The government of a small but important country has decided that the alphabet needs to be streamlined and reordered. Uppercase letters will be eliminated. They will issue a royal decree in the form of a String of B and A characters. The first character in the decree specifies whether a must come ( B )Before b in the new alphabet or ( A )After b . The second character determines the relative placement of b and c , etc. So, for example, "BAA" means that a must come Before b , b must come After c , and c must come After d . Any letters beyond these requirements are to be excluded, so if the decree specifies k comparisons then the new alphabet will contain the first k+1 lowercase letters of the current alphabet. Create a class Alphabet that contains the method choices that takes the decree as input and returns the number of possible new alphabets that conform to the decree. If more than 1,000,000,000 are possible, return -1. Definition
標簽： government streamline important alphabet
上傳時間： 2015-06-09
上傳用戶：weixiao99
* 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
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
(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
選擇映射法（SLM）和概率類算法都可以降低OFDM （Orthogonal Frequency Division Multiplexing）系統的PAPR（Peak to Average Power Ratio），傳統SLM算法自身較為復雜，但由于其優良的性能，棄之可惜。研究表明，SLM算法和限幅類算法在性能上具有一定的互補性。任何一個算法未必能達到抑制PAPR的理想效果，在深入研究了兩個算法的基礎上，將其優點聯合起來，以達到降低OFDM系統PAPR的目的。最后對聯合改進算法進行了分析與仿真，并驗證了聯合改進算法的有效性和可行性
上傳時間： 2013-11-22
上傳用戶：xinhaoshan2016
TIMER.ASM ********* [ milindhp@tifrvax.tifr.res.in ] Set Processor configuration word as = 0000 0000 1010 b. a] -MCLR tied to VDD (internally). b] Code protection off. c] WDT disabled. d] Internal RC oscillator [4 MHZ].
標簽： configuration Processor milindhp tifrvax
上傳時間： 2015-05-24
上傳用戶：wqxstar
The XML Toolbox converts MATLAB data types (such as double, char, struct, complex, sparse, logical) of any level of nesting to XML format and vice versa. For example, >> project.name = MyProject >> project.id = 1234 >> project.param.a = 3.1415 >> project.param.b = 42 becomes with str=xml_format(project, off ) "<project> <name>MyProject</name> <id>1234</id> <param> <a>3.1415</a> <b>42</b> </param> </project>" On the other hand, if an XML string XStr is given, this can be converted easily to a MATLAB data type or structure V with the command V=xml_parse(XStr).
標簽： converts Toolbox complex logical
上傳時間： 2016-02-12
上傳用戶：a673761058
Algorithm that find the peaks of a signal. You can choose the number of peaks you want to find and the valour of peak to begin to find.
上傳時間： 2013-12-24
上傳用戶：cc1915
