November 2016

[ICPC2016Dalian Onsite] Official explanation

• Author: rin
• Time: November 6, 2016
• Classification: Algo
• 7 comments

<trans data-src="[PDF版](">[PDF version]（</trans> <trans data-src="http://pan.baidu.com/s/1bpkkoTD">http://pan.baidu.com/s/1bpkkoTD</trans> <trans data-src=")">)</trans> <trans data-src="> 添加 `-D__STDC_FORMAT_MACROS`以支持`PRId64`">>Add '- D_STDC_FORMAT_MACROS' to support' PRId64`</trans> <trans data-src="#Problem A Wrestling Match">#Problem A Wrestling Match</trans> <trans data-src="##提交情况：170/434">##Submission: 170/434</trans> <trans data-src="##解题思路">##Solutions</trans> <trans data-src="本题做法比较随意，此处说两种：">The method of this question is quite casual. Here are two ways:</trans> <trans data-src="1. 读入数据，建好图以后，以每一个点为开始遍历的起始节点，搜索整个图，在搜索的过程中判断是否出现冲突">1. After reading data and building a graph, search the whole graph with each point as the starting node, and judge whether there is conflict in the process of searching</trans> <trans data-src="2. 用并查集维护题目中所谓『好选手』以及『坏选手』的关系，在建立并查集的过程中，就可判断冲突">2. Maintain the relationship between the so-called "good players" and "bad players" in the topic by using the combined search set. In the process of establishing and searching the set, conflicts can be judged</trans> <trans data-src="总的来说，第2种做法需要处理的特殊情况较少，并且计算量照第1种做法少，所以标程采用第2种做法。">In general, the second method needs to deal with fewer special cases, and the calculation amount is less according to the first method, so the second method is adopted for the scale.</trans>

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[CCPC2016Hefei] I / HDU5969

• Author: rin
• Time: November 5, 2016
• Classification: Algo
• 4 comments

<trans data-src="[">[</trans> <trans data-src="http://acm.hdu.edu.cn/showproblem.php?pid=5969">http://acm.hdu.edu.cn/showproblem.php?pid=5969</trans> <trans data-src="](">](</trans> <trans data-src="http://acm.hdu.edu.cn/showproblem.php?pid=5969">http://acm.hdu.edu.cn/showproblem.php?pid=5969</trans> <trans data-src=" ""> "</trans> <trans data-src="http://acm.hdu.edu.cn/showproblem.php?pid=5969">http://acm.hdu.edu.cn/showproblem.php?pid=5969</trans> <trans data-src="")">")</trans> <trans data-src=">###Problem Description">>###Problem Description</trans> <trans data-src=">B君和G君聊天的时候想到了如下的问题。">>When chatting with Mr. G, Mr. B thought of the following questions.</trans> <trans data-src="给定自然数$$l$$和$$r$$ ,选取2个整数$$x,y$$满足$$l \leq x \leq y \leq r $$,使得$$x|y$$最大。">Given the natural number $$l $$and $$r $$, select two integers $$x, and y $$satisfies $$l leq x leq y leq r $$, making $$x | y $$the largest.</trans> <trans data-src="其中|表示按位或，即C、 C++、 Java中的|运算。">Where | represents bitwise OR, that is, the | operation in C, C++, and Java.</trans> <trans data-src="###Input">###Input</trans> <trans data-src="包含至多10001组测试数据。">Contains up to 10001 sets of test data.</trans> <trans data-src="第一行有一个正整数，表示数据的组数。">The first line has a positive integer, representing the number of groups of data.</trans> <trans data-src="接下来每一行表示一组数据，包含两个整数$$l$$,$$r$$。">Next, each row represents a set of data, including two integers $$l $$, $$r $$.</trans> <trans data-src="保证 $$0 \leq l \leq r \leq 10^{18}$$。">Guarantee $$0 leq l leq r leq 10 ^ {18} $$.</trans> <trans data-src="###Output">###Output</trans> <trans data-src="对于每组数据输出一行，表示最大的位或。">Output one row for each set of data, representing the largest bit or.</trans> <trans data-src="1.钦定上界就是$$y=r$$ :hj-huamoji52:">1. The imperial upper boundary is $$y=r $$: hj-huamoji52:</trans> <trans data-src="2.判断$$l$$和$$r$$二进制位数的大小关系$$lenL, lenR$$">2. Judge the size relationship between $$l $$and $$r $$binary digits $$lenL, lenR$$</trans> <trans data-src="3.如果不相等，那么$$x=(111111\ldots )_{2}$$（$$lenR-1$$个1）一定大于等于$$l$$，小于等于$$r$$，那么$$l|r$$的结果就是$$2^">3. If they are not equal, then $$x=(111111 ldots) _ {2} $$($$lenR-1 $$1) must be greater than or equal to $$l $$, and less than or equal to $$r $$, then the result of $$l | r $$is $$2^</trans> <trans data-src="{lenR}-1">{lenR}-1</trans> <trans data-src="$$">$$</trans> <trans data-src="4.如果相等，令$$x=l$$，将$$x,r$$的最高位对齐，从高位往低位一位一位比较，把碰到的第一个不同的位$$i$$，置$$x[i]=0$$，再将$$x$$剩下的都置1">4. If equal, make $$x=l $$, align the highest order of $$x, r $$, compare from high order to low order, set $$i $$, the first different bit encountered, $$x [i]=0 $$, and then set the rest of $$x $$to 1</trans>

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