acm HDOJ 1051WoodenSticks Description: There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows: (a) The setup time for the first wooden stick is 1 minute. (b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l and weight w if l<=l and w<=w . Otherwise, it will need 1 minute for setup.
标签： WoodenSticks Description length wooden
上传时间： 2014-03-08
上传用户：netwolf
DirectX not only provides fast access to the hardware and therefore incredibly speedy performance, but it also makes it much easier for hardware developers to produce new devices that work well in the Windows environment. The DirectX APIs take away the necessity of writing your own low-level, device-specific code to access hardware such as the display adapter and network card, making it much easier for you to write programs that take full advantage of the computer s multimedia capabilities.
标签： performance incredibly therefore hardware
上传时间： 2016-02-16
上传用户：秦莞尔w
Generate 100 samples of a zero-mean white noise sequence with variance , by using a uniform random number generator. a Compute the autocorrelation of for . b Compute the periodogram estimate and plot it. c Generate 10 different realizations of , and compute the corresponding sample autocorrelation sequences , and . Compute the average autocorrelation sequence as and the corresponding periodogram for . d Compute and plot the average periodogram using the Bartlett method. e Comment on the results in parts (a) through (d).
标签： zero-mean Generate sequence variance
上传时间： 2016-03-04
上传用户：朗朗乾坤
Atheros无线芯片AR-6000系列wince 6驱动源代码(这东西我也没用过别问我,我是搜别的wince资源搜到的) AR6K SDIO support. Requires firmware 1.1 on SD13 cards. readme: Atheros Communications AR6001 WLAN Driver for SDIO installation Read Me March 26,2007 (based on k14 fw1.1) Windows CE Embedded CE 6.0 driver installation. 1. Unzip the installation file onto your system (called installation directory below) 2. Create an OS design or open an existing OS design in Platform Builder 6.0. a. The OS must support the SD bus driver and have an SD Host Controller driver (add these from Catalog Items). b. Run image size should be set to allow greater than 32MB. 3. a. From the Project menu select Add Existing Subproject... b. select AR6K_DRV.pbxml c. select open This should create a subproject within your OS Design project for the AR6K_DRV driver. 4. Build the solution. 转自Tony嵌入式,原文地址：http://www.cevx.com/bbs/dispbbs.asp?boardID=4&ID=11762&page=1
标签： wince Requires firmware Atheros
上传时间： 2014-11-11
上传用户：bibirnovis
Swfdec still is development software, but has also followed a rigid no-crashes-allowed policy. I believe it s stable enough now to be installed as a default plugin for people that can live with occasional crashes of their browser. But don t blame me if it does crash. File a bug at https://bugs.freedesktop.org/enter_bug.cgi?product=swfdec
标签： no-crashes-allowed development followed software
上传时间： 2016-04-14
上传用户：franktu
1) Write a function reverse(A) which takes a matrix A of arbitrary dimensions as input and returns a matrix B consisting of the columns of A in reverse order. Thus for example, if A = 1 2 3 then B = 3 2 1 4 5 6 6 5 4 7 8 9 9 8 7 Write a main program to call reverse(A) for the matrix A = magic(5). Print to the screen both A and reverse(A). 2) Write a program which accepts an input k from the keyboard, and which prints out the smallest fibonacci number that is at least as large as k. The program should also print out its position in the fibonacci sequence. Here is a sample of input and output: Enter k>0: 100 144 is the smallest fibonacci number greater than or equal to 100. It is the 12th fibonacci number.
标签： dimensions arbitrary function reverse
上传时间： 2016-04-16
上传用户：waitingfy
CRC16算法的Java实现，使用方法如下： CRC16 crc16 = new CRC16() byte[] b = new byte[] { // (byte) 0xF0,(byte)0xF0,(byte)0xF0,(byte)0x72 (byte) 0x2C, (byte) 0x00, (byte) 0xFF, (byte) 0xFE, (byte) 0xFE, (byte) 0x04, (byte) 0x00, (byte) 0x00, (byte) 0x00, (byte) 0x00 } for (int k = 0 k < b.length k++) { crc16.update(b[k]) } System.out.println(Integer.toHexString(crc16.getValue())) System.out.println(Integer.toHexString(b.length))
上传时间： 2014-12-20
上传用户：ve3344
The task of clustering Web sessions is to group Web sessions based on similarity and consists of maximizing the intra- group similarity while minimizing the inter-group similarity． The first and foremost question needed to be considered in clustering W b sessions is how to measure the similarity between Web sessions．However．there are many shortcomings in traditiona1 measurements．This paper introduces a new method for measuring similarities between Web pages that takes into account not only the URL but also the viewing time of the visited web page．Yhen we give a new method to measure the similarity of Web sessions using sequence alignment and the similarity of W eb page access in detail Experiments have proved that our method is valid and e币cient．
标签： sessions clustering similarity Web
上传时间： 2014-01-11
上传用户：songrui
《Java手機程式設計入門》／王森 書號：29014 頁數：約 492 頁 ISBN：957-200-527-8 出版日期：2001年08月25日 出版廠商：知城數位科技股份有限公司 訂價：380 第一章 Java 2 Micro Edition概論阵 第二章 Java程式設計簡介阵 第三章 撰寫您的第一個手機程式阵 第四章 在實體機器上執行MIDlet阵 第五章 J2ME Wireless Toolkit阵 第六章 Motorola A6288手機程式開發阵 第七章 JBuilder MobileSet阵 第八章 MIDP for Palm 第九章 MIDlet的事件處理阵 第十章 MIDP圖形使用者介面程式設計阵 第十一章 MIDP圖形處理阵 第十二章 MIDP資料庫程式設計阵 第十三章 MIDP網路程式設計阵 附錄A MID其他參考資源總整理阵 附錄B Motorola J2ME SDK
上传时间： 2016-12-01
上传用户：coeus
Instead of finding the longest common subsequence, let us try to determine the length of the LCS. Then tracking back to find the LCS. Consider a1a2…am and b1b2…bn. Case 1: am=bn. The LCS must contain am, we have to find the LCS of a1a2…am-1 and b1b2…bn-1. Case 2: am≠bn. Wehave to find the LCS of a1a2…am-1 and b1b2…bn, and a1a2…am and b b b b1b2…bn-1 Let A = a1 a2 … am and B = b1 b2 … bn Let Li j denote the length of the longest i,g g common subsequence of a1 a2 … ai and b1 b2 … bj. Li,j = Li-1,j-1 + 1 if ai=bj max{ L L } a≠b i-1,j, i,j-1 if ai≠j L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
标签： the subsequence determine Instead
上传时间： 2013-12-17
上传用户：evil
