<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>水 on 111qqz's blog</title><link>https://111qqz.com/en/tags/%E6%B0%B4/</link><description>Recent content in 水 on 111qqz's blog</description><generator>Hugo -- gohugo.io</generator><language>en</language><copyright>© 2015-2026 111qqz</copyright><lastBuildDate>Sat, 27 Aug 2016 06:52:38 +0000</lastBuildDate><atom:link href="https://111qqz.com/en/tags/%E6%B0%B4/index.xml" rel="self" type="application/rss+xml"/><item><title>seerc 2014 D - Frame (傻逼题)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-08-27-seerc-2014-d-frame-e582bbe980bce9a298/</link><pubDate>Sat, 27 Aug 2016 06:52:38 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-08-27-seerc-2014-d-frame-e582bbe980bce9a298/</guid><description>&lt;p&gt;&lt;a href="http://acm.hust.edu.cn/vjudge/contest/130303#problem/D" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;思路：注意xy-(x-2)*(y-2)=2x+2y-4，一定被2整除。因此siz为2的也是合法的。这个比较容易忘掉。&lt;/p&gt;</description></item><item><title>bc #77 div 2 1001 ||hdu 5650 so easy (傻逼题)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-27-hdu5650/</link><pubDate>Sun, 27 Mar 2016 02:34:38 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-27-hdu5650/</guid><description>&lt;p&gt;&lt;a href="http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=682&amp;amp;pid=1001" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
題意：已知一个包含 nn 个元素的正整数集合 SS，设 f(S)f(S) 为集合 SS 中所有元素的异或(XOR)的结果。
如：S={1,2,3}, 则 f(S) = 0f(S)=0。&lt;/p&gt;</description></item><item><title>uva 152 Tree's a Crowd</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-01-21-uva152/</link><pubDate>Thu, 21 Jan 2016 06:33:22 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-01-21-uva152/</guid><description>&lt;p&gt;题意：题意：给你一组三维空间中的点，每个点到其它点都有个距离，其中有个最小距离，如果这个最小距离小于10，就将对应的距离的点个数加１，最后输出距离为0,1,2&amp;hellip;8,9的点的个数。（from 百度）
老实说，上面这题意也讲的不明不白，其实这题非常水，就是对每个点进行判断，找出和其他点最短的距离，在下标为该距离的数组上+1,最后输出数组下标0-9的数。
trick：其实最小距离大于9的就不用存放了，只要开个大小10的数组。（不会概括。。。抄的别人的）&lt;/p&gt;</description></item><item><title>codeforces #322 div 2 A. Vasya the Hipster(纱布题)</title><link>https://111qqz.com/en/post/acm-icpc/2015/2015-09-28-codeforces322div2a-vasyathehipstere99a8fe4bebfe6909e/</link><pubDate>Mon, 28 Sep 2015 17:05:00 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2015/2015-09-28-codeforces322div2a-vasyathehipstere99a8fe4bebfe6909e/</guid><description>&lt;p&gt;A. Vasya the Hipster&lt;/p&gt;
&lt;p&gt;time limit per test&lt;/p&gt;
&lt;p&gt;1 second&lt;/p&gt;
&lt;p&gt;memory limit per test&lt;/p&gt;
&lt;p&gt;256 megabytes&lt;/p&gt;
&lt;p&gt;input&lt;/p&gt;
&lt;p&gt;standard input&lt;/p&gt;
&lt;p&gt;output&lt;/p&gt;
&lt;p&gt;standard output&lt;/p&gt;
&lt;p&gt;One day Vasya the Hipster decided to count how many socks he had. It turned out that he had &lt;em&gt;a&lt;/em&gt; red socks and &lt;em&gt;b&lt;/em&gt; blue socks.&lt;/p&gt;</description></item><item><title>uva 489</title><link>https://111qqz.com/en/post/acm-icpc/2015/2015-09-28-uva489/</link><pubDate>Mon, 28 Sep 2015 10:25:00 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2015/2015-09-28-uva489/</guid><description>&lt;p&gt;In ``Hangman Judge,&amp;rsquo;&amp;rsquo; you are to write a program that judges a series of Hangman games. For each game, the answer to the puzzle is given as well as the guesses. Rules are the same as the classic game of hangman, and are given as follows:&lt;/p&gt;</description></item></channel></rss>