Resharper 7.0.97.60 for visual studio
Resharper 7.0.97.60 for visual studio (2020-04-08, WINDOWS, 41956KB, 下载0次)
HttpHelper万能框架V2.2HttpHelper万能框架V2.2
Httphelper universal framework V2.2 (2020-02-12, C#, 85KB, 下载9次)
某列车调度站的铁道联接结构如图所示。
其中，A为入口，B为出口，S为中转盲端。所有铁道均为单轨单向式：列车行驶的方向只能是从A到S，再从S到B；也可以不在S中驻留，直接从A驶向B；另外，不允许超车。因为车厢可在S中驻留，所以它们从B端驶出的次序，可能与从A端驶入的次序不同。不过S的容量有限，同时驻留的车厢不得超过m节。
设某列车由编号依次为{a1, a2, ..., an}的n节车厢组成。调度员希望知道，按照以上交通规则，这些车厢能否以{1, 2, ..., n}的次序，重新排列后从B端驶出。
In this station, A is the entrance for each train and B is the exit. S is the transfer end. All single tracks are one-way, which means that the train can enter the station from A to S, and pull out from S to B. Note that the overtaking is not allowed. Because the compartments can reside in S, the order that they pull out at B may differ from that they enter at A. However, because of the limited capacity of S, no more that m compartments can reside at S simultaneously.
Assume that a train consist of n compartments labeled {1, 2, , n}. A dispatcher wants to know whether these compartments can pull out at B in the order of {a1, a2, , an} (a sequence). If can, in what order he should operate it? (2019-04-21, Visual C++, 1106KB, 下载1次)
某列车调度站的铁道联接结构如图所示。其中，A为入口，B为出口，S为中转盲端。所有铁道均为单轨单向式：列车行驶的方向只能是从A到S，再从S到B；也可以不在S中驻留，直接从A驶向B；另外，不允许超车。因为车厢可在S中驻留，所以它们从B端驶出的次序，可能与从A端驶入的次序不同。不过S的容量有限，同时驻留的车厢不得超过m节。设某列车由编号依次为{a1, a2, ..., an}的n节车厢组成。调度员希望知道，按照以上交通规则，这些车厢能否以{1, 2, ..., n}的次序，重新排列后从B端驶出。
In this station A is the entrance for each train and B is the exit S is the transfer end. All single tracks are one-way which means that the train can enter the station from A to S and pull out from S to B Note that the overtaking is not allowed Because the compartments can reside in S the order that they pull out at B may differ from that they enter at A However because of the limited capacity of S no more that m compartments can reside at S simultaneously Assume that a train consist of n compartments labeled 1 2 n A dispatcher wants to know whether these compartments can pull out at B in the order of a1 a2 an a sequence If can in what order he should operate it (2019-04-21, Visual C++, 1106KB, 下载0次)
可以通过登陆学生账号进行教室、实验室、会议室等活动区域的预约和查看，但是预约只能预约未来三天，且按预约时间先后审核。预约时间早的可以预约成功，
Reservations and checks can be made in classrooms, laboratories, conference rooms and other activity areas by landing the student account, but reservations can only be made for the next three days, and checked according to the appointment time. Early appointment can make an appointment successfully. (2018-11-18, C#, 107KB, 下载9次)
Fahrenheit to Celsius convertor using c
Fahrenheit to Celsius convertor using c (2017-02-11, C++ Builder, 2KB, 下载1次)
实现字符串的解码，读取一串字符，将其转换成另一种形式输出。
a c program (2013-10-22, C#, 1KB, 下载5次)
c#编码和解码
例如：" E8 A7 A3 E7 A0 81"解码后为"解码"；编码为逆过程。
c# coding and decoding such as: " E8 A7 A3 E7 A0 81" decoded " decoding" coding for the reverse process. (2013-04-24, C#, 39KB, 下载5次)
本程序用来画立方体的正轴测，画出消隐图，画出视野中能够看到的部分
when the program is running it will draw a cub,and the cub is a hideplot
(2011-01-11, C#, 41KB, 下载5次)
本程序可以画出立体图形的三视图：主视图、侧视图、侧视图；还可以画出正轴测
when the program is running,it can draw a three view drawing of the object
(2011-01-11, C#, 38KB, 下载11次)
EXCEL数据处理程序，用于分析EXCEL中指定数据之间的关系，比如A1=A2+A3等。
EXCEL data processing program for the analysis of EXCEL in the relationship between the specified data, such as A1 = A2+ A3 and so on. (2010-11-29, C#, 84KB, 下载15次)
a c language in a7 segment display counter
a c language in a7 segment display counter (2010-11-15, Visual Basic, 17KB, 下载4次)
我做的程序是用来控制一个温度梯度的变化，具体的是从97度到54度再到72度
I have done the procedure is used to control a temperature gradient change, specifically from 97 degrees to 54 degrees then 72 degrees (2009-11-25, C/C++, 2KB, 下载5次)
模拟动态分区存储管理方式的主存分配与回收
（1）确定主存空间分配表（空闲分区，已分配分区）；
（2）采用一种分配方法（BF，WF或FF等）完成主存空间分配和回收；
（3）编写主函数对所有工作进行测试。
Simulate dynamic partition management for main memory storage allocation and recovery (1) determine the main memory space allocation table (free partitions have been allocated partition) (2) using a distribution method (BF, WF, or FF, etc.) to complete main memory space allocation and recycling (3) to prepare the main function of all the work to be tested. (2009-11-08, Visual C++, 3KB, 下载18次)
问题的提出：日历的编排是每400年一个大循环周期，即今年的月、日、星期几和400年前的完全一样。现行天文历法根据天体运行规律，取每年365.2425天。这样，每400年共有365.2425×400＝146097天。如果以365天作为一年，每400年就少了0.2425×400＝97天。这97天要靠设置闰年（每年366）天来凑齐，所以，每400年要设置97个闰
Of the problem: the structure of the calendar every 400 years is a big cycle, that is, this year' s month, day, week and 400 a few years ago, exactly the same. Existing astronomical calendar to run in accordance with the laws of celestial bodies, 365.2425 days a year access. In this way, every 400 years a total of 365.2425 × 400 = 146097 days. As if 365 days a year, every 400 years less 0.2425 × 400 = 97 days. This is a leap year 97 days depends on the set (366 a year) days to put together, so every 400 years a leap to set up 97 (2009-04-10, C++ Builder, 3KB, 下载4次)
生命蛋白质是由若干种氨基酸经不同的方式组合而成。在实验中，为了分析某个生命蛋白质的分子组成，通常用质谱实验测定其分子量x (正整数)，然后将分子量x分解为n个已知分子量a[i](i=1,.......,n)氨基酸的和的形式。某实验室所研究的问题中：
n=18, x 1000
a[i](i=1,.......,18)分别为57, 71, 87, 97, 99, 101, 103, 113, 114, 115, 128, 129, 131, 137, 147, 156, 163, 186
要求针对该实验室拥有或不拥有计算机的情况作出解答。
该题目的解答，将输出所有解及其个数。
Life by a number of protein amino acids by a combination of different ways. In the experiment, in order to analyze the life of a molecule composed of protein, usually measured by mass spectrometry molecular weight of x (positive integer), then the molecular weight x is decomposed into n known molecular weight of a [i] (i = 1,. ......, n) of amino acids and form. A laboratory study of the question: n = 18, x 1000a [i] (i = 1 ,......., 18) were 57, 71, 87, 97, 99, 101, 103, 113 , 114, 115, 128, 129, 131, 137, 147, 156, 163, 186 requests for the laboratory owned or not owned computer to answer. The subject answers, the output and the number of all solutions. (2008-08-08, C/C++, 1KB, 下载8次)