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A 数组推导大意$B$ 数组是 $A$ 数组的前缀最大值。给定 $B$ 数组,求 $A$ 数组元素和的最大值与最小值。
解析最小值:在改变处为该值,其他处为 $0$。最大值:总是目前最大值,即和 $B$ 数组相同。
代码cpp12345678910111213141 ...</div></div></div><div class="recent-post-item article-container"><div class="post_cover right_radius"><a href="/blog/2021/08/30/ccpc2021net/" title="2021 CCPC 网络赛"><img class="post_bg lazyload" data-src="/blog/img/pictures_for_blog/18.jpg" onerror="this.onerror=null;this.src='/blog/img/404.jpg'" alt="2021 CCPC 网络赛"></a></div><div class="recent-post-info"> <a class="article-title" href="/blog/2021/08/30/ccpc2021net/" title="2021 CCPC 网络赛">2021 CCPC 网络赛</a><time class="post-meta__date"><i class="fa fa-calendar" aria-hidden="true"></i> 2021-08-30</time><span class="article-meta"><span class="article-meta__separator">|</span><i class="fa fa-inbox article-meta__icon" aria-hidden="true"></i><a class="article-meta__categories" href="/blog/categories/%E4%BB%8E%E8%92%9F%E8%92%BB%E5%88%B0%E8%92%9F%E8%92%BB%E7%9A%84%E4%BF%AE%E7%82%BC/">从蒟蒻到蒟蒻的修炼</a></span><div class="content">部分题解(ABFGIK)题目链接:2021中国大学生程序设计竞赛(CCPC)- 网络选拔赛
A. Cut The Wire大意有无穷多个点编号 $1,2,3,\dots$。对于编号为 $x$ 的点满足:
如果 $x$ 是偶数,则 $x$ 与 $\frac{x}{2}$ 之间有一条边。
如果 $x$ ...</div></div></div><div class="recent-post-item article-container"><div class="post_cover left_radius"><a href="/blog/2020/03/28/luoguP4796/" title="洛谷 P4796 [BalticOI 2018]路径"><img class="post_bg lazyload" data-src="/blog/img/pictures_for_blog/15.jpg" onerror="this.onerror=null;this.src='/blog/img/404.jpg'" alt="洛谷 P4796 [BalticOI 2018]路径"></a></div><div class="recent-post-info"> <a class="article-title" href="/blog/2020/03/28/luoguP4796/" title="洛谷 P4796 [BalticOI 2018]路径">洛谷 P4796 [BalticOI 2018]路径</a><time class="post-meta__date"><i class="fa fa-calendar" aria-hidden="true"></i> 2020-03-28</time><span class="article-meta"><span class="article-meta__separator">|</span><i class="fa fa-inbox article-meta__icon" aria-hidden="true"></i><a class="article-meta__categories" href="/blog/categories/%E4%BB%8E%E8%92%9F%E8%92%BB%E5%88%B0%E8%92%9F%E8%92%BB%E7%9A%84%E4%BF%AE%E7%82%BC/">从蒟蒻到蒟蒻的修炼</a></span><div class="content">题目相关题目链接:P4796 [BalticOI 2018]路径题目描述:  给定一张 $N$ 个点 $M$ 条边的无向图,每个点有一个颜色,所有点的颜色共有 $K$ 种,编号为 $1\ldots K$。求图上有多少条长度至少为 $2$ 的简单路径,满足路径上的每一个点的颜色互不 ...</div></div></div><div class="recent-post-item article-container"><div class="post_cover right_radius"><a href="/blog/2020/02/22/2019nanchang/" title="2019 ICPC Asia Nanchang Regional"><img class="post_bg lazyload" data-src="/blog/img/pictures_for_blog/14.jpg" onerror="this.onerror=null;this.src='/blog/img/404.jpg'" alt="2019 ICPC Asia Nanchang Regional"></a></div><div class="recent-post-info"> <a class="article-title" href="/blog/2020/02/22/2019nanchang/" title="2019 ICPC Asia Nanchang Regional">2019 ICPC Asia Nanchang Regional</a><time class="post-meta__date"><i class="fa fa-calendar" aria-hidden="true"></i> 2020-02-22</time><span class="article-meta"><span class="article-meta__separator">|</span><i class="fa fa-inbox article-meta__icon" aria-hidden="true"></i><a class="article-meta__categories" href="/blog/categories/%E4%BB%8E%E8%92%9F%E8%92%BB%E5%88%B0%E8%92%9F%E8%92%BB%E7%9A%84%E4%BF%AE%E7%82%BC/">从蒟蒻到蒟蒻的修炼</a></span><div class="content">2019 ICPC Asia Nanchang Regionalsolved:ACEGL 5/13A. 9102题意:  让你搞一个可持久化并查集,而且资瓷删除。不要求强制在线。
解析:  对于有时间戳要求但是不要求强制在线的,我们可以按照时间排序,依次处理。 ...</div></div></div><div class="recent-post-item article-container"><div class="post_cover left_radius"><a href="/blog/2020/02/02/luoguP2000/" title="洛谷 P2000 拯救世界"><img class="post_bg lazyload" data-src="/blog/img/pictures_for_blog/13.jpg" onerror="this.onerror=null;this.src='/blog/img/404.jpg'" alt="洛谷 P2000 拯救世界"></a></div><div class="recent-post-info"> <a class="article-title" href="/blog/2020/02/02/luoguP2000/" title="洛谷 P2000 拯救世界">洛谷 P2000 拯救世界</a><time class="post-meta__date"><i class="fa fa-calendar" aria-hidden="true"></i> 2020-02-02</time><span class="article-meta"><span class="article-meta__separator">|</span><i class="fa fa-inbox article-meta__icon" aria-hidden="true"></i><a class="article-meta__categories" href="/blog/categories/%E4%BB%8E%E8%92%9F%E8%92%BB%E5%88%B0%E8%92%9F%E8%92%BB%E7%9A%84%E4%BF%AE%E7%82%BC/">从蒟蒻到蒟蒻的修炼</a></span><div class="content">题目相关题目链接:P2000 拯救世界题目背景:  公元 2000 年,根据预言家诺查丹玛斯的预言,世界就要毁灭了!!!
题目描述:  为了拯救世界,小 a 和 uim 决定召唤出 kkksc03 大神和 lzn 大神。根据古籍记载,召唤出任何一位大神,都需要 ...</div></div></div><div class="recent-post-item article-container"><div class="post_cover right_radius"><a href="/blog/2020/01/31/luoguP6034/" title="洛谷 P6034 Ryoku 与最初之人笔记"><img class="post_bg lazyload" data-src="/blog/img/pictures_for_blog/12.jpg" onerror="this.onerror=null;this.src='/blog/img/404.jpg'" alt="洛谷 P6034 Ryoku 与最初之人笔记"></a></div><div class="recent-post-info"> <a class="article-title" href="/blog/2020/01/31/luoguP6034/" title="洛谷 P6034 Ryoku 与最初之人笔记">洛谷 P6034 Ryoku 与最初之人笔记</a><time class="post-meta__date"><i class="fa fa-calendar" aria-hidden="true"></i> 2020-01-31</time><span class="article-meta"><span class="article-meta__separator">|</span><i class="fa fa-inbox article-meta__icon" aria-hidden="true"></i><a class="article-meta__categories" href="/blog/categories/%E4%BB%8E%E8%92%9F%E8%92%BB%E5%88%B0%E8%92%9F%E8%92%BB%E7%9A%84%E4%BF%AE%E7%82%BC/">从蒟蒻到蒟蒻的修炼</a></span><div class="content">题目相关题目链接:Ryoku 与最初之人笔记题目背景:  Ryoku 在阅读「最初之人」的笔记的时候,发现了一个有趣的运算:$xor$,这个运算的输入是两个数,输出是一个数,对应的运算时将输入的两个数化为二进制,再把每一位进行比较,若相同则输出的二进制中的这一位为 $0$,否则为 ...</div></div></div><div class="recent-post-item article-container"><div class="post_cover left_radius"><a href="/blog/2020/01/13/luoguP5093/" title="洛谷 P5093 [USACO04OPEN]The Cow Lineup"><img class="post_bg lazyload" data-src="/blog/img/pictures_for_blog/11.jpg" onerror="this.onerror=null;this.src='/blog/img/404.jpg'" alt="洛谷 P5093 [USACO04OPEN]The Cow Lineup"></a></div><div class="recent-post-info"> <a class="article-title" href="/blog/2020/01/13/luoguP5093/" title="洛谷 P5093 [USACO04OPEN]The Cow Lineup">洛谷 P5093 [USACO04OPEN]The Cow Lineup</a><time class="post-meta__date"><i class="fa fa-calendar" aria-hidden="true"></i> 2020-01-13</time><span class="article-meta"><span class="article-meta__separator">|</span><i class="fa fa-inbox article-meta__icon" aria-hidden="true"></i><a class="article-meta__categories" href="/blog/categories/%E4%BB%8E%E8%92%9F%E8%92%BB%E5%88%B0%E8%92%9F%E8%92%BB%E7%9A%84%E4%BF%AE%E7%82%BC/">从蒟蒻到蒟蒻的修炼</a></span><div class="content">题目相关题目链接:洛谷 P5093 [USACO04OPEN]The Cow Lineup题目描述:  约翰的 $N$ ( $1 \leq N \leq 100000 $ )只奶牛站成了一列。每只奶牛都写有一个号牌,表示她的品种,号牌上的号码在 $1 \ldots K $ ( $ ...</div></div></div><div class="recent-post-item article-container"><div class="post_cover right_radius"><a href="/blog/2019/12/19/luoguP4683/" title="洛谷 P4683 [IOI2008] Type Printer 打印机"><img class="post_bg lazyload" data-src="/blog/img/pictures_for_blog/10.jpg" onerror="this.onerror=null;this.src='/blog/img/404.jpg'" alt="洛谷 P4683 [IOI2008] Type Printer 打印机"></a></div><div class="recent-post-info"> <a class="article-title" href="/blog/2019/12/19/luoguP4683/" title="洛谷 P4683 [IOI2008] Type Printer 打印机">洛谷 P4683 [IOI2008] Type Printer 打印机</a><time class="post-meta__date"><i class="fa fa-calendar" aria-hidden="true"></i> 2019-12-19</time><span class="article-meta"><span class="article-meta__separator">|</span><i class="fa fa-inbox article-meta__icon" aria-hidden="true"></i><a class="article-meta__categories" href="/blog/categories/%E4%BB%8E%E8%92%9F%E8%92%BB%E5%88%B0%E8%92%9F%E8%92%BB%E7%9A%84%E4%BF%AE%E7%82%BC/">从蒟蒻到蒟蒻的修炼</a></span><div class="content">题目相关题目链接:洛谷 P4683 [IOI2008] Type Printer 打印机题目描述:  你需要利用一台可移动的打印机打印出 $N$ 个单词。这种可移动式打印机是一种老式打印机,它需要你将一些小的金属块(每个包含一个字母)放到打印机上以组成单词。然后将这些小金属块压在 ...</div></div></div><div class="recent-post-item article-container"><div class="post_cover left_radius"><a href="/blog/2019/12/08/cf1254b/" title="Codeforces Round #601 (Div. 1) B2. Send Boxes to Alice (Hard Version)"><img class="post_bg lazyload" data-src="/blog/img/pictures_for_blog/9.jpg" onerror="this.onerror=null;this.src='/blog/img/404.jpg'" alt="Codeforces Round #601 (Div. 1) B2. Send Boxes to Alice (Hard Version)"></a></div><div class="recent-post-info"> <a class="article-title" href="/blog/2019/12/08/cf1254b/" title="Codeforces Round #601 (Div. 1) B2. Send Boxes to Alice (Hard Version)">Codeforces Round #601 (Div. 1) B2. Send Boxes to Alice (Hard Version)</a><time class="post-meta__date"><i class="fa fa-calendar" aria-hidden="true"></i> 2019-12-08</time><span class="article-meta"><span class="article-meta__separator">|</span><i class="fa fa-inbox article-meta__icon" aria-hidden="true"></i><a class="article-meta__categories" href="/blog/categories/%E4%BB%8E%E8%92%9F%E8%92%BB%E5%88%B0%E8%92%9F%E8%92%BB%E7%9A%84%E4%BF%AE%E7%82%BC/">从蒟蒻到蒟蒻的修炼</a></span><div class="content">题目相关题目链接(镜像):Codeforces Round #601 (Div. 1) B2. Send Boxes to Alice (Hard Version)题目描述:
  英文原文我就不打了这里只放个翻译。。。  给定 $n$ 个盒子,第 $i$ 个盒 ...</div></div></div><div class="recent-post-item article-container"><div class="post_cover right_radius"><a href="/blog/2019/12/04/luoguP2522/" title="洛谷 P2522 [HAOI2011]Problem b"><img class="post_bg lazyload" data-src="/blog/img/pictures_for_blog/8.jpg" onerror="this.onerror=null;this.src='/blog/img/404.jpg'" alt="洛谷 P2522 [HAOI2011]Problem b"></a></div><div class="recent-post-info"> <a class="article-title" href="/blog/2019/12/04/luoguP2522/" title="洛谷 P2522 [HAOI2011]Problem b">洛谷 P2522 [HAOI2011]Problem b</a><time class="post-meta__date"><i class="fa fa-calendar" aria-hidden="true"></i> 2019-12-04</time><span class="article-meta"><span class="article-meta__separator">|</span><i class="fa fa-inbox article-meta__icon" aria-hidden="true"></i><a class="article-meta__categories" href="/blog/categories/%E4%BB%8E%E8%92%9F%E8%92%BB%E5%88%B0%E8%92%9F%E8%92%BB%E7%9A%84%E4%BF%AE%E7%82%BC/">从蒟蒻到蒟蒻的修炼</a></span><div class="content">题目相关题目链接:洛谷 P2522 [HAOI2011]Problem b题目描述:
  对于给出的 $n$ 个询问,每次求有多少个数对 $(x,y)$ ,满足 $a≤x≤b$ , $c≤y≤d$ ,且 $gcd(x,y)=k$ ,$gcd(x,y)$ 函数为x和y的最大公约数。 ...</div></div></div><nav id="pagination"><div class="pagination"><span class="page-number current">1</span><a class="page-number" href="/blog/page/2/">2</a><a class="extend next" rel="next" href="/blog/page/2/"><i class="fa fa-fw fa-chevron-right" aria-hidden="true"></i></a></div></nav></div><div class="aside_content" id="aside_content"><div class="card-widget card-info"><div class="card-content"><div class="card-info-avatar is-center"><img class="avatar-img" src="/blog/img/avatar_2.jpg" onerror="this.onerror=null;this.src='/blog/img/friend_404.gif'" alt="avatar"><div class="author-info__name">lonlyn</div><div class="author-info__description"></div></div><div class="card-info-data"><div class="card-info-data-item is-center"><a href="/blog/archives/"><div class="headline">文章</div><div class="length_num">18</div></a></div><div class="card-info-data-item is-center"> <a 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