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poj2411 basic algorithm 0ms, more exchanges
poj2411 基本0ms的算法，多交流-poj2411 basic algorithm 0ms, more exchanges
- 2023-08-06 10:20:02下载
- 积分：1
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stevens classic advanced programming in the unix environment (2nd edition) book...
stevens的经典著作advanced programming in the unix environment（2nd edition）一书的例程代码，在类unix环境（如linux）中编译使用。-stevens classic advanced programming in the unix environment (2nd edition) book routine code in the class unix environment (such as linux) compiled to use.
- 2022-06-02 09:38:24下载
- 积分：1
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该多媒体播放器使用Qt4编写，视频，音频播放，图片浏览，放大缩小；系统设置(万年历，时间设置等)...
该多媒体播放器使用Qt4编写，视频，音频播放，图片浏览，放大缩小；系统设置(万年历，时间设置等)-This multimedia player uses Qt4 writing, video, audio playback, photo viewing, zoom system settings (calendar, time setting, etc.)
- 2022-02-11 21:55:46下载
- 积分：1
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Network congestion in the PI algorithm source code, you can use NS
网络拥塞中的PI算法源码，可以在用NS-2仿真软件方针，也可以用于其它算法开发-Network congestion in the PI algorithm source code, you can use NS-2 simulation software approach, can also be used for the development of other algorithms
- 2023-07-07 02:10:02下载
- 积分：1
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这是一个简单的模拟银行存取款的例子,目的是了解如何开发RPC服务程序.代码在LINUX下测试运行通过...
这是一个简单的模拟银行存取款的例子,目的是了解如何开发RPC服务程序.代码在LINUX下测试运行通过-This is a simple bank deposit simulation example, purpose is to understand how to develop RPC service procedures. the code in the Linux operating through testing
- 2022-02-04 02:51:55下载
- 积分：1
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《深入浅出Linux设备驱动编程》，学习嵌入式LINUX驱动编程的好资料...
《深入浅出Linux设备驱动编程》，学习嵌入式LINUX驱动编程的好资料- Layman s language and Linux device driver programming, Study of Embedded LINUX good data-driven programming
- 2022-01-26 06:07:44下载
- 积分：1
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清华大学－电子信息工程系－实验用ARM
清华大学－电子信息工程系－实验用ARM-linux－源代码－NAND篇
- 2022-02-03 14:05:35下载
- 积分：1
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金丝雀
Canary-1.2.tar.bz2.rar Canary-1.2.tar.bz2.rar
- 2022-02-06 11:44:59下载
- 积分：1
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Very classic linux driver development articles to learn linux driver development...
非常经典的linux驱动开发文章
对于学习linux驱动开发有很好帮助
-Very classic linux driver development articles to learn linux driver development for a very good help
- 2022-11-29 07:05:03下载
- 积分：1
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int show_char(int n, const char *name, chtype code)
{
const int height =...
int show_char(int n, const char *name, chtype code)
{
const int height = 16
int row = 4 + (n height)
int col = (n / height) * COLS / 2
mvprintw(row, col, " *s : ", COLS/4, name)
addch(code)
return n + 1
}- int show_char(int n, const char*name, chtype code)
{
const int height = 16
int row = 4+ (n height)
int col = (n/height)* COLS/2
mvprintw(row, col, "*s : ", COLS/4, name)
addch(code)
return n+ 1
}
- 2023-01-13 11:50:04下载
- 积分：1
