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Blue<b>To</b>oth

  • * 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(shù) * a----矩陣A * m----矩陣B的列數(shù) * b----矩

    * 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(shù) * a----矩陣A * m----矩陣B的列數(shù) * 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

    標(biāo)簽: 矩陣 AX 高斯 元素

    上傳時(shí)間: 2015-07-26

    上傳用戶:xauthu

  • Implemented BFS, DFS and A* To compile this project, use the following command: g++ -o search ma

    Implemented BFS, DFS and A* To compile this project, use the following command: g++ -o search main.cpp Then you can run it: ./search The input is loaded from a input file in.txt Here is the format of the input file: The first line of the input file shoud contain two chars indicate the source and destination city for breadth first and depth first algorithm. The second line of input file shoud be an integer m indicate the number of connections for the map. Following m lines describe the map, each line represents to one connection in this form: dist city1 city2, which means there is a connection between city1 and city2 with the distance dist. The following input are for A* The following line contains two chars indicate the source and destination city for A* algorithm. Then there is an integer h indicate the number of heuristic. The following h lines is in the form: city dist which means the straight-line distance from the city to B is dist.

    標(biāo)簽: Implemented following compile command

    上傳時(shí)間: 2014-01-01

    上傳用戶:lhc9102

  • We have a group of N items (represented by integers from 1 to N), and we know that there is some tot

    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(biāo)簽: represented integers group items

    上傳時(shí)間: 2016-01-17

    上傳用戶:jeffery

  • (1) 、用下述兩條具體規(guī)則和規(guī)則形式實(shí)現(xiàn).設(shè)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

    (1) 、用下述兩條具體規(guī)則和規(guī)則形式實(shí)現(xiàn).設(shè)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(gè)字母對(duì)應(yīng)下列的語言.

    標(biāo)簽: 字母 tAdA 語言 詞匯

    上傳時(shí)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(dòng)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(jīng)過研究發(fā)現(xiàn)

    1.有三根桿子A,B,C。A桿上有若干碟子 2.每次移動(dòng)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(jīng)過研究發(fā)現(xiàn),漢諾塔的破解很簡(jiǎn)單,就是按照移動(dòng)規(guī)則向一個(gè)方向移動(dòng)金片: 如3階漢諾塔的移動(dòng):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(shè)計(jì)中的經(jīng)典遞歸問題

    標(biāo)簽: 移動(dòng) 發(fā)現(xiàn)

    上傳時(shí)間: 2016-07-25

    上傳用戶:gxrui1991

  • 1. 下列說法正確的是 ( ) A. Java語言不區(qū)分大小寫 B. Java程序以類為基本單位 C. JVM為Java虛擬機(jī)JVM的英文縮寫 D. 運(yùn)行Java程序需要先安裝JDK

    1. 下列說法正確的是 ( ) A. Java語言不區(qū)分大小寫 B. Java程序以類為基本單位 C. JVM為Java虛擬機(jī)JVM的英文縮寫 D. 運(yùn)行Java程序需要先安裝JDK 2. 下列說法中錯(cuò)誤的是 ( ) A. Java語言是編譯執(zhí)行的 B. Java中使用了多進(jìn)程技術(shù) 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(diǎn)的一項(xiàng)是( ) A. 安全性 B. 分布式 C. 移植性 D. 編譯執(zhí)行 4. 下列語句中,正確的項(xiàng)是 ( ) A . int $e,a,b=10 B. char c,d=’a’ C. float e=0.0d D. double c=0.0f

    標(biāo)簽: Java A. B. C.

    上傳時(shí)間: 2017-01-04

    上傳用戶:netwolf

  • PCA9549 Octal bus switch with

    The PCA9549 provides eight bits of high speed TTL-compatible bus switching controlledby the I2C-bus. The low ON-state resistance of the switch allows connections to be madewith minimal propagation delay. Any individual A to B channel or combination of channelscan be selected via the I2C-bus, determined by the contents of the programmable Controlregister. When the I2C-bus bit is HIGH (logic 1), the switch is on and data can flow fromPort A to Port B, or vice versa. When the I2C-bus bit is LOW (logic 0), the switch is open,creating a high-impedance state between the two ports, which stops the data flow.An active LOW reset input (RESET) allows the PCA9549 to recover from a situationwhere the I2C-bus is stuck in a LOW state. Pulling the RESET pin LOW resets the I2C-busstate machine and causes all the bits to be open, as does the internal power-on resetfunction.

    標(biāo)簽: switch Octal 9549 with

    上傳時(shí)間: 2014-11-22

    上傳用戶:xcy122677

  • the calculator s usage! after you have inputed 2 operators,choose + - * / function! But the only

    the calculator s usage! after you have inputed 2 operators,choose + - * / function! But the only situation I did t deal with is that when you choos + fuction ,and the operaters signs is like this -A+B,just turn it to B-A!

    標(biāo)簽: calculator the operators function

    上傳時(shí)間: 2016-02-12

    上傳用戶:lili123

  • The XML Toolbox converts MATLAB data types (such as double, char, struct, complex, sparse, logical)

    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(biāo)簽: converts Toolbox complex logical

    上傳時(shí)間: 2016-02-12

    上傳用戶:a673761058

  • Implement the following integer methods: a) Method celsius returns the Celsius equivalent of a Fahr

    Implement the following integer methods: a) Method celsius returns the Celsius equivalent of a Fahrenheit calculation celsius = 5.0 / 9.0 * ( fahrenheit - 32 ) b) Method fahrenheit returns the Fahrenheit equivalent of a Celsius the calculation fahrenheit = 9.0 / 5.0 * celsius + 32 c) Use the methods from parts (a) and (b) to write an application either to enter a Fahrenheit temperature and display the Celsius or to enter a Celsius temperature and display the Fahrenheit equivalent.

    標(biāo)簽: equivalent Implement the following

    上傳時(shí)間: 2014-01-19

    上傳用戶:jackgao

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