BP算法拟合f(x1,x2)=sin((x1)/x1)*sin((x2)/x2)*函数,样本数为21*21对

该程序实现家谱系统的管理功能，包括动态添加，删除，修改家谱成员-Realize the program genealogical system management functions, including dynamically add, delete, modify members Genealogy

资源描述基于XML技术的服务器目录索引工具      > 程序调用文件基于asp 

程序被开发作为操作系统工作，本科计算机科学的问题。 程序模拟两个进程调度程序： 第一份工作更短和轮叫调度。

这是一个文档序列化程序，是用Vc++ 6.0开发的。-This is a sequence of procedures document is the use of vitamin C 6.0.

应用背景Ｃ语言实现的windows任务管理器，界面与windows自带的任务管理器差不多，能停止正在运行的任务，能显示CPU使用等功能。。。关键技术C language implementation of the Windows Task Manager, the interface and the windows comes with the task manager, can stop being running tasks, can display the use of CPU and other functions...

Gauss-Jordan消去法 实现矩阵求逆 有详细算法解释-Gauss-Jordan elimination method to achieve matrix inverse Algorithm is explained in detail

关于猎人过河问题 要解决这个问题就要使过河时载两个过河，返回时尽量只有一个回来。用一个字符串数组来存猎人，狼，羊，白菜；下标依次为0，1，2，3；但他们都有河这边和那边两种状态；为方便则定义一个结构，只含一个int型变量n；当在河这边时n设为0；在河那边时n设为1。由于每次过河与返回都要考虑狼能否吃羊或羊能否吃白菜；则需要一个函数来判断每次选择是否满足条件。-Hunters across the river on the issues to resolve this issue so we should cross the river when the two set out across the river to return to when there is only one back as far as possible. Using a string array to keep hunters, the wolf, sheep, and cabbage subscript followed by 0,1,2,3 but they have a river here and there two states To facilitate the definition of a structure containing only an int variable n When in the river here when n is set to 0 in the river when n is set to one side. Due to cross the river and return each time to consider whether the wolf can eat the sheep吃羊or cabbage you need a function to determine each choose whether or not to meet the conditions.

封装非终结符集合的类Vn，可以求出 FIRST集和FOLLOW集。-Package of non-terminal symbol set of classes Vn, you can find the FIRST sets and FOLLOW sets.

gdgdfgdfgdfg 自卫队 sd 当日 自卫队 dsf dsf ds 消防处 f fdsfsdfeffffffffffffffffffffffffffffffffffffdsfsdfsdfds 消防处 f df df dfs df dsf d f sdfds 消防处 fd f dsf d fddsfdfdsfdsfs