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声音技术
说明： 在dos下声音的处理pc喇叭发声,声卡技术,WAV文件的播放技术(voice processing pc speakers audible, audio technology, WAV file playback technology)
- 2005-12-02 21:38:44下载
- 积分：1
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medo
设X[ 0 : n - 1]和Y[ 0 : n – 1 ]为两个数组，每个数组中含有n个已排好序的数。找出X和Y的2n个数的中位数。 编程任务 利用分治策略试设计一个O (log n)时间的算法求出这2n个数的中位数。 数据输入 由文件input.txt提供输入数据。文件的第1行中有1个正整数n（n<=200），表示每个数组有n个数。接下来的两行分别是X，Y数组的元素。结果输出 程序运行结束时，将计算出的中位数输出到文件output.txt中(Let X [0: n- 1] and Y [0: n- 1] for the two arrays, each array containing the n number has been sorted. 2n X and Y to identify the number of digits. programming tasks using the divide and conquer strategy try to design an O (log n) time algorithm to calculate this median number 2n. Data input by the input data provided input.txt file. The first line in the file has a positive integer n (n < = 200), that there are n numbers of each array. The next two lines are the X, Y array elements. The end result is output program runs, the calculated median output to file output.txt)
- 2021-03-22 16:29:16下载
- 积分：1
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49808171
用冲激响应不变法和双线性变换法设计IIR的MATLAB程序(Design of MATLAB Program for IIR by Impulse response Invariant method and Bilinear Transformation method)
- 2019-01-05 03:30:43下载
- 积分：1
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travelingsalesman
用动态规划法求解旅行商问题 已经加入注释 欢迎批评指正(dynamic programming method for the traveling salesman problem has joined Notes welcome criticism correction)
- 2005-06-09 15:40:17下载
- 积分：1
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AdminTest
to check the user has the administration rights is more complicated in windows 7, this is the sample to check it.
- 2011-06-06 23:33:03下载
- 积分：1
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Win32Timer
基于Visual C++ 的win32控制台的程序，实现使用定时器的功能，(Visual C++ win32 console-based program, using the timer function,)
- 2012-03-06 14:20:34下载
- 积分：1
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Hufferman
使用哈弗曼编码技术进行数据压缩和解压缩，缺点在于只能用于文本(Havermann coding techniques using data compression and decompression, disadvantage is that the text can only be used)
- 2010-01-03 19:23:10下载
- 积分：1
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347013
说明： OPNET 无线仿真 PPT + 源码 好好学习吧(OPNET wireless simulation PPT source code)
- 2019-01-04 05:14:51下载
- 积分：1
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codlwhqchsource
这是基于systemview平台的cdma下行链路仿真源码, 能满足cdma开发过程中的仿真需要,(This is the cdma downlink simulation source code based on systemview platform, which can meet the simulation needs in the cdma development process.)
- 2019-01-05 07:37:36下载
- 积分：1
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medo
设X[ 0 : n - 1]和Y[ 0 : n – 1 ]为两个数组，每个数组中含有n个已排好序的数。找出X和Y的2n个数的中位数。 编程任务 利用分治策略试设计一个O (log n)时间的算法求出这2n个数的中位数。 数据输入 由文件input.txt提供输入数据。文件的第1行中有1个正整数n（n<=200），表示每个数组有n个数。接下来的两行分别是X，Y数组的元素。结果输出 程序运行结束时，将计算出的中位数输出到文件output.txt中(Let X [0: n- 1] and Y [0: n- 1] for the two arrays, each array containing the n number has been sorted. 2n X and Y to identify the number of digits. programming tasks using the divide and conquer strategy try to design an O (log n) time algorithm to calculate this median number 2n. Data input by the input data provided input.txt file. The first line in the file has a positive integer n (n < = 200), that there are n numbers of each array. The next two lines are the X, Y array elements. The end result is output program runs, the calculated median output to file output.txt)
- 2021-03-22 16:29:16下载
- 积分：1
