资源列表
[matlab例程] Eprom
说明:EProm 试验程序,调试成功,解释详细-EProm testing procedures, testing success, to explain in detail<liushuiqing> 在 2025-10-09 上传 | 大小:17kb | 下载:0
[人工智能/神经网络/遗传算法] 11
说明:一篇关于bp神经网络与遗传算法相结合的硕士论文,里边讲的很清楚,希望对大家有帮助-bpnetandGA<李艳> 在 2025-10-09 上传 | 大小:417kb | 下载:0
[matlab例程] 0470723173
说明:Cellular Authentication for Mobile and Internet Services introduces the reader into the field of secure communication for mobile applications, including secure web browsing with a phone or PC, Single Sign-On (SSO), mobile broadcast content protection<anna> 在 2025-10-09 上传 | 大小:1.48mb | 下载:0
[matlab例程] global-positioning-systems-inertial-navigation-an
说明:This book constitutes an excellent source for understanding the basic operation of the Global Positioning Systems (GPS). It also allows the reader to take the first steps towards understanding how the Kalman Filter is implemented in the estimation of<anna> 在 2025-10-09 上传 | 大小:3.78mb | 下载:0
[matlab例程] map-based-mobile-services-theories-methods-and-im
说明:The book is divided into three parts: theory, method and implementation. Starting with a summary of the state-of-the-art in mobile technologies, the first part analyses their impacts on cartography and pinpoints the missing theories concerned with th<anna> 在 2025-10-09 上传 | 大小:3.97mb | 下载:0
[数据结构常用算法] Expression-evaluation
说明:利用栈来求解表达式求值问题。用的是算数优先法,设置了两个栈:OPTR,用来寄存运算符;OPND,用以寄存操作数或运算结果。-Usage of the expression evaluation stack to solve the problem. Count method using a priority, set up two stacks: OPTR, for storage operator OPND, for the storage operand or operation resu<englishteam> 在 2025-10-09 上传 | 大小:1kb | 下载:0
[数据结构常用算法] Maze(Stack)
说明:应用栈求迷宫通路,用的是“穷举求解”的方法,即从入口出发,顺某一方向向前探索,若能走通,则继续往前走;否则沿原路退回,换一个方向再继续探索,直至探索到通路为止。-Application stack requirements maze path, using a "brute-force solving" approach, starting from the entrance, along a forward direction to explore, if we go through, the<englishteam> 在 2025-10-09 上传 | 大小:2kb | 下载:0
[数据结构常用算法] Expression-evaluation(suffix-method)
说明:在栈的思想下,利用后缀表达式解决表达式求值的问题。类似的,可以推广到前缀表达式求值。-Thinking in the stack, using postfix expression evaluation expression to solve the problem. Similarly, the prefix can be extended to expression evaluation.<englishteam> 在 2025-10-09 上传 | 大小:2kb | 下载:0
[数据结构常用算法] Huffmancode1
说明:按照一定的规则构造哈弗曼树,使得输入的元素都作为叶子节点,然后从叶子到根逆向球每个字符的哈弗曼编码-According to certain rules of construction Havermann tree, making the input element as a leaf node, and then reverse the ball from the leaf to the root code for each character Havermann<englishteam> 在 2025-10-09 上传 | 大小:1kb | 下载:0