<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Kaipeng Liu's Column]]></title><description><![CDATA[BEYOND REDEMPTION]]></description><link>https://blog.youkuaiyun.com/liukaipeng</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; liukaipeng]]></copyright><item><title><![CDATA[iPhone4离线地图制作]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/6068212</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/6068212</guid><author>liukaipeng</author><pubDate>Fri, 10 Dec 2010 17:07:00 +0800</pubDate><description><![CDATA[iPhone4离线地图制作]]></description><category></category></item><item><title><![CDATA[Conference List for Information Retrieval and Data Mining]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/5077808</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/5077808</guid><author>liukaipeng</author><pubDate>Fri, 25 Dec 2009 20:51:00 +0800</pubDate><description><![CDATA[SIGIR-20102010.1.15 : Abstracts for full research papers due2010.1.22 : Full research paper submissions due2010.2.12 : Posters, demonstration, and tutorial proposals du]]></description><category></category></item><item><title><![CDATA[Rubik's Cube Beijing Spring Open 2009]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3931522</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3931522</guid><author>liukaipeng</author><pubDate>Tue, 24 Feb 2009 11:34:00 +0800</pubDate><description><![CDATA[ Beijing Spring Open 2009 was held on Feb. 21st, 2009. I played in the game and passed the first round with a rank of 16th. Unfortunately, I ranked 24th in the second round and was not qualified to ]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 861 - Little Bishops]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3901412</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3901412</guid><author>liukaipeng</author><pubDate>Tue, 17 Feb 2009 17:02:00 +0800</pubDate><description><![CDATA[ Imagine that a chessboard with black and white squares. If a bishop is on a white square, can it attack bishops on black squares? No.So you can divide the chessboard into two independent ones. Th]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 10089 - Repackaging]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3887843</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3887843</guid><author>liukaipeng</author><pubDate>Fri, 13 Feb 2009 17:32:00 +0800</pubDate><description><![CDATA[ First, convert each package p = (s1, s2, s3) to a vector (s2 - s1, s3 - s1). All the packages forms convex hull. If point (0, 0) is in this convex hull, then there will be a solution, otherwise not]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 10090 - Marbles]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3837938</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3837938</guid><author>liukaipeng</author><pubDate>Mon, 19 Jan 2009 18:56:00 +0800</pubDate><description><![CDATA[ This problem requires us to find two non-negative numbers m1 and m2 that satisfying  n1*m1 + n2*m2 = nand minimize the cost  c = c1*m1 + c2*m2. First, we can use extended Euclids algorit]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 10042 - Smith Numbers]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3735867</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3735867</guid><author>liukaipeng</author><pubDate>Thu, 08 Jan 2009 18:09:00 +0800</pubDate><description><![CDATA[Follow the problems statements and youll get it.Code:/*************************************************************************** *   Copyright (C) 2008 by Liu Kaipeng                               ]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 10168 - Summation of Four Primes]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3735300</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3735300</guid><author>liukaipeng</author><pubDate>Thu, 08 Jan 2009 16:45:00 +0800</pubDate><description><![CDATA[According to the Goldbachs Conjecture, every even number that is not less than 4 is the sum of two primes. So we can just reduce the number to an even number by subtracting 4=2+2 or 5=2+3 from it, an]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 10139 - Factovisors]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3734813</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3734813</guid><author>liukaipeng</author><pubDate>Thu, 08 Jan 2009 15:35:00 +0800</pubDate><description><![CDATA[We check each of the prime factors of m to see if n! has enough prime factors of the same value. If so, m can divide n!, otherwise m can not divide n!.Code:/*******************************************]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 10104 - Euclid Problem]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3705001</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3705001</guid><author>liukaipeng</author><pubDate>Sun, 04 Jan 2009 16:18:00 +0800</pubDate><description><![CDATA[For the detail explanation of the algorithm, refer to Programming Challenges.Code:/*************************************************************************** *   Copyright (C) 2008 by Liu Kaipeng    ]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 10006 - Carmichael Numbers]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3588617</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3588617</guid><author>liukaipeng</author><pubDate>Tue, 23 Dec 2008 12:06:00 +0800</pubDate><description><![CDATA[Compute the remainder with the following equation:    (x*y)%n = (x%n)*(y%n)%nCode:/*************************************************************************** *   Copyright (C) 2008 by Liu Kaipeng    ]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 10110 - Light, More Light]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3549304</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3549304</guid><author>liukaipeng</author><pubDate>Thu, 18 Dec 2008 13:40:00 +0800</pubDate><description><![CDATA[ The final state of the last bulb is determined by the number of factors of n. Only the number which is the square of a number has an odd number of factors, thus it has a state of "yes".Code:/********]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 846 - Steps]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3447226</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3447226</guid><author>liukaipeng</author><pubDate>Thu, 04 Dec 2008 21:55:00 +0800</pubDate><description><![CDATA[Let f(n) denote the furthest distance we can go with n steps, then we can derive this formula easily:   f(n) = ((n+1)/2)^2 if n is odd, (1, 2, 3, 4, 3, ,2, 1)    f(n) = (n/2)^2 + n/2 if n is even. (1,]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 10049 - Self-describing Sequence]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3446558</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3446558</guid><author>liukaipeng</author><pubDate>Thu, 04 Dec 2008 20:01:00 +0800</pubDate><description><![CDATA[Let g(n) denote the nth number that satifies f(k) > f(k-1). We can know that     g(0) = 1, g(1) = 2, g(2) = 4, ... From the definition of g(n), we can know that all the values of f(k) where k belongs]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 10254 - The Priest Mathematician]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3444016</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3444016</guid><author>liukaipeng</author><pubDate>Thu, 04 Dec 2008 13:18:00 +0800</pubDate><description><![CDATA[From the description of this problem we can easily draw the recurrence of   f(n) = 2*f(n-k) + 2^k - 1and the base cases of   f(0) = 0 and f(1) = 1However, how can we determine which k would give the m]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 10247 - Complete Tree Labeling]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3441413</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3441413</guid><author>liukaipeng</author><pubDate>Thu, 04 Dec 2008 00:25:00 +0800</pubDate><description><![CDATA[Let f(k, d) denote the number of labelings of the k-ary tree of depth d. Moreover, let n(k, d) denote the number of nodes of the tree. The tree contains k subtrees, each contains n(k, d-1) nodes. Ther]]></description><category></category></item><item><title><![CDATA[UVa Online Jugde Automatic Posting Script]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3439268</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3439268</guid><author>liukaipeng</author><pubDate>Wed, 03 Dec 2008 18:03:00 +0800</pubDate><description><![CDATA[I develop a script to post my code to the UVa online judgeautomatically. This tool will also replace the local #includes withits real content when uploading the code to the server. So, I havent ]]></description><category></category></item><item><title><![CDATA[I Love Big Numbers!]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3433002</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3433002</guid><author>liukaipeng</author><pubDate>Wed, 03 Dec 2008 01:26:00 +0800</pubDate><description><![CDATA[My big integer class is finally finished. I have not tested it extensively yet. Anybody who is willing to find a tiny and clearly implemented big number class would find it worthy to have a try. Bug r]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 10157 - Expressions]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3432957</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3432957</guid><author>liukaipeng</author><pubDate>Wed, 03 Dec 2008 01:20:00 +0800</pubDate><description><![CDATA[Let n denotes the length of the expression, d denotes the depth at most, and the count is f(n,d). Obviously, f(n,d) = 0 when n is odd. For other conditions, let us consider where we can find the corre]]></description><category></category></item><item><title><![CDATA[UVa Problem Solution: 10198 - Counting]]></title><link>https://blog.youkuaiyun.com/liukaipeng/article/details/3392288</link><guid>https://blog.youkuaiyun.com/liukaipeng/article/details/3392288</guid><author>liukaipeng</author><pubDate>Thu, 27 Nov 2008 16:36:00 +0800</pubDate><description><![CDATA[ Let f(n) denote the number of ways to make the sum of n. When n > 3, the number of digits is greater than 1. The first digit d used must be one of 1, 2, 3 or 4. Remove the first digit, then the remai]]></description><category></category></item></channel></rss>