dos下多任务C源程序

五子棋 算法 谁先谁胜 ，用的是opengl 以及mfc 做的，只是写了大致的一些思路，具体的细节部分没有完整的呈现出来处理不好请勿见怪。将自己的 一些所得尽可能的于大家分享。不服来站啊 啊啊啊啊

浙江大学硕士学位论文自由曲面接触式测量方法研究与原型系统研制-Zhejiang University, a master s degree thesis of free-form surface-contact measurement methods research and development of the prototype system

高精度定时器，精度为毫秒，基于VC++开发的一个定时器，可用于高精度定时以及其他-High-precision timer, an accuracy of milliseconds, based on the VC++ Developed a timer, can be used for high precision timing, as well as other

说多功能刀 afah gadfhgadrh sh whj-son of bitch shite stupid gadhwtrjjjeh agkngn fgahoahiao gadhojh re gjoasdir a gji

关于编辑框和richtext编辑框的详细使用代码，告诉你很多不为人知的用法．-On the edit box and edit box richtext detailed the use of code, tell you a lot of unknown use.

工资与delphi 。该方案是在法国朗格。简单的管理，其中包含了工资的高度基本操作

一个游戏修改器源代码-a game for source code changes

对USB进行编程,如何读取USB的接口,包括USB的一些驱动程序.-right to program, how to read the USB interface, including some USB driver.

还有一个缺陷就是在只知道密文 x 及公钥（n,e）的情况下，只要将 (x^e) mod n 所得余数 s 再不断地循环操作 s = s^e mod n，此运算不断地循环 e 次之后，很多情况下都可以循环出原文，只是计算量过余多一些罢了。不过有不少情况下，根本都无须循环 e 次，不过对于1024位的 n 级别来说，e 也是一个相当大的数值，所以循环密文的余数以解得原文是有些不现实。 以上内容仅供参考，如有不实，请予更正-there is a defect in only know that the secret and public key-x (n, e) the circumstances, as long as (x ^ e) mod n from the remaining s to continuously cycle operation s = s ^ e mod n, this constant cycle of Operational e occasion, the very many circumstances can be recycled from the original, but I calculated the volume more than just. There are, however, many instances, simply do not need e cycle times, but for 1024 the level n, e is a very large figure, so secret circle the remainder of the text was obtained in the original is a bit unrealistic. The above is for reference only, if not true, I corrected

8051 controller based code in which SMS can control devices connected to Micro controller