數據結構課程設計 數據結構B+樹 B+ tree Library
上傳時間： 2013-12-31
上傳用戶：semi1981
* 高斯列主元素消去法求解矩陣方程AX=B,其中A是N*N的矩陣,B是N*M矩陣 * 輸入: n----方陣A的行數 * a----矩陣A * m----矩陣B的列數 * b----矩陣B * 輸出: det----矩陣A的行列式值 * a----A消元后的上三角矩陣 * b----矩陣方程的解X
上傳時間： 2015-07-26
上傳用戶：xauthu
The book "The Finite Difference Time Domain for Electromagnetics" by Karl Kunz and Raymond Luebbers, CRC Press, 1993, contains an FDTD code and output files in Appendix B. The same code and output files are contained in this directory.
標簽： Electromagnetics Difference The Luebbers
上傳時間： 2013-12-29
上傳用戶：chongcongying
We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.
標簽： represented integers group items
上傳時間： 2016-01-17
上傳用戶：jeffery
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.
標簽： performance incredibly therefore hardware
上傳時間： 2016-02-16
上傳用戶：秦莞爾w
(1) 、用下述兩條具體規則和規則形式實現.設大寫字母表示魔王語言的詞匯 小寫字母表示人的語言詞匯 希臘字母表示可以用大寫字母或小寫字母代換的變量.魔王語言可含人的詞匯. (2) 、B→tAdA A→sae (3) 、將魔王語言B(ehnxgz)B解釋成人的語言.每個字母對應下列的語言.
上傳時間： 2013-12-30
上傳用戶：ayfeixiao
1.有三根桿子A,B,C。A桿上有若干碟子 2.每次移動一塊碟子,小的只能疊在大的上面 3.把所有碟子從A桿全部移到C桿上 經過研究發現，漢諾塔的破解很簡單，就是按照移動規則向一個方向移動金片： 如3階漢諾塔的移動：A→C,A→B,C→B,A→C,B→A,B→C,A→C 此外，漢諾塔問題也是程序設計中的經典遞歸問題
上傳時間： 2016-07-25
上傳用戶：gxrui1991
1. 下列說法正確的是 （ ） A. Java語言不區分大小寫 B. Java程序以類為基本單位 C. JVM為Java虛擬機JVM的英文縮寫 D. 運行Java程序需要先安裝JDK 2. 下列說法中錯誤的是 ( ) A. Java語言是編譯執行的 B. Java中使用了多進程技術 C. Java的單行注視以//開頭 D. Java語言具有很高的安全性 3. 下面不屬于Java語言特點的一項是( ) A. 安全性 B. 分布式 C. 移植性 D. 編譯執行 4. 下列語句中，正確的項是 ( ) A . int $e,a,b=10 B. char c,d=’a’ C. float e=0.0d D. double c=0.0f
上傳時間： 2017-01-04
上傳用戶：netwolf
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.
標簽： the subsequence determine Instead
上傳時間： 2013-12-17
上傳用戶：evil
特點(FEATURES) 精確度0.1%滿刻度 (Accuracy 0.1%F.S.) 可作各式數學演算式功能如:A+B/A-B/AxB/A/B/A&B(Hi or Lo)/|A| (Math functioA+B/A-B/AxB/A/B/A&B(Hi&Lo)/|A|/etc.....) 16 BIT 類比輸出功能(16 bit DAC isolating analog output function) 輸入/輸出1/輸出2絕緣耐壓2仟伏特/1分鐘(Dielectric strength 2KVac/1min. (input/output1/output2/power)) 寬范圍交直流兩用電源設計(Wide input range for auxiliary power) 尺寸小,穩定性高(Dimension small and High stability)
上傳時間： 2013-11-24
上傳用戶：541657925
