一维信号小波的ＶＣ＋＋程序　　希望能帮到快要毕业的人们...

数据结构的两张上机题目,全部通过,没有错误.-data on the structure of the two-topic, all passed, no errors.

百万富翁与陌生人的换钱计划windows Visul C++源程序(电大版教材）-millionaire with strangers for money, the program windows Visul C source files (TVU version Aids)

简单的计算器，实现了最基础的加减乘除和平方等数学运算的一个小程序-Simple calculator, to achieve the most basic addition and subtraction, multiplication and division and square, such as a small mathematical procedures

Scrolabale text in 2@ME

编号为1，2，…,n的n个人按顺时针方向围坐一圈，每人持有一个密码（正整数），一开始人选一个正整数作为报数上限m,从第一个人开始按顺时针方向从自1开始顺序报数，报道m时停止报数。报m的人出列，将他的密码作为新的m值，从他的顺时针方向上的下一个人开始重新从1报数，如此下去，直至所有人全部出列为止，设计一个程序求出出列顺序。采用单向循环链表模拟此过程，按照出列的顺序印出各人的编号测试数据：m的初值为20；n=7,7个人的密码依次为：3，1，7，2，4，8，4，首先m的值为6（正确的出列顺序应为6,1,4,7,2,3,5）-No. 1, 2, ..., n n clockwise direction by individuals sitting around a circle, each holding a password (positive integers), a candidate started as a positive integers m ceiling on the number of reported from the first individuals to embark on the clockwise direction from the start sequence from a few newspaper reports m reportedly stopped a few. M reported out of the list of his password as a new value m, from the clockwise direction on the next re-started from a few reported so on, until all out all out, the design of a procedure sought out the order out. Listless one-cycle simulation process and follow the sequence shown in the numbers printed each test data : m for the initial 20; N = 7,7 personal passwords were : 3,1,7,2,4,8

一个c语言写的牛顿迭代功能的程序，程序结构很清晰，一看就懂。-A c language function in the Newton iteration process, program structure is very clear, look for one to understand.

实例08-TUSPY-08-TUSPY

VC++编程技术与难点剖析 出版社: 清华大学出版社 装帧: 平装 出版年: 2002-4-1 随书光盘-VC++ Programming technology and difficult problem of analysis Publisher: Tsinghua University Press Binding: Paperback Publication Year: 2002-4-1 book with CD-ROM

这个一个托盘程序，效果自己试下就知道了！-A tray of this procedure, the effect of their own under the test will know!

html编辑器，没有网页的朋友可以用他自己做-html editor, not a friend of the website can be used to do his own