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UltraVNC_1053_src

于 2010-05-19 发布 文件大小:3684KB
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下载积分: 1 下载次数: 11

代码说明:

  UltraVnc 1.0.5.3 src

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    集合的运算:交、并、补(难度系数:1.1) 全集:大写字母 ‘A’~’Z’ 要求实现以下功能: 1、集合的输入:自动去掉重复和非法的字符 2、集合的显示:输出集合的全部元素 3、输出一个给定集合的补集 4、输出两个给定集合的交集和并集 (jiao bing bu)
    2013-11-29 20:49:25下载
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    嵌入式Web服务器(以太网络远程控制开关)EA02使用说明书(Embedded server,remote control switch)
    2015-06-08 18:23:27下载
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  • farContral
    远程控制小系统,用VC实现,在VC6.0下编译通过(small remote control system, with VC in the next compile VC6.0)
    2007-04-17 08:44:19下载
    积分:1
  • gh0stdan-exe-ban
    gh0st想必大家都知道不用多说了。 这个不知道是那个牛人改的单exe无DLL不版(gh0st surely we all know rest is history. The cattle do not know that people have switched to single-exe version without DLL does not)
    2011-08-20 21:49:29下载
    积分:1
  • B06040324_B2
    模拟电信计费系统的设计与实现 要求:(1) 计费功能。根据存放在源数据文件中的通话记录和长途费率文件对每一条通话记录计算其通话费用,并将结果保存在费用文件中。其中: 通话费的计算方法如下: 通话费=长途电话费+本地电话费 长途电话费=费率(元/分钟)×通话时长(分钟) (通话时长不满1分钟的按1分钟计算) 本地电话费为:3分钟以内0.5元,以后每3分钟递增0.2元。 (2) 话费查询。输入一个电话号码,从费用文件中统计该电话号码的所有本地话费、长途话费,并从用户文件中查找其用户名,最后在屏幕上显示: 用户名 电话号码 本地话费 长途话费 话费总计 (3) 话单查询。输入一个电话号码,查询并在屏幕显示该用户的所有通话记录,格式为: 用户名 主叫电话号码 被叫电话号码 通话时长 (Design and implementation of simulation of telecommunication billing system Requirements: (1) the charging function. According to the calculation of the call charges stored in the source data file in the call records and the long rate documents for each call record, and save the results in cost file. Among them: The following conversation method to calculate the fee: Long distance telephone calls =+ local telephone fee Long distance telephone fee = rate (yuan/minute)* call time (minutes) (call time less than 1 minutes to 1 minutes) Local call: 3 minutes 0.5 yuan, increasing 0.2 yuan every 3 minutes later. (2) calls inquiries. Enter a phone number, all local calls, long-distance calls statistics the phone number from the cost in the file, and find the user name from user file, and finally displayed on the screen: The user name phone number local calls long distance charges charges totaling (3) if a single query. Enter a phone number, query and display all the call records of th)
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    积分:1
  • 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)
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