资源列表
进程调度算法.RAR
- 操作系统实验的进程调度算法实现：提供了四种重要的算法（FIFS、SFJ、HRN）-operating systems experimental process scheduling algorithm : to provide four important algorithm (FIFS, SFJ, 010-001)
hex2bin
- hex文件转为bin的源程序,我字迹一致再用没有长度限制-hex file into a bin of the source, I am writing and then there is no consistent length
20hao
- 这是一个入门级VC＋＋源码，代源码的作用是求100以内偶数之和-This is an entry-level VC source code, source code on behalf of the role is to strive for even less than 100, and the
sfp_exc
- crack sysfileprotect for a min.
siglebus
- DS1820.C ， 单总线DS1820.C希望对学习C51者有帮助-DS1820.C, single bus DS1820.C who wish to learn C51 help
moving
- 关于移动温度载荷，这里是它的瞬态分析的ansys例子，大家共同进步-a program for moving temperature in ansys
OFDM_TRail
- OFDM系统，子载波数目1024，前缀数目256，循环数目256，加入了高斯白噪声-OFDM system, the number of subcarriers 1024, the number prefix 256, the cycle number of 256, by adding a Gaussian white noise
index
- 根据数据生成3D的百分比图，并用不同的颜色标识出来，看上去直观、清淅，而且程序也很方便移植，如果有朋友经常需要用到这种功能，可以考虑将其转换为函数，这样用起来更方便了-Data generated in accordance with the percentage of 3D maps, and marked in different colors, seems intuitive, clear mission, and the transplant procedure is also ver
huiwen
- 这是一个回文验证程序，自己写的.希望可以对你们有一定帮助喽！-This is a huiwen program
guohe
- 过河问题：有一人要将自己的兔子、蔬菜和狐狸等三件物品运过河。过河所用的船每次只能装其中的两件，而这三件物品之间又存在一定的制约关系：兔子不能单独和狐狸以及不能和蔬菜在一起，因为狐狸要吃兔子，兔子也能吃蔬菜。试构造出问题模型，并编程实现这一问题的求解。-guohe
prepha_conj
- 本代码用VHDL实现了复数的卷积，也许对你会有用-this code written by VHDL completes the convolution of complex numbers,which may be quite helpful to you.
cy_dealnmap
- 处理nmap扫描结果,获取扫描结果中的IP地址、端口号以及服务名。-Deal with nmap scanning result, to get the port and service information