graspForth is my humble attempt at a Forth-in-C that has the following goals: GCC ......... to support all 32-bit micros that GCC cross-compiles to. Relocatable . to be able to run in-place in either Flash or Ram. Fast ........ to be "not much" slower than an assembly based native Forth. Small ....... to fit-in approx 300 words in less than 25Kbytes on a 32-bit machine. Portable .... to achieve a 5 minute port to a new 32bit micro-processor, or micro-controller.
标签： graspForth Forth-in-C following attempt
上传时间： 2015-05-23
上传用户：tfyt
These instructions assume that the 1.4 versions of the java and appletviewer commands are in your path. If they aren t, then you should either specify the complete path to the commands or update your PATH environment variable as described in the installation instructions for the Java 2 SDK.
标签： instructions appletviewer the commands
上传时间： 2015-06-01
上传用户：3到15
LCS（最长公共子序列）问题可以简单地描述如下： 一个给定序列的子序列是在该序列中删去若干元素后得到的序列。给定两个序列X和Y，当另一序列Z既是X的子序列又是Y的子序列时，称Z是序列X和Y的公共子序列。例如，若X={A，B，C，B，D，B，A}，Y={B，D，C，A，B，A}，则序列{B，C，A}是X和Y的一个公共子序列，但它不是X和Y的一个最长公共子序列。序列{B，C，B，A}也是X和Y的一个公共子序列，它的长度为4，而且它是X和Y的一个最长公共子序列，因为X和Y没有长度大于4的公共子序列。 最长公共子序列问题就是给定两个序列X={x1,x2,...xm}和Y={y1,y2,...yn}，找出X和Y的一个最长公共子序列。对于这个问题比较容易想到的算法是穷举，对X的所有子序列，检查它是否也是Y的子序列，从而确定它是否为X和Y的公共子序列，并且在检查过程中记录最长的公共子序列。X的所有子序列都检查过后即可求出X和Y的最长公共子序列。X的每个子序列相应于下标集{1,2,...,m}的一个子集。因此，共有2^m个不同子序列，从而穷举搜索法需要指数时间。
上传时间： 2015-06-09
上传用户：气温达上千万的
c语言版的多项式曲线拟合。 用最小二乘法进行曲线拟合. 用p－1 次多项式进行拟合，p<= 10 x,y 的第0个域x[0],y[0],没有用，有效数据从x[1],y[1] 开始 nNodeNum,有效数据节点的个数。 b,为输出的多项式系数，b[i] 为b[i-1]次项。b[0],没有用。 b,有10个元素ok。
上传时间： 2014-01-12
上传用户：变形金刚
高精度乘法基本思想和加法一样。其基本流程如下： ①读入被乘数s1，乘数s2 ②把s1、s2分成4位一段，转成数值存在数组a,b中；记下a,b的长度k1,k2； ③i赋为b中的最低位； ④从b中取出第i位与a相乘，累加到另一数组c中；（注意：累加时错开的位数应是多少位 ？） ⑤i:=i-1；检测i值：小于k2则转⑥,否则转④ ⑥打印结果
上传时间： 2015-08-16
上传用户：源弋弋
Listed below are the typographical conventions used in this guide. – Example C++ code and commands to be typed by the user are in non-bold characters in typewriter font. – Items where the user has to supply a name or number are given in lower-case italic characters in typewriter font. – Sections marked with a ‡ describe features that are also available in ANSI C.
标签： typographical conventions commands Example
上传时间： 2013-12-20
上传用户：xiaoxiang
The initial planning and thinking about this book began during a discussion of SQL Server futures in July 2001. The discussion was with Rob Howard during a trip to Microsoft to discuss the first book I was working on at that time. After that, I stayed involved in what was happening in ADO.NET by going to the SQL Server Yukon Technical Preview in Bellevue, Washington, in February 2002 and by working with the ASP.NET and SQL Server teams at Microsoft since July 2003.
标签： discussion planning thinking initial
上传时间： 2014-01-08
上传用户：cjf0304
Using Gaussian elimination to solve linear equations. // In this version, we allow matrix of any size. This is done by treating // the name of a 2-dimensional array as pointer to the beginning of the // array. This makes use of the fact that arrays in C are stored in // row-major order.
标签： elimination equations Gaussian version
上传时间： 2016-02-14
上传用户：hxy200501
分数是两个整数的比，通常表示为 (或b/a)的形式，其中b称为分子，a称为分母，分母不能为0。分数在计算机中以整数或浮点数（有限小数）的形式表示，大多数情况下都是近似表示，具有较大的误差，例如 ，在计算机中用整数表示为0，用浮点数表示为0.333333。本实例就是要设计一个Fraction (分数) 类类型，该类型的对象可以像基本类型数据一样进行运算，结果仍为分数，运算包括四则运算，关系运算，及求一元一次分式方程的解，输入输出要求按分数方式进行。
上传时间： 2016-02-18
上传用户：zhoujunzhen
This document represents the first stage in a process of taking the National Strategy for Police Information Systems (NSPIS) forward. It defines the mechanisms to ensure that we (and our partners) have access to the right information, in the right form, in the right time at an appropriate cost. The Strategy will ensure the Police Service has a collective understanding of the value of information and that we are able to exploit National Information Assets in support of local policing.
标签： represents the National Strategy
上传时间： 2016-03-08
上传用户：wangdean1101
