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该代码模拟建设银行的取款机,这是我的首次需求分析
该代码模拟建设银行的取款机,这是我的首次需求分析-the bank"s teller machines, this is my first demand analysis
- 2022-01-26 03:10:12下载
- 积分:1
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词法编译程序,不是太长
词法编译程序,不是太长-accidence compiler, not too long
- 2022-02-05 01:03:01下载
- 积分:1
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采用RS232串口通信
serial communication using RS232
- 2022-03-19 10:28:50下载
- 积分:1
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origin软件学习资料,相信大家都知道吧
origin软件学习资料,相信大家都知道吧-origin software, learning materials, I believe we all know
- 2022-02-15 21:04:37下载
- 积分:1
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假设有13位评委参与评分工作。方法是:去掉一个最高分和一个最低
分,然后计算剩余11个分数的平均值,...
假设有13位评委参与评分工作。方法是:去掉一个最高分和一个最低
分,然后计算剩余11个分数的平均值,
-Assumptions have score 13 judges involved in the work. Methods are: removal of a maximum points and minimum points, and then calculating the remaining 11 scores on average,
- 2022-04-17 23:40:53下载
- 积分:1
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本程序用C++语言实现了HTTP协议的请求消息,包括GET,POST方法,在windows下编译通过,可以运行。...
本程序用C++语言实现了HTTP协议的请求消息,包括GET,POST方法,在windows下编译通过,可以运行。-This procedure using C++ language implementation of the HTTP protocol" s request message, including GET, POST method, compiled through the windows, you can run.
- 2023-04-27 23:10:03下载
- 积分:1
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OCaml
----OCaml----
Define the test whether a list is empty.
- 2022-02-09 16:14:33下载
- 积分:1
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C#, the clock applet, simple implementation, beginners learn
C#编写的时钟小程序,简单的实现,初学者借鉴-C#, the clock applet, simple implementation, beginners learn
- 2022-06-27 13:56:19下载
- 积分:1
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动态规划
动态规划----矩阵连乘问题 动态规划法是解决问题的一种方法。它不规定为了得到结果需如何将问题划分为子问题的固定方法,而是按不同输入给出问题的具体实例的子问题划分方法,然后再进行运算、解答问题。 矩阵连乘问题的主要思想如下: 1)设置大小为连乘个数的方阵 2)主对角线上方各元素Di,j(ij)记录获得该最小工作量矩阵分组的第一组的最后一个矩阵的序列号 最后通过下方元素可知最终结果的分组方式。-dynamic programming matrix continually multiply-dynamic programming problem is a problem-solving method. It does not require the results need to be how to divide the problems of the sub- problems fixed, but different input given by the specific example of the problem partition method, and then calculate, and answer questions. Matrix continually multiply the main idea is as follows : 1) the installation of the size of continually multiply the number phalanx 2) above the main diagonal elements Di, j (ilt; J) Matrix Mi continually multiply to the smallest workload 3) below elements Di, j (IGT; J) the record was the smallest workload of a matrix of the first group of a matrix of the final sequence, followed by the final element of the final results of the
- 2022-06-17 10:27:59下载
- 积分:1
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旅行售货员问题的解是一棵排列树.在给出的图中寻找一条费用最优的回路...
旅行售货员问题的解是一棵排列树.在给出的图中寻找一条费用最优的回路-travel salesman problem is the solution with a tree. The plan is to find an optimal cost of the Loop
- 2022-02-04 07:16:44下载
- 积分:1