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本程序需在DOS下运行。每运行一次可执行程序random.exe，即可显示任意两个 16进制数的加法或减法表达式及其运算结果。在减法运算中，如果被减数小于减数，...
本程序需在DOS下运行。每运行一次可执行程序random.exe，即可显示任意两个 16进制数的加法或减法表达式及其运算结果。在减法运算中，如果被减数小于减数，显示 “Divide Overflow”的提示信息。-the process required under the DOS operating. Every time running executable random.exe can display any number of two 16-band addition or subtraction of expression and the results of operations. The subtraction operation, if less than subtrahend minuend, shows that the "Divide Overflow" message.
- 2022-01-26 07:00:58下载
- 积分：1
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跑动的男孩
跑动的男孩-running boy
- 2022-03-14 10:48:35下载
- 积分：1
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用汇编写的哦!其他也没什么说的，只是他一定要20个字，没办法拉...
用汇编写的哦!其他也没什么说的，只是他一定要20个字，没办法拉-compilation was made with the Oh! Others have nothing to say is that he must 20 characters, not to pull the ball! !
- 2022-05-05 14:01:09下载
- 积分：1
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LED显示屏驱动程序，接收字母符号汉字的内码，显示到显示屏上。...
LED显示屏驱动程序，接收字母符号汉字的内码，显示到显示屏上。-err
- 2022-06-02 07:40:38下载
- 积分：1
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汇编语言超浓缩教程~~！适合初学者，让你尽快进入汇编的领域。...
汇编语言超浓缩教程~~！适合初学者，让你尽快进入汇编的领域。-assembly language-enrichment Guide ~ ~! For beginners, so as soon as you enter the field of compilation.
- 2022-05-15 02:45:42下载
- 积分：1
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单源最短路径问题
单源最短路径问题-single source shortest path problem
- 2023-01-05 19:20:03下载
- 积分：1
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Dining philosophers problem is presented and solved Dijkatra typical synchroniza...
哲学家就餐问题是由Dijkatra提出并解决的典型同步问题。该问题描述的是有五个哲学家共用一张圆桌，分别坐在周围的五张椅子上，圆桌上有五个碗和五只筷子，他们的生活方式是交替的进行思考和就餐。平时，一个哲学家进行思考，饥饿时便试图取用其左右最靠近他的筷子，只有在他拿到两只筷子时才能进餐。进餐完，放下筷子继续思考。-Dining philosophers problem is presented and solved Dijkatra typical synchronization problems. Description of the problem is that there are five philosophers share a round-table were sitting around the five chair, round table has five bowl and five chopsticks, their way of life is the turn of thinking and eating. Normally, a philosopher to think, hunger when trying to access their nearest him about the chopsticks, only if he can get two chopsticks when eating. End meals, to lay down their chopsticks to continue thinking.
- 2023-06-13 12:15:03下载
- 积分：1
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这是一个关于嵌入试开发的简单例子，大家可以参考一下，为以后的学习打基础...
这是一个关于嵌入试开发的简单例子，大家可以参考一下，为以后的学习打基础-This is a test of embedded development simple example, we can take a look to the future to build up the foundation of learning
- 2022-05-07 06:53:39下载
- 积分：1
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最高响应比，先进先出算法的周转时间，平均周转时间
最高响应比，先进先出算法的周转时间，平均周转时间-The highest response ratio, FIFO algorithm turnaround time
- 2022-02-02 04:54:54下载
- 积分：1
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用汇编语言,并且递归求菲波那契函数FIB(N)
用汇编语言,并且递归求菲波那契函数FIB(N)---(N from 1 to 24) (huangyujie)-assembly language, and for recursive function Fibonacci FIB (N )---( N from 1 to 24) (huangyujie)
- 2022-07-03 02:48:47下载
- 积分：1
