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  • Grammatica is a C# and Java parser generator (compiler compiler). It improves upon simlar tools (lik

    Grammatica is a C# and Java parser generator (compiler compiler). It improves upon simlar tools (like yacc and ANTLR) by creating well-commented and readable source code, by having automatic error recovery and detailed error messages, and by support for testing and debugging grammars without generating source code. It supports LL(k) grammars with an unlimited number of look-ahead tokens.

    标签: compiler Grammatica generator improves

    上传时间: 2015-01-11

    上传用户:stella2015

  • 电力系统在台稳定计算式电力系统不正常运行方式的一种计算。它的任务是已知电力系统某一正常运行状态和受到某种扰动

    电力系统在台稳定计算式电力系统不正常运行方式的一种计算。它的任务是已知电力系统某一正常运行状态和受到某种扰动,计算电力系统所有发电机能否同步运行 1运行说明: 请输入初始功率S0,形如a+bi 请输入无限大系统母线电压V0 请输入系统等值电抗矩阵B 矩阵B有以下元素组成的行矩阵 1正常运行时的系统直轴等值电抗Xd 2故障运行时的系统直轴等值电抗X d 3故障切除后的系统直轴等值电抗 请输入惯性时间常数Tj 请输入时段数N 请输入哪个时段发生故障Ni 请输入每时段间隔的时间dt

    标签: 电力系统 计算 运行

    上传时间: 2015-06-13

    上传用户:it男一枚

  • The concept of sustainable development has received growing recognition, but it is a new idea for m

    The concept of sustainable development has received growing recognition, but it is a new idea for many business executives. For most, the concept remains abstract and theoretical. Protecting an organization’s capital base is a well-accepted business principle. Yet organizations do not generally recognize

    标签: development sustainable recognition received

    上传时间: 2014-09-04

    上传用户:498732662

  • The concept of sustainable development has received growing recognition, but it is a new idea for m

    The concept of sustainable development has received growing recognition, but it is a new idea for many business executives. For most, the concept remains abstract and theoretical. Protecting an organization’s capital base is a well-accepted business principle. Yet organizations do not generally recognize

    标签: development sustainable recognition received

    上传时间: 2013-12-12

    上传用户:362279997

  • 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

  • μC/OS-II Goals Probably the most important goal of μC/OS-II was to make it backward compatible with

    μC/OS-II Goals Probably the most important goal of μC/OS-II was to make it backward compatible with μC/OS (at least from an application’s standpoint). A μC/OS port might need to be modified to work with μC/OS-II but at least, the application code should require only minor changes (if any). Also, because μC/OS-II is based on the same core as μC/OS, it is just as reliable. I added conditional compilation to allow you to further reduce the amount of RAM (i.e. data space) needed by μC/OS-II. This is especially useful when you have resource limited products. I also added the feature described in the previous section and cleaned up the code. Where the book is concerned, I wanted to clarify some of the concepts described in the first edition and provide additional explanations about how μC/OS-II works. I had numerous requests about doing a chapter on how to port μC/OS and thus, such a chapter has been included in this book for μC/OS-II.

    标签: OS-II compatible important Probably

    上传时间: 2013-12-02

    上传用户:jkhjkh1982

  • Digital Signature Algorithm (DSA)是Schnorr和ElGamal签名算法的变种

    Digital Signature Algorithm (DSA)是Schnorr和ElGamal签名算法的变种,被美国NIST作为DSS(DigitalSignature Standard)。算法中应用了下述参数: p:L bits长的素数。L是64的倍数,范围是512到1024; q:p - 1的160bits的素因子; g:g = h^((p-1)/q) mod p,h满足h < p - 1, h^((p-1)/q) mod p > 1; x:x < q,x为私钥 ; y:y = g^x mod p ,( p, q, g, y )为公钥; H( x ):One-Way Hash函数。DSS中选用SHA( Secure Hash Algorithm )。 p, q, g可由一组用户共享,但在实际应用中,使用公共模数可能会带来一定的威胁。签名及验证协议如下: 1. P产生随机数k,k < q; 2. P计算 r = ( g^k mod p ) mod q s = ( k^(-1) (H(m) + xr)) mod q 签名结果是( m, r, s )。 3. 验证时计算 w = s^(-1)mod q u1 = ( H( m ) * w ) mod q u2 = ( r * w ) mod q v = (( g^u1 * y^u2 ) mod p ) mod q 若v = r,则认为签名有效。   DSA是基于整数有限域离散对数难题的,其安全性与RSA相比差不多。DSA的一个重要特点是两个素数公开,这样,当使用别人的p和q时,即使不知道私钥,你也能确认它们是否是随机产生的,还是作了手脚。RSA算法却作不到。

    标签: Algorithm Signature Digital Schnorr

    上传时间: 2014-01-01

    上传用户:qq521

  • 汉诺塔!!! Simulate the movement of the Towers of Hanoi puzzle Bonus is possible for using animation

    汉诺塔!!! Simulate the movement of the Towers of Hanoi puzzle Bonus is possible for using animation eg. if n = 2 A→B A→C B→C if n = 3 A→C A→B C→B A→C B→A B→C A→C

    标签: the animation Simulate movement

    上传时间: 2017-02-11

    上传用户:waizhang

  • 本代码为编码开关代码

    本代码为编码开关代码,编码开关也就是数字音响中的 360度旋转的数字音量以及显示器上用的(单键飞梭开 关)等类似鼠标滚轮的手动计数输入设备。 我使用的编码开关为5个引脚的,其中2个引脚为按下 转轮开关(也就相当于鼠标中键)。另外3个引脚用来 检测旋转方向以及旋转步数的检测端。引脚分别为a,b,c b接地a,c分别接到P2.0和P2.1口并分别接两个10K上拉 电阻,并且a,c需要分别对地接一个104的电容,否则 因为编码开关的触点抖动会引起轻微误动作。本程序不 使用定时器,不占用中断,不使用延时代码,并对每个 细分步数进行判断,避免一切误动作,性能超级稳定。 我使用的编码器是APLS的EC11B可以参照附件的时序图 编码器控制流水灯最能说明问题,下面是以一段流水 灯来演示。

    标签: 代码 编码开关

    上传时间: 2017-07-03

    上传用户:gaojiao1999

  • 【问题描述】 在一个N*N的点阵中

    【问题描述】 在一个N*N的点阵中,如N=4,你现在站在(1,1),出口在(4,4)。你可以通过上、下、左、右四种移动方法,在迷宫内行走,但是同一个位置不可以访问两次,亦不可以越界。表格最上面的一行加黑数字A[1..4]分别表示迷宫第I列中需要访问并仅可以访问的格子数。右边一行加下划线数字B[1..4]则表示迷宫第I行需要访问并仅可以访问的格子数。如图中带括号红色数字就是一条符合条件的路线。 给定N,A[1..N] B[1..N]。输出一条符合条件的路线,若无解,输出NO ANSWER。(使用U,D,L,R分别表示上、下、左、右。) 2 2 1 2 (4,4) 1 (2,3) (3,3) (4,3) 3 (1,2) (2,2) 2 (1,1) 1 【输入格式】 第一行是数m (n < 6 )。第二行有n个数,表示a[1]..a[n]。第三行有n个数,表示b[1]..b[n]。 【输出格式】 仅有一行。若有解则输出一条可行路线,否则输出“NO ANSWER”。

    标签: 点阵

    上传时间: 2014-06-21

    上传用户:llandlu