资源列表
producter
- 多个生产者，消费者，在多个缓冲区下，多进程执行的代码-Multiple producers, consumers, multiple buffers, multi-process execution code
cPPproducter
- 单个生产者消费者，多个生产者消费者的代码，多进程多缓冲区-Code of individual producers and consumers more than producers and consumers, multi-process buffer
123
- 获取cpu序列号,硬盘ID,网卡MAC地址 -Get cpu serial number, hard disk ID card MAC address
shujujiegou
- 数据结构 代码集，包括顺序表，串，队列，图，等等十几的代码，内容详尽-Data structure code sets, including the sequence table, string, queue, Figure, and more than a dozen code detaile
4
- 输入一串序列，自动排序输出，且窗口颜色可修改-Input sequence ordering
zhongxinfa
- 基于C语言的重心法求解，详细的编程，运用于物流单设施选址-it is based on C language about how to establish a center ， or to found the most important point
1
- 在STR到STR+99(程序中以STR+10为例子)单元中存放着一个字符串，试编写程序测试该字符串中是否有数字，若有将CL置1，否则CL置0 （要求在debug中调试出该程序）-Stored a string in STR to STR+99 (the program to STR+10 for example) unit, the test program to test whether the string figures, if CL will set 1, otherwise the C
2
- 在字节数组中找出第一个负数，并将该负数存入RES单元中；假设该数组中包含20个带符号数，且至少有1个负数（要求在debug中调试出该程序）-20 with the number of symbols to identify the first negative number in the byte array, and the negative deposit the RES unit assumptions contained in the array, and at least one n
3
- 试编写一个汇编程序，能对键盘输入的小写字母用大写字母显示出来（要求采用子程序格式，即采用子程序完成将小写字母转化成大写字母）-Try to write an assembler uppercase letters, lowercase letters of the keyboard input (requires a subroutine format using subroutine completed lowercase letters into uppercase letters disp
4
- 有2个数组： ary1 db 12,-35,0,126,-90,-5,68,120,1,-19 ary2 db 24,25,0,-38,-89,99,68,100,2,-20 比较两个数组的对应位，将大的数放在ary1数组中，小的数放在ary2中（要求采用子程序格式）-There are two arrays: ary1 db 12,-35,0,126,-90,-5,68,120,1,-19 ary2 db 24,25,0,-38,-89,99,68,100,2,-20
5
- 用DOS的10号功能调用输入一个字符串，并用DOS的9号功能调用将这个字符串输出到屏幕上显示-Enter a string, called DOS 10 DOS the 9th call this string output to the screen display
ansydualiefenxi
- ANSYS水压试验疲劳apdl，对学习apdl的人有很打帮助-ANSYS the hydrostatic test fatigue apdl, learning apdl there are playing to help