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实例288——创建MFC扩展DLL,实现圆形按钮类
创建MFC扩展DLL,实现圆形按钮类。
- 2022-08-17 13:37:55下载
- 积分:1
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通过设计一个《学生成绩统计管理》,进一步熟悉C++中类的概念、类的封装、继承的实现方式。了解系统开发的需求分析、类层次设计、模块分解、编码测试、模块组装与整体调...
通过设计一个《学生成绩统计管理》,进一步熟悉C++中类的概念、类的封装、继承的实现方式。了解系统开发的需求分析、类层次设计、模块分解、编码测试、模块组装与整体调试的全过程,加深对C++的理解与Visual C++环境的使用:逐步熟悉程序设计的方法,并养成良好的编程习惯。-Through the design of a " student achievement statistics management," become more familiar with C++, the concept of classes, the class encapsulation, inheritance of implementation. Understanding of system development, needs analysis, class-level design, module decomposition, code testing, module assembly and commissioning of the entire process as a whole, to deepen the understanding of C++ and Visual C++ environment to use: become familiar with program design methods, and to develop good programming habits.
- 2022-03-18 01:44:35下载
- 积分:1
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VBA在Word 2000中的信息
VBA For Word 2000 资料-Information of VBA For Word 2000
- 2022-02-13 15:13:25下载
- 积分:1
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User Name: Legal Program Code License: 49EGT4Y4DDXE8GN329AM2NPVJ
用户名:Legal Program
- 2023-09-08 06:00:03下载
- 积分:1
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三星6400 onenand 测试代码 ,三星6400原厂开发板带
三星6400 onenand 测试代码 ,三星6400原厂开发板带-Samsung 6400 onenand test code, Samsung original 6400 development board with
- 2022-01-31 09:24:12下载
- 积分:1
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一个很强的五子棋哦,希望大家能够喜欢..里面有具体的算法,主要用于人机对战使用的...大家满慢用吧!...
一个很强的五子棋哦,希望大家能够喜欢..里面有具体的算法,主要用于人机对战使用的...大家满慢用吧!-this is a five chess game.
- 2022-03-04 02:48:05下载
- 积分:1
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一个可以自由设定关机时间的小程序
一个可以自由设定关机时间的小程序-a free set of small-time shutdown procedures
- 2022-05-26 19:34:53下载
- 积分:1
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seamless scroller, auto insert div element
无缝滚动,自己写的,效率很高,使用方便,自动建立div层-seamless scroller, auto insert div element
- 2022-08-26 03:27:04下载
- 积分:1
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The source implementation of the BCH error correction encoding and decoding proc...
该源码实现了BCH纠错编解码程序,并附上水印处理程序-The source implementation of the BCH error correction encoding and decoding process, together with a watermark processing
- 2022-07-28 10:29:08下载
- 积分:1
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there is a defect in only know that the secret and public key
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
- 2022-03-22 01:18:28下载
- 积分:1