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Java案例精编源代码
- Java课程设计案例精编源代码,欢迎大家查询下载 -Java curriculum design intensive case Provision source code, we welcome inquiries Download
1316-How-Many-Fibs
- HDU ACM 1316 Problem Descr iption Recall the definition of the Fibonacci numbers: f1 := 1 f2 := 2 fn := fn-1 + fn-2 (n >= 3) Given two numbers a and b, calculate how many Fibonacci numbers are in the range [a, b].
EventTimer
- 本程序实现一个基于事件的定时器，不需要MFC的支持，解决了没有MFC时的定时问题，定时准确，简单易用。-This procedure of the case based on the implementation of a timer, do not need the support of MFC, MFC has not solved the problem when the timing, timing is accurate, easy to use.
jiaocheng
- 10852C++案例教程源代码案例教程课件，需要者可下载。-10852C++ case Tutorial Tutorial courseware case the source code, available for download for those who need it.
DateEdit
- 日期控件显示大小写 控件主要属性： ShowSize：值为bsBig/bsSmall 用于设置显示日期是大写还是小写 ShowType: 值为stYear/stMonth/stDay 用于设置控件显示的是日期的年／月／日 CanMove: 值为Boolean型： 用于设置控件是否可以在运行期移动 PrintSet: 值为psPrint/psNoPrint 用于设置打印与否标志位,现没有用到-Case-control control displays the
MemoryPool_src
- 内存池的使用，里面很详细的介绍了在软件开发过程中关于内存分配，new/del的情况。-The use of the memory pool, which is described in detail in the software development process on the memory allocation, new/del case.
C_languageclassic_case
- 一些C语言里面经常用的经典案例，C语言经典案例-Some C language which often use the classic case, C language classic case
bufengtouzhen
- 布冯投针 找一根铁丝弯成一个圆圈，使其直径恰恰等于平行线间的距离d。可以想象得到，对于这样的圆圈来说，不管怎么扔下，都将和平行线有两个交点。因此，如果圆圈扔下的次数为n次，那么相交的交点总数必为2n。 现在设想把圆圈拉直，变成一条长为πd的铁丝。显然，这样的铁丝扔下时与平行线相交的情形要比圆圈复杂些，可能有4个交点，3个交点，2个交点，1个交点，甚至于都不相交。 由于圆圈和直线的长度同为πd，根据机会均等的原理，当它们投掷次数较多，且相等时，两者与平行线组交点的总数可望也是一样的。这就是说，
3.5.2
- 特殊情况下的素数分类。考虑了内存与cpu双重因素。-Prime number in the special case of classification. Consider the memory with cpu dual factors.
AVLTree
- 数据结构中重要的AVL树，例采用模板类实现，实现了AVL 中情况的判别以及对树的旋转操作，是大二数据结构课程设计的一部分 虽然是采用的控制台方式显示但是转成图像MFC显示或是其他显示也是很方便的-AVL tree data structure, for example, using the template class implements a the AVL case of discrimination, and the tree rotation operation, the sophomo
lowpbe
- * Alternatively, the contents of this file may be used under the * terms of the GNU General Public License Version 2 or later (the * "GPL"), in which case the provisions of the GPL are applicable- * Alternatively, the contents of this file may
personal-manage-system
- 人事管理系统案例，mdb数据库。具备人员查询，修改，添加等功能，MFC编程-Personnel Management System Case mdb database. Have personnel query, modify, add features, MFC programming
Finger
- URU4000指纹仪开发,指纹采集，指纹匹配。-URU4000 fingerprint development, 1, setup.exe fingerprint scanner driver, first in the fingerprint device is not inserted in the case, install the driver and restart the computer 2, URUSDK Development Kit 3, fingerprint scanne
liulantupian
- 在程序中需要处理大幅图片时候，本例实现利用滚动条浏览大幅图片的功能-In the program to deal with large picture, in this case, we can realize the function of using the scroll bar to browse for pictures sharply
String
- boost library - string test Case conversion Predicates and Classification Trimming find algorithms Replace Algorithms Find Iterator Split
Reversible_primes
- 素数幻方 求四阶的素数幻方。即在一个4X4 的矩阵中，每一个格填 入一个数字，使每一行、每一列和两条对角线上的4 个数字所组成的四位数，均为可逆素数。-Prime magic square For fourth order prime magic square.Namely in a 4 x4 matrix, fill in each case a number, make each row, each column and two of the four Numbers on the
cname
- 一个输入名字的控制台程序，能将输入名字的首字母大写，能识别空格-A C++ programme reads in a person’s full name (first names and family name separated by spaces) from the keyboard, changes the first letter of the names into upper case if it is not in, and then displays the name o
5name
- 输入五个姓名，并将几段名字的首字母大写，能识别空格。最后将输入的名字输出到data文件-Write a programme that reads in the name (Given name+ space+ family name) of 5 persons and writes them to a disk file. Each name in the disk file always starts on a new line. Note that you need consider
management-system-for-student
- 题目 学生健康情况管理系统 【问题描述】 实现学生健康情况管理的几个操作功能（新建、插入、删除、从文件读取、写入文件和查询、屏幕输出等功能）。健康表中学生的信息有学号、姓名、出生日期、性别、身体状况等。 【实验内容】 必做内容 利用顺序存储结构来实现 系统的菜单功能项如下： 1------新建学生健康表 2------向学生健康表插入学生信息 3------在健康表删除学生信息 4------从文件中读取健康表信息 5------向文件写入学生健康表
circlecenter-radius
- 用于求解二维情况下的三点外接圆半径及圆心，功能完备，适用各种情况-Is used to solve the two-dimensional case of three circumscribed circle radius and the center of the circle, fully functional and applicable situations