ioi.cf/post/i-ak-noi-wc2021/">
</link>
<updated>2021-02-06T11:05:22.000Z</updated>
<content type="html"><![CDATA[<h1 id="前言">前言</h1>
<p><strong>CCF 宣传片语录镇楼：</strong></p>
<p><strong>假设有 l 到 r，这是 l，这是 r。这些边我不加。这些边我不加！暴力怎么做？暴力是不是，加边、加边、加边，然后，并查集查询。</strong></p>
<ul>
<li>NOI WC2020 “冬”令营 & 夏令营，<s>嫌太贵</s>没有报名，看到有些同学拿到了铜牌非常羡慕。</li>
<li>APIO2020 报名了 B 类参赛，但是因为做题策略原因爆零打铁，有些同学又拿到了铜牌。<a href="/blog/ak-ioi/i-ak-apio2020">游记</a></li>
<li>CSP2020 提高级挂了将近 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>100</mn></mrow><annotation encoding="application/x-tex">100</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span><span class="mord">0</span><span class="mord">0</span></span></span></span> 分，只获得了六级分数线+一等奖。</li>
<li>NOIP2020 挂了不知道多少分，沦为 2= 人。</li>
</ul>
<p>（备注：初二年级，第一次 WC）</p>
<h1 id="20201224-2459208">2020.12.24 | 2459208</h1>
<p>看到了官网的通知，准备报名。问了半天老师说过几天发一个问卷统一报名。</p>
<h1 id="20201226-2459210">2020.12.26(?) | 2459210(?)</h1>
<p>催完老师了，报名成功 /cy。</p>
<h1 id="202111-2459216">2021.1.1 | 2459216</h1>
<p>报名成功了，进微信群了 qaq，祝我成功拿到 Cu。</p>
<h1 id="2021129-2459244">2021.1.29 | 2459244</h1>
<p>下午 15:00 进会议室，看了亿遍 CCF 宣传片，还截到了 3M 原则的三张图片（</p>
<h1 id="2021131-2459246">2021.1.31 | 2459246</h1>
<p>被 ClCN 在群里拍了，于是就知道了 ClCN 的真名（危</p>
<p>然后在犇犇说了一下，于是不知道多少人知道了我的真名（当然也可能没注意）</p>
<p>然后转眼就看到 ClCN 拍了何琳（危</p>
<h1 id="202121-2459247">2021.2.1 | 2459247</h1>
<p>本来以为开幕式会半小时，结果只有 7 分钟。。。</p>
<p>想进第一课堂但发现课件都看不懂，果断去第二课堂。</p>
<p>上午听懂了一些 dp 模型和优化，这很好。</p>
<p><s>下午听懂了如何从同学/家长那里骗钱，这非常好！！！11</s></p>
<h1 id="202122-2459248">2021.2.2 | 2459248</h1>
<p>上午听课把网络流复习了一遍，感觉还不错。</p>
<p><s>记住了老师名言：我们是不是只要在下午的课前下课就行了？</s></p>
<p>下午加边老师（？）讲课，听懂了基础的字符串，exKMP 和 SA 后面的有点晕，但还是学会了一些东西的。</p>
<h1 id="202123-2459249">2021.2.3 | 2459249</h1>
<p>上午讲树论，于是前面听懂了，后面定义和板子都不讲就一句“打板子就可以了”直接带过，成功的导致了我原来会的依然会、原来不会的依然不会，于是听到点分治实在忍不住了直接退直播间。</p>
<p>下午第一课堂 IOI+，第二课堂 pj-，于是啥都没听就跑去<s>颓废</s>写文化课。</p>
<h1 id="202124-2459250">2021.2.4 | 2459250</h1>
<p>上午跑去第一课堂听了半天 MO 课 & 希腊语课，下午讲多线程和并行计算懒得听。</p>
<h1 id="202125-2459251">2021.2.5 | 2459251</h1>
<p>早上 7:30 起来了，又去看了一眼 CCF 通知，才发现考试是 8:30 开始而不是 8:00，于是就来洛谷背板子，背了一下网络流、模拟退火、二维凸包，结果啥都没用上。</p>
<p>先把所有题看了一遍，最后决定打暴力顺序 T2→T3→T1。T2、T3 直接打的暴力，T1 暴力打的一点都不顺利，调了好久搞了个玄学做法，但一直 hack 不掉（可能因为我手算太麻烦搞得全特别小），最后放弃。</p>
<p>线上赛，具体过程没啥好描述的，放张图：</p>
<figure data-type="image" tabindex="1"><img src="https://cdn.luogu.com.cn/upload/image_hosting/31i46gsk.png" alt="" loading="lazy"></figure>
<p>最后跟同学说了一下估分，发现我们几个好像差不多，这是我的垃圾估分：</p>
<ul>
<li><span class="katex"><span class="katex-mathml"><math><semantics><mrow><mrow><mo fence="true">(</mo><mn>0</mn><mo>∼</mo><mn>32</mn><mo fence="true">)</mo></mrow><mo>+</mo><mrow><mo fence="true">(</mo><mn>35</mn><mo>∼</mo><mn>50</mn><mo fence="true">)</mo></mrow><mo>+</mo><mn>20</mn><mo>=</mo><mn>55</mn><mo>∼</mo><mn>102</mn></mrow><annotation encoding="application/x-tex">\left(0\sim 32\right)+\left(35\sim 50\right)+20=55\sim 102</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="minner"><span class="mopen delimcenter" style="top:0em;">(</span><span class="mord">0</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">∼</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mord">3</span><span class="mord">2</span><span class="mclose delimcenter" style="top:0em;">)</span></span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="minner"><span class="mopen delimcenter" style="top:0em;">(</span><span class="mord">3</span><span class="mord">5</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">∼</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mord">5</span><span class="mord">0</span><span class="mclose delimcenter" style="top:0em;">)</span></span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">2</span><span class="mord">0</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">5</span><span class="mord">5</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">∼</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span><span class="mord">0</span><span class="mord">2</span></span></span></span></li>
</ul>
<p>不知道有没有写挂，希望能拿个牌牌 qaq。</p>
<p>然后下午听讲题，说 T1 思路就是加边加边加边并查集查询，<s>CCF 宣传片泄题建议禁三！！！11</s></p>
<p>其他的大概听了下，果然我还是太菜了。</p>
<p>第一次 NOIWC 完结撒花，成绩等出来再说（</p>
<h1 id="202126-2459252">2021.2.6 | 2459252</h1>
<p>分数线 70/118/158，打铁祭（（（虽然还不知道我多少</p>
<p>upd：<span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>32</mn><mo>+</mo><mn>35</mn><mo>+</mo><mn>20</mn><mo>=</mo><mn>87</mn></mrow><annotation encoding="application/x-tex">32+35+20=87</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">3</span><span class="mord">2</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">3</span><span class="mord">5</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">2</span><span class="mord">0</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">8</span><span class="mord">7</span></span></span></span>，拿到了铜牌牌 awa</p>
<p>吐槽一句，今年分数线咋比去年高那么多啊 /kel</p>
<p>总结：</p>
<ul>
<li>第一次 WC 拿到 Cu 还是很惊喜的，但是跟同学最高分还是差了 18 分，果然还是我太菜了，【同学真名】txdy！！！！！！！11</li>
<li>暴力出奇迹，要仔细审题，不要漏掉取模啥的，把能拿到的部分分都拿了，少失误（</li>
<li>没了，因为我太菜了。</li>
</ul>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[题解 P1175@洛谷 【表达式的转换】]]></title>
<id>https://blog.ak-ioi.cf/post/solution-luogu-p1175/</id>
<link href="https://blog.ak-ioi.cf/post/solution-luogu-p1175/">
</link>
<updated>2021-01-22T14:17:38.000Z</updated>
<content type="html"><![CDATA[<p>本题题解貌似已经比较多了，但是我在本题卡了一段时间，写出了好几处错误，因此写一篇题解梳理一下思路。</p>
<p>题意：模拟 <code>+-*/^()</code> 以及数字初始为一位数的后缀表达式运算过程。</p>
<p><strong>注意这里有一个运算符 <code>^</code>，我一开始就没有注意到，然后样例也没有，硬是对着代码和题面看了半天才发现！</strong></p>
<hr>
<p>思路：</p>
<p>模拟。先转化为后缀表达式，再模拟运算过程，逐步求解。</p>
<p><strong>子问题一：中缀表达式转后缀表达式。</strong></p>
<p>我们考虑运算符的优先级，可以对于每个运算符给定一个数为优先级，方便后续比较。这里，我设置的优先级为：</p>
<pre><code class="language-cpp">int priority(char c) {
if(c == '^') return 3;
if(c == '*' || c == '/') return 2;
if(c == '+' || c == '-') return 1;
if(c == '(' || c == ')') return 0;
throw "WA! Unexpected operator";
}
</code></pre>
<p>其中 throw 行是为了防止出现错误添加的异常，实际上也并没有进行处理，可以忽略。</p>
<p>然后考虑转化，遍历原字符串每一位，按照如下规则处理：</p>
<ul>
<li>如果是数字，直接输出到后缀表达式。</li>
<li>如果是左括号 <code>(</code>，可以压进符号栈中。</li>
<li>如果是右括号 <code>)</code>，将符号栈中符号弹出到后缀表达式，直到遇到第一个左括号 <code>(</code>，并将其弹出。</li>
<li>如果是乘方 <code>^</code>，可以压进符号栈中。</li>
<li>如果是加减乘除 <code>+-*/</code>，利用上面的优先级，将符号栈中符号弹出到后缀表达式，直到遇到第一个优先级低于（不含）它的符号，然后将其压进符号栈。</li>
</ul>
<p>那么问题来了，怎么存储后缀表达式呢？这里我使用了一个结构体 <code>struct</code>，一个指示变量表示这个是数字还是符号，一个联合体 <code>union</code> 来存储具体内容，然后使用向量容器 <code>vector</code> 来按顺序存储后缀表达式。</p>
<p>结构体定义和转换后缀表达式部分代码如下：</p>
<pre><code class="language-cpp">struct Node {
int type;
union {
int x;
char op;
}data;
Node() {}
Node(int x) : type(1) {data.x = x;}
Node(char c) : type(0) {data.op = c;}
};
void toSuffix() {
rep(i, 0, n-1) {
if(isdigit(c[i])) v.push_back(Node(int(c[i]^'0')));
else {
if(c[i] == '(') op.push(c[i]);
else if(c[i] == ')') {
while(op.top() != '(') {
v.push_back(Node(char(op.top())));
op.pop();
}
op.pop();
}
else if(c[i] == '^') op.push(c[i]);
else {
while(!op.empty() && priority(op.top()) >= priority(c[i])) {
v.push_back(Node(char(op.top())));
op.pop();
}
op.push(c[i]);
}
}
}
while(!op.empty()) {
v.push_back(Node(char(op.top())));
op.pop();
}
}
</code></pre>
<p><strong>子问题二：模拟运算步骤得出结果。</strong></p>
<p>这部分比较简单，做过 <a href="/problem/P1449">P1449 后缀表达式</a> 的同学都应该基本会了。我们按顺序读取后缀表达式内存储的数据，如果是数字则压入数字栈，否则弹出两个数字做运算后压回。<strong>注意运算的顺序，应该是第二个弹出的数作为被减数（加数、乘数、被除数、底数），第一个弹出的数作为减数（加数、乘数、除数、指数）！</strong></p>
<p>然后每处理一个运算符，输出一次即可。</p>
<p>这部分代码：</p>
<pre><code class="language-cpp">void prtall() {
int sz = v.size();
rep(i, 0, sz-1) {
if(v[i].type) printf("%d%c", v[i].data.x, " \n"[i==sz-1]);
else printf("%c%c", v[i].data.op, " \n"[i==sz-1]);
}
}
void prtsec(int u) {
int sz = v.size();
while(!s.empty()) {t.push(s.top()); s.pop();}
while(!t.empty()) {printf("%d ", t.top()); s.push(t.top()); t.pop();}
rep(i, u, sz-1) {
if(v[i].type) printf("%d ", v[i].data.x);
else printf("%c ", v[i].data.op);
}
puts("");
}
void calc() {
prtall();
int sz = v.size();
rep(i, 0, sz-1) {
if(v[i].type) s.push(v[i].data.x);
else {
int a, b; char _;
b = s.top(); s.pop();
a = s.top(); s.pop();
_ = v[i].data.op;
if(_ == '+') s.push(a+b);
else if(_ == '-') s.push(a-b);
else if(_ == '*') s.push(a*b);
else if(_ == '/') s.push(a/b);
else s.push(qpow(a, b));
prtsec(i+1);
}
}
}
</code></pre>
<p>这就是本题的主体代码，我也取得了 AC，代码长度 107 行、共 2441 个字符（本地统计），可以算是一个中模拟。</p>
<p>由于主要代码在上面都贴过了，为了防止影响篇幅和版面，将完整代码放到 <a href="/paste/713luey5">云剪贴板</a>。</p>
<p>完结撒花！</p>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[题解 CF156B@洛谷 | 156B@Codeforces 【Suspects】]]></title>
<id>https://blog.ak-ioi.cf/post/solution-luogu-cf156b-codeforces-156b/</id>
<link href="https://blog.ak-ioi.cf/post/solution-luogu-cf156b-codeforces-156b/">
</link>
<updated>2021-01-20T11:11:05.000Z</updated>
<content type="html"><![CDATA[<p><span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">n</span></span></span></span> 个人中有 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span> 个罪犯，每个人说一句话，说第 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi></mrow><annotation encoding="application/x-tex">i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.65952em;vertical-align:0em;"></span><span class="mord mathdefault">i</span></span></span></span> 个人是/不是罪犯，其中有 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>m</mi></mrow><annotation encoding="application/x-tex">m</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">m</span></span></span></span> 句真话，要判断这些话的真假。</p>
<p>思路：</p>
<p>首先假设所有人都不是罪犯，统计出真话条数，然后枚举每一个人，在这个真话条数的基础上加上这个人是罪犯时对答案的贡献，如果刚好有 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>m</mi></mrow><annotation encoding="application/x-tex">m</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">m</span></span></span></span> 句真话，标记有嫌疑，最后判断即可。</p>
<p>但是这个贡献怎么求呢？</p>
<p>很简单，如果所有人都不是罪犯，则真话条数就是 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>a</mi><mi>i</mi></msub><mo><</mo><mn>0</mn></mrow><annotation encoding="application/x-tex">a_i\lt 0</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6891em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel"><</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">0</span></span></span></span> 的数量；当 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>x</mi></mrow><annotation encoding="application/x-tex">x</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">x</span></span></span></span> 是罪犯时，减去 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>a</mi><mi>i</mi></msub><mo>=</mo><mo>−</mo><mi>x</mi></mrow><annotation encoding="application/x-tex">a_i=-x</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.58056em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.66666em;vertical-align:-0.08333em;"></span><span class="mord">−</span><span class="mord mathdefault">x</span></span></span></span> 的数量，加上 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>a</mi><mi>i</mi></msub><mo>=</mo><mi>x</mi></mrow><annotation encoding="application/x-tex">a_i=x</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.58056em;vertical-align:-0.15em;"></span><span class="mord"><span class="mord mathdefault">a</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.31166399999999994em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">x</span></span></span></span> 的数量，即为最新的贡献，这里在读入时也处理出来。</p>
<p>最后话的真假的判断方式也很简单：</p>
<p>如果 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi></mrow><annotation encoding="application/x-tex">i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.65952em;vertical-align:0em;"></span><span class="mord mathdefault">i</span></span></span></span> 指控一个人是罪犯：经过如上处理，发现他不可能是罪犯，则为假话；如果他可能是罪犯且嫌疑人数为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span>，则为真话；否则不确定。说一个人不是罪犯的情况类似，不展开细说，可以参考代码。</p>
<hr>
<p>代码：</p>
<pre><code class="language-cpp">//By: Luogu@rui_er(122461)
#include <bits/stdc++.h>
#define loop while(true)
#define rep(x,y,z) for(int x=y;x<=z;x++)
#define per(x,y,z) for(int x=y;x>=z;x--)
#define fil(x,y) memset(x, y, sizeof(x))
#define mulT0 int T; for(scanf("%d", &T);T;T--)
#define mulT1 int T, rnds; for(scanf("%d", &T),rnds=1;rnds<=T;rnds++)
using namespace std;
typedef long long ll;
const int N = 1e5+5;
int n, m, a[N], contribution[N], suspect[N], tot, s;
int main() {
scanf("%d%d", &n, &m);
rep(i, 1, n) {
scanf("%d", &a[i]);
if(a[i] > 0) ++contribution[a[i]];
else --contribution[-a[i]], ++s;
}
rep(i, 1, n) if(s + contribution[i] == m) suspect[i] = 1, ++tot;
rep(i, 1, n) {
if(a[i] > 0) {
if(!suspect[a[i]]) puts("Lie");
else if(tot == 1) puts("Truth");
else puts("Not defined");
}
else {
if(!suspect[-a[i]]) puts("Truth");
else if(tot == 1) puts("Lie");
else puts("Not defined");
}
}
return 0;
}
</code></pre>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[题解 CF128D@洛谷 | 128D@Codeforces 【Numbers】]]></title>
<id>https://blog.ak-ioi.cf/post/solution-luogu-cf128d-codeforces-128d/</id>
<link href="https://blog.ak-ioi.cf/post/solution-luogu-cf128d-codeforces-128d/">
</link>
<updated>2021-01-18T10:51:17.000Z</updated>
<content type="html"><![CDATA[<p>简单构造题。</p>
<p>题意：给定 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">n</span></span></span></span> 个数，是否能排成一圈，使相邻两个数差 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span>。</p>
<hr>
<p>根据题意，为了使相邻两个数差 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span>，圆圈中 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mo>+</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">+1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">+</span><span class="mord">1</span></span></span></span> 和 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mo>−</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">-1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">−</span><span class="mord">1</span></span></span></span> 的次数必定相等才能组成闭环，因此 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">n</span></span></span></span> 必须是偶数，对于奇数可以直接输出无解（代码注释 <code>{1}</code>）。</p>
<p>然后同样根据相邻两个数差 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span> 的性质，我们将原数列排序，枚举相邻两个下标的数，如果差大于 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span> 也输出无解（代码注释 <code>{2}</code>）。</p>
<p>之后找到数列最小值，把每个数减去最小值再加 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span>，即可得到一串与原数列等效的、最小值为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span> 的数列，经过前面两个特判，可以保证最大值不超过 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn><msup><mn>0</mn><mn>5</mn></msup></mrow><annotation encoding="application/x-tex">10^5</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8141079999999999em;vertical-align:0em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141079999999999em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">5</span></span></span></span></span></span></span></span></span></span></span>。然后从最小值开始，可以 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mo>+</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">+1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">+</span><span class="mord">1</span></span></span></span> 时选择 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mo>+</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">+1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">+</span><span class="mord">1</span></span></span></span>，否则看看是否可以 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mo>−</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">-1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">−</span><span class="mord">1</span></span></span></span>，若不可以就跳出循环。</p>
<p>由于闭环性质，要形成环则最后一个数必定是 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>2</mn></mrow><annotation encoding="application/x-tex">2</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">2</span></span></span></span>，如果不是 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>2</mn></mrow><annotation encoding="application/x-tex">2</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">2</span></span></span></span> 则不能成环，输出无解（代码注释 <code>{3}</code>）。</p>
<p>否则将整个桶扫一遍，如果有地方不为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>0</mn></mrow><annotation encoding="application/x-tex">0</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">0</span></span></span></span> 则代表成环失败，输出无解（代码注释 <code>{4}</code>）。</p>
<p>否则即代表我们构造出了一个符合题意的环，输出有解即可。</p>
<p>这样就水过了一道 CF <code>*2000</code> 的构造题（</p>
<hr>
<p>代码：</p>
<pre><code class="language-cpp">//By: Luogu@rui_er(122461)
#include <bits/stdc++.h>
#define loop while(true)
#define rep(x,y,z) for(int x=y;x<=z;x++)
#define per(x,y,z) for(int x=y;x>=z;x--)
#define fil(x,y) memset(x, y, sizeof(x))
#define mulT0 int T; for(scanf("%d", &T);T;T--)
#define mulT1 int T, rnds; for(scanf("%d", &T),rnds=1;rnds<=T;rnds++)
using namespace std;
typedef long long ll;
const int N = 1e5+5, inf = 1e9;
int n, a[N], buc[N], mi = inf;
int main() {
scanf("%d", &n);
if(n & 1) return puts("NO"), 0; // {1}
rep(i, 1, n) scanf("%d", &a[i]), mi = min(mi, a[i]);
sort(a+1, a+1+n);
rep(i, 2, n) if(a[i] - a[i-1] > 1) return puts("NO"), 0; // {2}
rep(i, 1, n) ++buc[a[i]-mi+1];
int u = 1; --buc[1];
loop {
if(buc[u+1]) --buc[++u];
else if(buc[u-1]) --buc[--u];
else break;
}
if(u != 2) return puts("NO"), 0; // {3}
rep(i, 0, N-1) if(buc[i]) return puts("NO"), 0; // {4}
return puts("YES"), 0;
}
</code></pre>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[题解 CF1365E@洛谷 | 1365E@Codeforces 【Maximum Subsequence Value】]]></title>
<id>https://blog.ak-ioi.cf/post/solution-luogu-cf1365e-codeforces-1365e/</id>
<link href="https://blog.ak-ioi.cf/post/solution-luogu-cf1365e-codeforces-1365e/">
</link>
<updated>2021-01-17T15:01:35.000Z</updated>
<content type="html"><![CDATA[<p>思路：</p>
<p>考虑如何使取出的 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>k</mi></mrow><annotation encoding="application/x-tex">k</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.69444em;vertical-align:0em;"></span><span class="mord mathdefault" style="margin-right:0.03148em;">k</span></span></span></span> 个数价值最大。</p>
<p>对于 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>k</mi><mo>∈</mo><mrow><mo fence="true">{</mo><mn>1</mn><mo separator="true">,</mo><mn>2</mn><mo separator="true">,</mo><mn>3</mn><mo fence="true">}</mo></mrow></mrow><annotation encoding="application/x-tex">k\in\left\{1,2,3\right\}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.73354em;vertical-align:-0.0391em;"></span><span class="mord mathdefault" style="margin-right:0.03148em;">k</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">∈</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="minner"><span class="mopen delimcenter" style="top:0em;">{</span><span class="mord">1</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord">2</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord">3</span><span class="mclose delimcenter" style="top:0em;">}</span></span></span></span></span>，根据定义，若 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>k</mi></mrow><annotation encoding="application/x-tex">k</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.69444em;vertical-align:0em;"></span><span class="mord mathdefault" style="margin-right:0.03148em;">k</span></span></span></span> 个数中至少有一个数的二进制位 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi></mrow><annotation encoding="application/x-tex">i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.65952em;vertical-align:0em;"></span><span class="mord mathdefault">i</span></span></span></span> 上是 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span>，则价值增加 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msup><mn>2</mn><mi>i</mi></msup></mrow><annotation encoding="application/x-tex">2^i</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.824664em;vertical-align:0em;"></span><span class="mord"><span class="mord">2</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.824664em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight">i</span></span></span></span></span></span></span></span></span></span></span>，这个定义十分眼熟，发现就是按位或的结果。对于 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>k</mi><mo>≥</mo><mn>4</mn></mrow><annotation encoding="application/x-tex">k\ge 4</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.83041em;vertical-align:-0.13597em;"></span><span class="mord mathdefault" style="margin-right:0.03148em;">k</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">≥</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">4</span></span></span></span>，<span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>k</mi></mrow><annotation encoding="application/x-tex">k</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.69444em;vertical-align:0em;"></span><span class="mord mathdefault" style="margin-right:0.03148em;">k</span></span></span></span> 每增加 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span>，就会使 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>max</mi><mo></mo><mrow><mo fence="true">(</mo><mn>1</mn><mo separator="true">,</mo><mi>k</mi><mo>−</mo><mn>2</mn><mo fence="true">)</mo></mrow></mrow><annotation encoding="application/x-tex">\max\left(1,k-2\right)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mop">max</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="minner"><span class="mopen delimcenter" style="top:0em;">(</span><span class="mord">1</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault" style="margin-right:0.03148em;">k</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">−</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mord">2</span><span class="mclose delimcenter" style="top:0em;">)</span></span></span></span></span> 增加 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span>，因此每个二进制位的出现次数都需要加 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span> 才能使答案更优，显然不如 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>k</mi><mo>=</mo><mn>3</mn></mrow><annotation encoding="application/x-tex">k=3</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.69444em;vertical-align:0em;"></span><span class="mord mathdefault" style="margin-right:0.03148em;">k</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">3</span></span></span></span> 的情况。</p>
<p>因此对于 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>n</mi><mo>∈</mo><mrow><mo fence="true">{</mo><mn>1</mn><mo separator="true">,</mo><mn>2</mn><mo fence="true">}</mo></mrow></mrow><annotation encoding="application/x-tex">n\in\left\{1,2\right\}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.5782em;vertical-align:-0.0391em;"></span><span class="mord mathdefault">n</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">∈</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="minner"><span class="mopen delimcenter" style="top:0em;">{</span><span class="mord">1</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord">2</span><span class="mclose delimcenter" style="top:0em;">}</span></span></span></span></span>，答案即为这些数的或；对于 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>n</mi><mo>≥</mo><mn>3</mn></mrow><annotation encoding="application/x-tex">n\ge 3</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.7719400000000001em;vertical-align:-0.13597em;"></span><span class="mord mathdefault">n</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">≥</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">3</span></span></span></span>，答案为数列任取三个数能得到的最大的或。可以花费 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="script">O</mi><mo>(</mo><msup><mi>n</mi><mn>3</mn></msup><mo>)</mo></mrow><annotation encoding="application/x-tex">\mathcal{O}(n^3)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.064108em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathcal" style="margin-right:0.02778em;">O</span></span><span class="mopen">(</span><span class="mord"><span class="mord mathdefault">n</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141079999999999em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">3</span></span></span></span></span></span></span></span><span class="mclose">)</span></span></span></span> 枚举取的三个数得到答案。</p>
<p>因为懒得特判前者，我选择牺牲了一定的代码常数，将 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>i</mi><mo separator="true">,</mo><mi>j</mi><mo separator="true">,</mo><mi>k</mi></mrow><annotation encoding="application/x-tex">i,j,k</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8888799999999999em;vertical-align:-0.19444em;"></span><span class="mord mathdefault">i</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault" style="margin-right:0.05724em;">j</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault" style="margin-right:0.03148em;">k</span></span></span></span> 都从 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span> 枚举到 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">n</span></span></span></span>，但是对于前者也能得到正确的结果。</p>
<hr>
<p>代码：</p>
<pre><code class="language-cpp">//By: Luogu@rui_er(122461)
#include <bits/stdc++.h>
#define loop while(true)
#define rep(x,y,z) for(ll x=y;x<=z;x++)
#define per(x,y,z) for(ll x=y;x>=z;x--)
#define fil(x,y) memset(x, y, sizeof(x))
#define mulT0 ll T; for(scanf("%lld", &T);T;T--)
#define mulT1 ll T, rnds; for(scanf("%lld", &T),rnds=1;rnds<=T;rnds++)
using namespace std;
typedef long long ll;
const ll N = 505;
ll n, a[N], ans;
int main() {
scanf("%lld", &n);
rep(i, 1, n) scanf("%lld", &a[i]);
rep(i, 1, n) rep(j, 1, n) rep(k, 1, n) ans = max(ans, a[i]|a[j]|a[k]);
printf("%lld\n", ans);
return 0;
}
</code></pre>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[题解 P7262@洛谷 【Get Your Wish】]]></title>
<id>https://blog.ak-ioi.cf/post/solution-luogu-p7262/</id>
<link href="https://blog.ak-ioi.cf/post/solution-luogu-p7262/">
</link>
<updated>2021-01-10T13:45:23.000Z</updated>
<content type="html"><![CDATA[<p>题意很清楚了这里就不过多解释了，说一下思路：</p>
<p>一看题，感觉很多水滴的时候较难模拟下落过程，因此考虑根据给出的重力方向和电子元件反推。我们使用 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>s</mi><mrow><mi>i</mi><mo separator="true">,</mo><mi>j</mi></mrow></msub></mrow><annotation encoding="application/x-tex">s_{i,j}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.716668em;vertical-align:-0.286108em;"></span><span class="mord"><span class="mord mathdefault">s</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.311664em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mathdefault mtight">i</span><span class="mpunct mtight">,</span><span class="mord mathdefault mtight" style="margin-right:0.05724em;">j</span></span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.286108em;"><span></span></span></span></span></span></span></span></span></span> 表示如果 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mo fence="true">(</mo><mi>i</mi><mo separator="true">,</mo><mi>j</mi><mo fence="true">)</mo></mrow><annotation encoding="application/x-tex">\left(i,j\right)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="minner"><span class="mopen delimcenter" style="top:0em;">(</span><span class="mord mathdefault">i</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathdefault" style="margin-right:0.05724em;">j</span><span class="mclose delimcenter" style="top:0em;">)</span></span></span></span></span> 位置有水滴，是否会影响到电子元件。则以方向竖直向下为例，<span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>s</mi><mrow><mi>i</mi><mo separator="true">,</mo><mi>j</mi></mrow></msub><mo>=</mo><msub><mi>s</mi><mrow><mi>i</mi><mo separator="true">,</mo><mi>j</mi><mo>+</mo><mn>1</mn></mrow></msub><mi mathvariant="normal">∣</mi><mrow><mo fence="true">(</mo><mtext>当前位置有水滴</mtext><mo fence="true">)</mo></mrow></mrow><annotation encoding="application/x-tex">s_{i,j}=s_{i,j+1}|\left(\text{当前位置有水滴}\right)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.716668em;vertical-align:-0.286108em;"></span><span class="mord"><span class="mord mathdefault">s</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.311664em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mathdefault mtight">i</span><span class="mpunct mtight">,</span><span class="mord mathdefault mtight" style="margin-right:0.05724em;">j</span></span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.286108em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:1.036108em;vertical-align:-0.286108em;"></span><span class="mord"><span class="mord mathdefault">s</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.311664em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mathdefault mtight">i</span><span class="mpunct mtight">,</span><span class="mord mathdefault mtight" style="margin-right:0.05724em;">j</span><span class="mbin mtight">+</span><span class="mord mtight">1</span></span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.286108em;"><span></span></span></span></span></span></span><span class="mord">∣</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="minner"><span class="mopen delimcenter" style="top:0em;">(</span><span class="mord text"><span class="mord cjk_fallback">当前位置有水滴</span></span><span class="mclose delimcenter" style="top:0em;">)</span></span></span></span></span>，其他三个方向以此类推，其中 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi mathvariant="normal">∣</mi></mrow><annotation encoding="application/x-tex">|</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord">∣</span></span></span></span> 表示或，即两个条件中任一满足则为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span>。</p>
<p>然后枚举整个 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>n</mi><mo>×</mo><mi>m</mi></mrow><annotation encoding="application/x-tex">n\times m</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.66666em;vertical-align:-0.08333em;"></span><span class="mord mathdefault">n</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathdefault">m</span></span></span></span> 的图：如果存在水滴使对应位置的 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>s</mi><mrow><mi>i</mi><mo separator="true">,</mo><mi>j</mi></mrow></msub><mo>=</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">s_{i,j}=1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.716668em;vertical-align:-0.286108em;"></span><span class="mord"><span class="mord mathdefault">s</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.311664em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mathdefault mtight">i</span><span class="mpunct mtight">,</span><span class="mord mathdefault mtight" style="margin-right:0.05724em;">j</span></span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.286108em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span>，则答案必定为 <code>GG</code>；若对于所有水滴对应位置的 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><msub><mi>s</mi><mrow><mi>i</mi><mo separator="true">,</mo><mi>j</mi></mrow></msub></mrow><annotation encoding="application/x-tex">s_{i,j}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.716668em;vertical-align:-0.286108em;"></span><span class="mord"><span class="mord mathdefault">s</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.311664em;"><span style="top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mathdefault mtight">i</span><span class="mpunct mtight">,</span><span class="mord mathdefault mtight" style="margin-right:0.05724em;">j</span></span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.286108em;"><span></span></span></span></span></span></span></span></span></span> 均为 <span class="katex"><span class="katex-mathml"><math><semantics><mrow><mn>0</mn></mrow><annotation encoding="application/x-tex">0</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">0</span></span></span></span>，则答案为 <code>OK</code>。</p>
<p>赛时代码：</p>
<pre><code class="language-cpp">//By: Luogu@rui_er(122461)
#include <bits/stdc++.h>
#define loop while(true)
#define rep(x,y,z) for(int x=y;x<=z;x++)
#define per(x,y,z) for(int x=y;x>=z;x--)
#define fil(x,y) memset(x, y, sizeof(x))
#define mulT0 int T; for(scanf("%d", &T);T;T--)
#define mulT1 int T, rnds; for(scanf("%d", &T),rnds=1;rnds<=T;rnds++)
using namespace std;
typedef long long ll;
const int N = 105;
//const char nxt[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
int n, m, pos, s[N][N], ans = 0;
char mp[N][N], poss[N];
int main() {
scanf("%d%d%s", &n, &m, poss+1);
rep(i, 1, n) scanf("%s", mp[i]+1);
pos = (poss[1] == 'v' ? 1 : (poss[1] == '<' ? 2 : (poss[1] == '^' ? 3 : 0)));
if(!pos) rep(i, 1, n) per(j, m, 1) s[i][j] = s[i][j+1] | (mp[i][j] == 'x');
else if(pos == 1) rep(j, 1, m) per(i, n, 1) s[i][j] = s[i+1][j] | (mp[i][j] == 'x');
else if(pos == 2) rep(i, 1, n) rep(j, 1, m) s[i][j] = s[i][j-1] | (mp[i][j] == 'x');
else rep(j, 1, m) rep(i, 1, n) s[i][j] = s[i-1][j] | (mp[i][j] == 'x');
rep(i, 1, n) rep(j, 1, m) ans |= (s[i][j] && (mp[i][j] == 'o'));
printf("%s\n", (char*[]){"OK", "GG"}[ans]);
return 0;
}
</code></pre>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[2020 年 10~12 月水题选做]]></title>
<id>https://blog.ak-ioi.cf/post/2020-10-to-12/</id>
<link href="https://blog.ak-ioi.cf/post/2020-10-to-12/">
</link>
<updated>2020-12-20T11:40:10.000Z</updated>
<content type="html"><![CDATA[<p>由于格式渲染和相对地址等原因，请前往 <a href="https://www.luogu.com.cn/blog/ak-ioi/2020-10-to-12">洛谷博客</a> 查看。</p>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[CSP2020-S & NOIP2020 & THUPC2021 游记]]></title>
<id>https://blog.ak-ioi.cf/post/i-ak-csp2020-s-noip2020-thupc2021/</id>
<link href="https://blog.ak-ioi.cf/post/i-ak-csp2020-s-noip2020-thupc2021/">
</link>
<updated>2020-12-15T09:55:21.000Z</updated>
<content type="html"><![CDATA[<ul>
<li>https://www.luogu.com.cn/blog/ak-ioi/i-ak-csp2020</li>
<li>https://www.luogu.com.cn/blog/ak-ioi/i-ak-noip2020-qwq</li>
<li>https://www.luogu.com.cn/blog/ak-ioi/i-ak-thupc2021</li>
</ul>
]]></content>
</entry>
</feed>