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chapter03

于 2010-12-22 发布 文件大小:7KB
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  Windows驱动开发技术详解 第三章源码(Detailed Windows driver development techniques Chapter Source)

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    2005-10-14 11:23:19下载
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    2013-12-01 19:53:05下载
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    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)
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