﻿<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:trackback="http://madskills.com/public/xml/rss/module/trackback/" xmlns:wfw="http://wellformedweb.org/CommentAPI/" xmlns:slash="http://purl.org/rss/1.0/modules/slash/"><channel><title>博客园-银河-随笔分类-算法</title><link>http://www.cnblogs.com/skyivben/category/141101.html</link><description>SKYIV STUDIO</description><language>zh-cn</language><lastBuildDate>Sat, 02 Aug 2008 08:32:43 GMT</lastBuildDate><pubDate>Sat, 02 Aug 2008 08:32:43 GMT</pubDate><ttl>60</ttl><item><title>Timus 1057. Amount of degrees</title><link>http://www.cnblogs.com/skyivben/archive/2008/08/01/1258443.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Fri, 01 Aug 2008 13:51:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/08/01/1258443.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1258443.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/08/01/1258443.html#Feedback</comments><slash:comments>2</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1258443.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1258443.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1057">Timus 1057. Amount of degrees</a> 要求计算指定范围内能够由 K 个不同的 B 的幂次之和组成的整数的个数。这个程序的关键在于第 41 到 54 行的 GetNth 方法，该方法返回满足以下条件的第 N 个 B 进制整数：该整数中有 K 个 1，其余数字全部都是 0。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/08/01/1258443.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1258443.html?type=1" width = "1" height = "1" />]]></description></item><item><title>使用快速傅里叶变换计算圆周率</title><link>http://www.cnblogs.com/skyivben/archive/2008/07/29/1255889.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Tue, 29 Jul 2008 13:34:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/07/29/1255889.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1255889.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/07/29/1255889.html#Feedback</comments><slash:comments>6</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1255889.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1255889.html</trackback:ping><description><![CDATA[摘要: <img src="http://tex.72pines.com/latex.php?latex=$\pi_{i %2B 1}=\pi_i\left(\frac{X_{i %2B 1} %2B 1}{Y_i %2B 1}\right)$" class="desc_img"/>前几天，我发表了一篇随笔：“<a target="_blank" href="http://www.cnblogs.com/skyivben/archive/2008/07/25/1250891.html">BigArithmetic - 提供任意精度的算术运算的静态类</a>”。现在，让我们使用 BigArithmetic 类来计算圆周率。我们需要一个计算 π 的分析算法。有用的算法是二次收敛的，即每一次迭代使有效位数增加一倍。计算 π 的二次收敛算法是基于 ACM 法(算术几何平均法)。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/07/29/1255889.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1255889.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1005. Stone pile</title><link>http://www.cnblogs.com/skyivben/archive/2008/07/26/1251907.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Sat, 26 Jul 2008 02:04:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/07/26/1251907.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1251907.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/07/26/1251907.html#Feedback</comments><slash:comments>3</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1251907.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1251907.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1005">Timus 1005. Stone pile</a> 要求将若干石头分为两堆使其重量差最小。这道题是说，给你一堆石头，总数在 1 到 20 之间，每个石头的重量在 1 到 100,000 之间。要求你将这堆石头分成两堆，使这两堆石头的重量差最小，并输出这个重量差。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/07/26/1251907.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1251907.html?type=1" width = "1" height = "1" />]]></description></item><item><title>再谈 BigInteger - 使用快速傅里叶变换</title><link>http://www.cnblogs.com/skyivben/archive/2008/07/25/1251697.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Fri, 25 Jul 2008 14:07:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/07/25/1251697.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1251697.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/07/25/1251697.html#Feedback</comments><slash:comments>13</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1251697.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1251697.html</trackback:ping><description><![CDATA[摘要: 我在“<a target="_blank" href="http://www.cnblogs.com/skyivben/archive/2008/07/13/1241681.html">浅谈 BigInteger</a>”的随笔中实现了一个 Skyiv.Numeric.BigInteger 类，那时乘法是使用常规的 O(N<sup>2</sup>) 的算法，所以比 .NET Framework 3.5 Base Class Library 中的 System.Numeric.BigInteger 类稍慢，后者的乘法是使用 Karatsuba 算法，其时间复杂度约为 O(N<sup>1.585</sup>)。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/07/25/1251697.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1251697.html?type=1" width = "1" height = "1" />]]></description></item><item><title>BigArithmetic - 提供任意精度的算术运算的静态类</title><link>http://www.cnblogs.com/skyivben/archive/2008/07/25/1250891.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Thu, 24 Jul 2008 16:06:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/07/25/1250891.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1250891.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/07/25/1250891.html#Feedback</comments><slash:comments>7</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1250891.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1250891.html</trackback:ping><description><![CDATA[摘要: 在上一篇随笔“<a target="_blank" href="http://www.cnblogs.com/skyivben/archive/2008/07/23/1248413.html">使用快速傅里叶变换计算大整数乘法</a>”中，已经讲述了使用快速傅里叶变换计算大整数乘法的原理。在这一篇随笔中，我们就使用快速傅里叶变换来实现一个提供任意精度的算术运算的静态类：BigArithmetic。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/07/25/1250891.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1250891.html?type=1" width = "1" height = "1" />]]></description></item><item><title>使用快速傅里叶变换计算大整数乘法</title><link>http://www.cnblogs.com/skyivben/archive/2008/07/23/1248413.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Wed, 23 Jul 2008 14:10:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/07/23/1248413.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1248413.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/07/23/1248413.html#Feedback</comments><slash:comments>30</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1248413.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1248413.html</trackback:ping><description><![CDATA[摘要: <p><img src="http://www.cnblogs.com/images/cnblogs_com/skyivben/FFT/axb2.jpg" class="desc_img"/>我们知道，两个 N 位数字的整数的乘法，如果使用常规的算法，时间复杂度是 O(N<sup>2</sup>)。然而，使用快速傅里叶变换，时间复杂度可以降低到 O(N logN loglogN)。 </p>&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/07/23/1248413.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1248413.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1603. Erudite</title><link>http://www.cnblogs.com/skyivben/archive/2008/07/16/1244637.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Wed, 16 Jul 2008 12:47:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/07/16/1244637.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1244637.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/07/16/1244637.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1244637.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1244637.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1603">Timus 1603. Erudite</a> 要求解一个字谜。这个题目首先给出一个 4x4 的字谜，然后给出许多单词，要求你判断这些单词是否能够由前面的字谜组成。规则是，你能够在字谜中上下左右行走，也能够拐弯，但是已经使用过的字母不允许再用。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/07/16/1244637.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1244637.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1318. Logarithm</title><link>http://www.cnblogs.com/skyivben/archive/2008/07/14/1242948.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Mon, 14 Jul 2008 14:41:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/07/14/1242948.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1242948.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/07/14/1242948.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1242948.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1242948.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1318">Timus 1318. Logarithm</a> 要求将一些 128-bit 数进行 XOR 运算后，取对数(只计算到整数部分)，然后求它们的和。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/07/14/1242948.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1242948.html?type=1" width = "1" height = "1" />]]></description></item><item><title>浅谈 BigInteger</title><link>http://www.cnblogs.com/skyivben/archive/2008/07/13/1241681.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Sun, 13 Jul 2008 13:25:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/07/13/1241681.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1241681.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/07/13/1241681.html#Feedback</comments><slash:comments>15</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1241681.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1241681.html</trackback:ping><description><![CDATA[摘要: 本文介绍了四种可用于 C# 的 BigInteger 类：.NET Framework 3.5 Base Class Library 中的 System.Numeric.BigInteger、Skyiv.Numeric.BigInteger、ChewKeongTAN.BigInteger 和 java.math.BigInteger。并对她们进行比较。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/07/13/1241681.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1241681.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1402. Cocktails</title><link>http://www.cnblogs.com/skyivben/archive/2008/07/10/1239710.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Thu, 10 Jul 2008 01:36:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/07/10/1239710.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1239710.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/07/10/1239710.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1239710.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1239710.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1402">Timus 1402. Cocktails</a> 要求计算出鸡尾酒共有多少种调法。这道题目是说，给定 N (1 ≤ N ≤ 21) 种不同的原料，要求计算出总共能够调制出多少种不同的鸡尾酒。每种原料最多只能使用一次，但是投放原料的次序不同调制出来的鸡尾酒也不同。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/07/10/1239710.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1239710.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1531. Zones on a plane</title><link>http://www.cnblogs.com/skyivben/archive/2008/07/06/1236843.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Sun, 06 Jul 2008 11:07:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/07/06/1236843.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1236843.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/07/06/1236843.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1236843.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1236843.html</trackback:ping><description><![CDATA[摘要: <p><img src="http://www.cnblogs.com/images/cnblogs_com/skyivben/Timus/T1531-half.jpg" align="right"/><a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1531">Timus 1531. Zones on a plane</a> 要求计算满足给定条件的简单多边形的个数。这道题目是说，在平原上有若干三角形组成的区域 Z<sub>i</sub>。Z<sub>1</sub> 包含两个等腰直角三角形，组成一个正方形。Z<sub>n+1</sub> 由 Z<sub>n</sub> 按以下方法生成：以 Z<sub>n</sub> 中的三角形的直角边作为新的等腰直角三角形的斜边，然后再移去 Z<sub>n</sub> 中的三角形，剩下的三角形就组成了 Z<sub>n+1</sub>。现在要求计算 Z<sub>n</sub> 中包含多少个简单多边形。</p>&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/07/06/1236843.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1236843.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1013. K-based numbers. Version 3</title><link>http://www.cnblogs.com/skyivben/archive/2008/07/04/1235062.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Fri, 04 Jul 2008 11:22:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/07/04/1235062.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1235062.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/07/04/1235062.html#Feedback</comments><slash:comments>1</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1235062.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1235062.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1013">Timus 1013. K-based numbers. Version 3</a> 要求计算出不包括相邻的零的 N 位 K-进制数共有多少个。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/07/04/1235062.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1235062.html?type=1" width = "1" height = "1" />]]></description></item><item><title>应用程序的性能: C# vs C/C++</title><link>http://www.cnblogs.com/skyivben/archive/2008/07/02/1234335.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Wed, 02 Jul 2008 15:15:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/07/02/1234335.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1234335.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/07/02/1234335.html#Feedback</comments><slash:comments>28</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1234335.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1234335.html</trackback:ping><description><![CDATA[摘要: 比较 C# 和 C/C++ 应用程序的性能&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/07/02/1234335.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1234335.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1219. Symbolic Sequence</title><link>http://www.cnblogs.com/skyivben/archive/2008/07/01/1233632.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Tue, 01 Jul 2008 15:41:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/07/01/1233632.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1233632.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/07/01/1233632.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1233632.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1233632.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1219">Timus 1219. Symbolic Sequence</a> 要求输出满足给定条件的一百万个小写拉丁字母。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/07/01/1233632.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1233632.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1081. Binary Lexicographic Sequence</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/30/1232721.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Mon, 30 Jun 2008 13:09:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/30/1232721.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1232721.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/30/1232721.html#Feedback</comments><slash:comments>5</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1232721.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1232721.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1081">Timus 1081. Binary Lexicographic Sequence</a> 要求给出第 K 个 N 位二进制数，该二进制数不得有相邻的“1”。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/30/1232721.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1232721.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1152. The False Mirrors</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/28/1231604.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Sat, 28 Jun 2008 06:00:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/28/1231604.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1231604.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/28/1231604.html#Feedback</comments><slash:comments>3</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1231604.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1231604.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1152">Timus 1152. The False Mirrors</a> 说述消灭怪物的故事，故事中的主角周围环绕着一圈 N (3 ≤ N ≤ 20) 个阳台，每个阳台上有若干个怪物。主角每开一枪能够摧毁相邻的三个阳台，然后幸存的怪物还击，假设每个怪物每次攻击对主角的伤害都是一个单位。然后开始新一轮枪战，直到消灭所有的怪物。你的任务就是计算出主角受到的最小伤害。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/28/1231604.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1231604.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1010. Discrete Function</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/26/1230752.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Thu, 26 Jun 2008 14:35:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/26/1230752.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1230752.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/26/1230752.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1230752.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1230752.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1010">Timus 1010. Discrete Function</a> 要求根据给定的离散函数在直角坐标系中的对应点的连线中找出倾角最大的。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/26/1230752.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1230752.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1150. Digits</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/25/1230049.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Wed, 25 Jun 2008 15:35:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/25/1230049.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1230049.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/25/1230049.html#Feedback</comments><slash:comments>1</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1230049.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1230049.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1150">Timus 1150. Digits</a> 要求计算 1 到 N 的正整数中包含 0 .. 9 的数目。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/25/1230049.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1230049.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1153. Supercomputer</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/24/1229260.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Tue, 24 Jun 2008 15:29:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/24/1229260.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1229260.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/24/1229260.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1229260.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1229260.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1153">Timus 1153. Supercomputer</a> 要求根据自然数列的前 N 项和求 N 值。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/24/1229260.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1229260.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1114. Boxes</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/24/1228651.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Tue, 24 Jun 2008 01:53:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/24/1228651.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1228651.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/24/1228651.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1228651.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1228651.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1114">Timus 1114. Boxes</a> 要求计算出将两种颜色的球放到盒子中的各种组合的数目。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/24/1228651.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1228651.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1108. Heritage</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/23/1228284.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Mon, 23 Jun 2008 08:34:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/23/1228284.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1228284.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/23/1228284.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1228284.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1228284.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1108">Timus 1108. Heritage</a> 要求对给定的若干继承人作出一个遗产分配方案。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/23/1228284.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1228284.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1028. Stars</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/22/1227543.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Sun, 22 Jun 2008 02:08:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/22/1227543.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1227543.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/22/1227543.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1227543.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1227543.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1028">Timus 1028. Stars</a> 要求计算直角坐标系中星星的等级。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/22/1227543.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1227543.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1015. Test the Difference!</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/21/1227370.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Sat, 21 Jun 2008 07:57:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/21/1227370.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1227370.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/21/1227370.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1227370.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1227370.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1015">Timus 1015. Test the Difference!</a> 要求将赌场中的骰子分类。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/21/1227370.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1227370.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1032. Find a multiple</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/21/1227218.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Sat, 21 Jun 2008 02:13:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/21/1227218.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1227218.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/21/1227218.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1227218.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1227218.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1032">Timus 1032. Find a multiple</a> 要求在 N 个正整数中找出其中若干个使得它们的和是 N 的倍数。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/21/1227218.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1227218.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1011. Conductors</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/20/1226461.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Fri, 20 Jun 2008 03:29:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/20/1226461.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1226461.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/20/1226461.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1226461.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1226461.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1011">Timus 1011. Conductors</a> 要求根据售票员所占比例范围计算叶卡特琳堡的最小总人口数。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/20/1226461.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1226461.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1079. Maximum</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/19/1225759.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Thu, 19 Jun 2008 03:53:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/19/1225759.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1225759.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/19/1225759.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1225759.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1225759.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1079">Timus 1079. Maximum</a> 要求输出指定数列中的最大值。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/19/1225759.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1225759.html?type=1" width = "1" height = "1" />]]></description></item><item><title>Timus 1601. AntiCAPS</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/18/1224730.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Wed, 18 Jun 2008 07:57:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/18/1224730.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1224730.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/18/1224730.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1224730.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1224730.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru/problem.aspx?space=1&num=1601">Timus 1601. AntiCAPS</a> 要求把一段大写英文字母的文本转换为适当的小写英文字母。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/18/1224730.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1224730.html?type=1" width = "1" height = "1" />]]></description></item><item><title>可以使用C#语言的在线ACM题库</title><link>http://www.cnblogs.com/skyivben/archive/2008/06/07/1215740.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Sat, 07 Jun 2008 12:43:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2008/06/07/1215740.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/1215740.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2008/06/07/1215740.html#Feedback</comments><slash:comments>32</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/1215740.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/1215740.html</trackback:ping><description><![CDATA[摘要: <a target="_blank" href="http://acm.timus.ru">俄罗斯乌拉尔大学在线题库</a> 和 <a target="_blank" href="http://www.spoj.pl">Sphere Onlile Judge (SPOJ)</a> 都是可以使用C#语言的在线ACM题库，有兴趣的朋友可以去试试。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2008/06/07/1215740.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/1215740.html?type=1" width = "1" height = "1" />]]></description></item><item><title>数独解算器(ASP.NET 2.0)</title><link>http://www.cnblogs.com/skyivben/archive/2007/01/26/631233.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Fri, 26 Jan 2007 07:09:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2007/01/26/631233.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/631233.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2007/01/26/631233.html#Feedback</comments><slash:comments>20</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/631233.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/631233.html</trackback:ping><description><![CDATA[摘要: 数独游戏 在9x9的方格内进行， 分为3x3的小方格，被称为“区”。<br>数独游戏首先从已经填入数字的格子开始。<br>数独游戏的目的是根据下列规则，用1至9之间的数字填满空格：<br>每个数字在每一行、每一列和每一区只能出现一次。<br>我在 Linux 服务器上用 ASP.NET 2.0 实现了一个数独解算器。<br>&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2007/01/26/631233.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/631233.html?type=1" width = "1" height = "1" />]]></description></item><item><title>计算圆周率的C#程序</title><link>http://www.cnblogs.com/skyivben/archive/2005/09/30/247225.html</link><dc:creator>银河</dc:creator><author>银河</author><pubDate>Fri, 30 Sep 2005 03:22:00 GMT</pubDate><guid>http://www.cnblogs.com/skyivben/archive/2005/09/30/247225.html</guid><wfw:comment>http://www.cnblogs.com/skyivben/comments/247225.html</wfw:comment><comments>http://www.cnblogs.com/skyivben/archive/2005/09/30/247225.html#Feedback</comments><slash:comments>15</slash:comments><wfw:commentRss>http://www.cnblogs.com/skyivben/comments/commentRss/247225.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/skyivben/services/trackbacks/247225.html</trackback:ping><description><![CDATA[摘要: 　关于圆周率，可参阅“<A href="http://zh.wikipedia.org/w/index.php?title=%E5%9C%93%E5%91%A8%E7%8E%87&amp;variant=zh-cn">维基百科-圆周率</A>”。我前天在博客园也发表了一篇随笔：“<A href="http://skyivben.cnblogs.com/archive/2005/09/28/245960.html">计算圆周率的C程序</A>” 。这个C#版的计算圆周率程序就是在C程序的基础上改写的。C#版的程序必须使用C#2.0编译，算法和C程序是一样的，都是利用圆周率的反正切展式的泰勒级数来计算，但C#程序充分使用面象对象的编程方法，并且程序中有适当的注释，比C程序容易理解多了。C#程序从配置文件中读取计算所用的公式，允许自己增加计算公式。&nbsp;&nbsp;<a href='http://www.cnblogs.com/skyivben/archive/2005/09/30/247225.html'>阅读全文</a><img src ="http://www.cnblogs.com/skyivben/aggbug/247225.html?type=1" width = "1" height = "1" />]]></description></item></channel></rss>