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  • 源代码用动态规划算法计算序列关系个数 用关系"<"和"="将3个数a

    源代码\用动态规划算法计算序列关系个数 用关系"<"和"="将3个数a,b,c依次序排列时,有13种不同的序列关系: a=b=c,a=b<c,a<b=v,a<b<c,a<c<b a=c<b,b<a=c,b<a<c,b<c<a,b=c<a c<a=b,c<a<b,c<b<a 若要将n个数依序列,设计一个动态规划算法,计算出有多少种不同的序列关系, 要求算法只占用O(n),只耗时O(n*n).

    标签: lt 源代码 动态规划 序列

    上传时间: 2013-12-26

    上传用户:siguazgb

  • The government of a small but important country has decided that the alphabet needs to be streamline

    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

    标签: government streamline important alphabet

    上传时间: 2015-06-09

    上传用户:weixiao99

  • CMAC网络最初主要用来求解机械手的关节运动。W.T.Miller等人把CMAC网络成功的运用到机器人的控制上

    CMAC网络最初主要用来求解机械手的关节运动。W.T.Miller等人把CMAC网络成功的运用到机器人的控制上,S.Cetinkunt等又将其运用到高精度机械工具的伺服控制。

    标签: CMAC Miller 网络 机械手

    上传时间: 2015-07-04

    上传用户:tianjinfan

  • 這是台灣鳥哥linux私房菜的電子書檔

    這是台灣鳥哥linux私房菜的電子書檔,內容是一些linux的基礎教學與架設伺服器的設定,希望大家會喜歡

    标签: linux

    上传时间: 2014-01-18

    上传用户:qq21508895

  • 在ATMEGA128单片机上开发的一个机器人控制程序

    在ATMEGA128单片机上开发的一个机器人控制程序,同时控制12个伺服马达的运动

    标签: ATMEGA 128 单片机 机器人控制

    上传时间: 2014-01-07

    上传用户:lanwei

  • 上下文无关文法(Context-Free Grammar, CFG)是一个4元组G=(V, T, S, P)

    上下文无关文法(Context-Free Grammar, CFG)是一个4元组G=(V, T, S, P),其中,V和T是不相交的有限集,S∈V,P是一组有限的产生式规则集,形如A→α,其中A∈V,且α∈(V∪T)*。V的元素称为非终结符,T的元素称为终结符,S是一个特殊的非终结符,称为文法开始符。 设G=(V, T, S, P)是一个CFG,则G产生的语言是所有可由G产生的字符串组成的集合,即L(G)={x∈T* | Sx}。一个语言L是上下文无关语言(Context-Free Language, CFL),当且仅当存在一个CFG G,使得L=L(G)。 *⇒ 例如,设文法G:S→AB A→aA|a B→bB|b 则L(G)={a^nb^m | n,m>=1} 其中非终结符都是大写字母,开始符都是S,终结符都是小写字母。

    标签: Context-Free Grammar CFG

    上传时间: 2013-12-10

    上传用户:gaojiao1999

  • dsp电机经典控制

    dsp电机经典控制,包括伺服电机的控制算法,特别事伺服电机控制

    标签: dsp 电机 控制

    上传时间: 2013-12-22

    上传用户:ma1301115706

  • GT400运动控制卡控制程序

    GT400运动控制卡控制程序,用于控制伺服电机

    标签: 400 GT 运动控制卡 控制

    上传时间: 2015-12-09

    上传用户:liansi

  • 8051入間程序

    8051入間程序,包括串并轉換、數碼管驅動、ADC、BCD、16x2 LCD、串口通訊、伺服器驅動等。由本人編寫。

    标签: 8051 程序

    上传时间: 2013-12-17

    上传用户:ommshaggar

  • 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.

    标签: represented integers group items

    上传时间: 2016-01-17

    上传用户:jeffery