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这是一个经过调试的液晶屏代码,可以给初学习提供参考,
这是一个经过调试的液晶屏代码,可以给初学习提供参考,-This is an LCD screen through the debugging code, you can provide a reference to the beginning of learning,
- 2022-09-14 04:55:03下载
- 积分:1
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一个可以美化滚动条的代码
一个可以美化滚动条的代码-a beauty of the rolling code
- 2023-03-11 18:50:04下载
- 积分:1
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windows底层的内存搜索
windows底层的内存搜索-windows bottom of the memory search
- 2023-08-31 14:15:03下载
- 积分:1
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more and more plug
多风格多插件美化版本同学录-more and more plug-style landscaping version website
- 2022-12-23 13:05:03下载
- 积分:1
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voice codec plugin for winamp.
selever api 帮助文档 大家来下啊-Help selever api to everyone under the ah
- 2022-11-22 15:35:03下载
- 积分:1
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侯俊杰翻译的windows95系统与编程,内容很深奥,但对于理解windows95是一本不可多得的好书...
侯俊杰翻译的windows95系统与编程,内容很深奥,但对于理解windows95是一本不可多得的好书-Translation侯俊杰the windows95 system and the programming is very difficult, but understanding is a windows95 rare books
- 2022-06-29 13:48:21下载
- 积分:1
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程序构造柱
The program contructs column-weight three girth-eight QC-LPDC codes given row-weight and sub-matrix size. row-weights and sub-matrices sizes are variable
- 2023-05-17 18:10:03下载
- 积分:1
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功能添加: 在保持原数据库结构及源码结构的基础上添加了新添API函数的功能...
功能添加: 在保持原数据库结构及源码结构的基础上添加了新添API函数的功能-Function add: While maintaining the original database structure and source code, based on the structure adds additional API function function
- 2023-08-29 06:40:02下载
- 积分:1
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ADS projects, three versions of 2410nboot, 1 months 2440nboot, used to guide the...
ADS工程,3个版本的2410nboot,1个2440nboot,用来在nand flash引导eboot或者CE操作系统镜像。其中的Nboot_2410(eboot)可以直接引导eboot,nboot位于sector 0,eboot位于sector 2,eboot大小256K(参数大小均可修改) Nboot_2410(toc)需要在sector 1中写入toc file,来引导eboot或者OS image。-ADS projects, three versions of 2410nboot, 1 months 2440nboot, used to guide the eboot in the nand flash or CE operating system image. Which Nboot_2410 (eboot) guide can be directly eboot, nboot located in sector 0, eboot located in sector 2, eboot size of 256K (the size parameter can be modified) Nboot_2410 (toc) need to write in the sector 1 in the toc file, to guide the eboot or OS image.
- 2022-02-24 12:26:06下载
- 积分:1
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有n个整数,使其前面各数顺序向后移m个位置,最后m个数变成最前面的m个数...
有n个整数,使其前面各数顺序向后移m个位置,最后m个数变成最前面的m个数-There are integers n, the number of its previous order of the backward shift position m, and finally the number m of m into the top of the number of
- 2022-03-25 22:59:30下载
- 积分:1