Data Structures and Program Design in C++ by Robert Kruse, Alexander Ryba 源代码

插值类 CInterpolate 一元全区间不等距插值 一元全区间等距插值 一元三点不等距插值 连分式不等距插值 埃尔米特不等距插值 光滑不等距插值 二元全区间插值-interpolation category CInterpolate one yuan is not equidistant between the region interpolation one yuan region equidistant between 3:00 interpolation one yuan is not even fractional offset interpolation is not equidistant interpolation Hermite"s not equidistant interpolation smooth interpolation is not equidistant between two yuan region Interpolation

用梯形公式和simpson公式求积分-with trapezoidal formula and simpson formula quadratures

数据库系统概念第五版课件及习题答案_PPTNotes-Database system concepts fifth edition of the courseware and the answer _PPTNotes Exercises

八 方块移动游戏要求从一个含8个数字（用1-8表示）的方块以及一个空格方块（用0表示）的3x3矩阵的起始状态开始，不断移动该空格方块以使其和相邻的方 块互换，直至达到所定义的目标状态。空格方块在中间位置时有上、下、左、右4个方向可移动，在四个角落上有2个方向可移动，在其他位置上有3个方向可移 动。例如，假设一个3x3矩阵的初始状态为： 8 0 3 2 1 4 7 6 5 目标状态为： 1 2 3 8 0 4 7 6 5-Eight-box requirements of mobile games from one containing eight numbers (1-8 with express) the box, as well as a space box (with 0) of the 3x3 matrix of the initial start, and constantly move the box to make space its adjacent box and swap until they reach their defined target state. Spaces in the middle position box has the upper and lower, left and right direction of four removable, in the four corners there are two directions can be mobile, in other locations has three movable direction. For example, suppose a 3x3 matrix of the initial state as follows: 803,214,765 target state as follows: 123,804,765

给定一个n位正整数a，去掉其中任意k

算法导论introduction to algorithm答案-Introduction to algorithms introduction to algorithm answers

随着市场经济的迅猛发展，国内医药企业的竞争也日趋激烈。企业内部的信息管理急需信息化。针对鑫诚药业有限公司的业务情况和企业发展的要求，设计并建立该公司的医药基础信息管理系统。解决企业对于药品信息、员工信息、客户信息以及供应商信息的管理问题。通过该系统，可以快速的获取有关药品、员工、客户及供应商的基础信息。-With the rapid development of market economy, the domestic pharmaceutical enterprises and competition is increasingly fierce. Enterprise information management in urgent need of information technology. Cheng Xin Pharmaceutical Co., Ltd. for the business situation and business requirements of the development, design and build the basis of the company

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用从c＋＋写的floyd最短路径算法，可以求邻接矩阵任意两点的最短路径。-from using c++ to write the Floyd shortest path algorithm, can help adjacency matrix arbitrary 2:00 Shortest Path.

数据结构――C语言版配套源代码-Data structure- C language version of matching source code