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时滞程序
时滞程序
- 2022-03-04 11:35:35下载
- 积分：1
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AppWizard has created this chess application for you. This application
not onl...
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StdAfx.h
StdAfx.cpp
这对文件由用于将一些预编译信息纳入程序。
chessDlg.h
chessDlg.cpp
这一对文件定义并实现了，象棋的主界面。
resource.h
chess.rc
这是整个工程中使用的Windows资源列表。包括置于res子目录下的图标，
位图以及光标等内容。
define.h
这是一个包含程序中的数据表示的定义的头文件。
NewGame.h
NewGame.cpp
这一对文件定义并实现用于新游戏的设置的对话框。
Eveluation.h
Eveluation.cpp
这一对文件定义并实现了估值核心类。
MoveGenerator.h
MoveGenerator.cpp
这一对文件定义并实现了走法产生器。
SearchEngine.h
SearchEngine.cpp
这一对文件定义了搜索引擎接口。
NegamaxEngine.h
NegamaxEngine.cpp
这一对文件定义并实现负极大值搜索引擎-
AppWizard has created this chess application for you. This application
not only demonstrates the basics of using the Microsoft Foundation classes
but is also a starting point for writing your application.
This file contains a summary of what you will find in each of the files that
make up your chess application.
chess.dsp
This file (the project file) contains information at the project level and
is used to bui
- 2022-02-03 16:50:45下载
- 积分：1
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there is a defect in only know that the secret and public key
还有一个缺陷就是在只知道密文 x 及公钥（n,e）的情况下，只要将 (x^e) mod n 所得余数 s 再不断地循环操作 s = s^e mod n，此运算不断地循环 e 次之后，很多情况下都可以循环出原文，只是计算量过余多一些罢了。不过有不少情况下，根本都无须循环 e 次，不过对于1024位的 n 级别来说，e 也是一个相当大的数值，所以循环密文的余数以解得原文是有些不现实。 以上内容仅供参考，如有不实，请予更正-there is a defect in only know that the secret and public key-x (n, e) the circumstances, as long as (x ^ e) mod n from the remaining s to continuously cycle operation s = s ^ e mod n, this constant cycle of Operational e occasion, the very many circumstances can be recycled from the original, but I calculated the volume more than just. There are, however, many instances, simply do not need e cycle times, but for 1024 the level n, e is a very large figure, so secret circle the remainder of the text was obtained in the original is a bit unrealistic. The above is for reference only, if not true, I corrected
- 2022-03-22 01:18:28下载
- 积分：1
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This the VB6 code for Dot Matrix in VB6,
This the VB6 code for Dot Matrix in VB6,-This is the VB6 code for Dot Matrix in VB6,
- 2022-01-26 00:53:32下载
- 积分：1
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一个飞机票订票和退票系统。假设某民航机场有m个航次的班机，每个航次都只到达一个目的地，试为该机场售票处设计一个自动订票和退票系统。
该系统需要实现以...
一个飞机票订票和退票系统。假设某民航机场有m个航次的班机，每个航次都只到达一个目的地，试为该机场售票处设计一个自动订票和退票系统。
该系统需要实现以下功能：
(1) 订票：若该航次余票数大于等于客订票数，则在该航次的乘客表（按乘客姓氏字母词典顺序排列）中，插入订票乘客的信息项，并修改该航次有关数据，否则给出相应提示信息；
(2) 退票：若退票数小于该乘客原订票数，则在相应的乘客表中找到该乘客项，修改该航次及乘客表中有关数据。当由于退票使得该乘客的订票数为零时，要从乘客表中撤销该乘客项；否则给出相应的提示信息；
(3) 将某航次的余票数恢复初值为该航次的最大客票数；
(4) 显示：显示该机场各航次班机的售票情况。-Booking air tickets and a refund system. Assuming there are m a civil airport voyage of the flights, each voyage only to reach a destination, the trial design for an automated airport ticket booking and refund system.
The system needs to achieve the following functions:
(1) booking: If the voyage than the number of votes greater than or equal to the number of passengers booking, the passenger in the voyage List (alphabetical by passengers dictionary order), insert the booking information of passengers, and modify the data of the voyage Otherwise, the corresponding message
(2) Refund: If the refund is less than the number of votes of the original passeng
- 2022-08-03 17:49:14下载
- 积分：1
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C language implementation Prim algorithm based on minimum spanning tree in undir...
C语言实现普里姆算法 根据无向网构造最小生成树-C language implementation Prim algorithm based on minimum spanning tree in undirected network structure
- 2022-08-24 00:31:09下载
- 积分：1
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这里的三维动画明星
here r 3D animated stars
- 2022-09-25 04:05:03下载
- 积分：1
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视点改变的效果的程序
视点改变的效果的程序-Perspective The effect of the change in procedures
- 2023-04-23 22:50:03下载
- 积分：1
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vfp试验素材，包含一般试验中涉及的源代码段，适用于初学者学习使用...
vfp试验素材，包含一般试验中涉及的源代码段，适用于初学者学习使用-vfp test material, including the general test in paragraph relating to the source code for beginners learning to use. .
- 2022-08-21 13:47:04下载
- 积分：1
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document control, the designated directory or document to change (such as to cha...
文件监控,指定目录或者文件发生变动(如改名字,删除),会报告-document control, the designated directory or document to change (such as to change the name, delete), the report
- 2022-09-27 17:35:03下载
- 积分：1
