登录
首页 » C# » Dry-eye

Dry-eye

于 2011-08-18 发布 文件大小:1KB
0 1131
下载积分: 1 下载次数: 6

代码说明:

  this is CLIPS code for eye disease

文件列表:

Dry eye.clp,10396,2011-08-08

下载说明:请别用迅雷下载,失败请重下,重下不扣分!

发表评论

0 个回复

  • b64
    base64 decoder to decode the original char from base64 encoded string(base64 decoder)
    2009-07-17 11:03:47下载
    积分:1
  • Chap11
    c plusplus beginner structures example source codes
    2012-06-11 01:46:02下载
    积分:1
  • codlwhqchsource
    说明:  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
  • 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
  • 抗积分饱和PID控制算法及仿真抗积分饱PIDsimulation2
    抗积分饱和PID控制算法及仿真抗积分饱和PID控制算法及仿真抗积分饱和PID控制算法及仿真抗积分饱和PID控制算法及仿真(Anti-integral saturation PID control algorithm and simulation)
    2020-07-07 19:38:56下载
    积分:1
  • ljsc
    说明:  OpenCF Bluetooth OpenNETCF Bluetooth Widco div hr div B 文件列表 B : div div(OpenCF Bluetooth OpenNETCF Bluetooth Widco div hr div B file list B: div div)
    2019-01-04 04:49:51下载
    积分:1
  • c_sdk
    c语言和windows SDK编写的遍历全磁盘搜索指定文件的源码。(c language and the windows SDK write traversing full disk search for the specified file source.)
    2020-12-09 14:59:19下载
    积分:1
  • 252616
    通信原理实验指导书,包括全部通信原理章节,原理分析,硬件解释,编程指导等(The communication principle experiment instruction book, including all the communication principle chapter, the principle analysis, the hardware explanation, the programming instruction and so on)
    2019-01-05 07:39:21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)
    2013-09-25 05:53:29下载
    积分:1
  • Dry-eye
    this is CLIPS code for eye disease
    2011-08-18 11:04:38下载
    积分:1
  • 696516资源总数
  • 106914会员总数
  • 0今日下载