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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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NovatelRange
上传程序用于解析Novatel输出range 信息，解析结果已经验证。(The upload process is used to resolve the Novatel output range information, analytical results have been verified.)
- 2013-05-17 16:24:27下载
- 积分：1
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reload_perf_counters
PowerShell script for reload perfomance counters in Exchange server
- 2017-05-12 20:00:40下载
- 积分：1
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ATM
编写一个控制台程序，模拟在ATM机的存款和取款过程，当输入给定的卡号和密码（初始卡号和密码到数据文件里查看）时，系统能登录ATM系统，根据用户的身份不同显示不同的程序界面；取款
每次取款金额只能为100的倍数，总额不超过2000元，不允许透支。且取款之后，要记录取款的金额和时间。
存款
每次存款金额只能为100的倍数，总额不超过2000元。
修改密码
新密码长度不小于6位，不允许出现6位完全相同的情况，只有旧密码正确，新密码符合要求，且两次输入相同的情况下才可以成功修改密码。(Write a console program to simulate the ATM deposits and withdrawals process, when you enter a given number and password (the initial number and password to the data file view), the system can log in ATM systems, according to the user' s identity different display different The program interface withdrawal amount each withdrawal can only be a multiple of 100, the total does not exceed 2,000 yuan, does not allow overdrafts. And withdrawals after the withdrawal of the amount to be recorded and time. Deposit Amount per Deposit only a multiple of 100, totaling no more than 2,000 yuan. Change Password New Password length not less than six, allowed six identical situation, only the old password is correct, the new password meets the requirements, and enter the same case twice before they can successfully change the password.)
- 2013-10-14 22:50:17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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gus
用全选主元法高斯消去法求解线性方程组AX B(A guauss solver for linear system AX B)
- 2017-04-21 08:51:02下载
- 积分：1
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template
template 模版函数 函数模版 深入了解(template template template-depth understanding of the function function)
- 2013-07-15 17:33:53下载
- 积分：1
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BY2003
说明： 在程序加载时出现，可以点击画面跳过。
2.登录窗口 只有在输入正确的用户名的密码时，点击登录，才能登录。
3.主体窗口 主体窗口中显示的是当天的话费清单。
4.显示窗口 显示窗口中显示的是当前的各路中断线的状态。
5.打印窗口 可以打印当天的话费清单。
6.退出程序 退出程序。
7.当前费用 可以结算出从当天的00:00:00到些时的费用总和，以及通话时第总和。
(in loading procedures there, click on the picture to skip. 2. Login window only in the correct user name password, click on Login can be logged. 3. The main window of the main window shows the list of the day's calls. 4. Show window display window shows the current rostrum were interrupted state. 5. Print window and print the same day the calls list. 6. Exit exit. 7. The current cost of clearing it out of 00:00:00 from the day when some of the costs of the total, and its total telephone conversation.)
- 2006-03-10 17:23:33下载
- 积分：1
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shiper
Воронежский гос университет, задача по сям
- 2009-10-11 01:30:49下载
- 积分：1
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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
