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MP085传感器的资料
这是MP085传感器的资料,资料齐全,有需要的可以下载哦(This is the data of MP085 sensor. The data is complete. You can download it if you need it.)
- 2020-06-19 08:00:02下载
- 积分:1
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calculator2
实现计算器功能,还有漂亮的界面 ,使用VC6.0 mfc实现(calculator)
- 2010-01-06 08:53:14下载
- 积分:1
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LodTerrain
该程序实现了多分辨率地形漫游系统。地形高程数据是采用随机函数生成的,然后使用LOD技术模拟地形。(The program achieved a multi-resolution terrain roaming system. Terrain elevation data is generated using random function and then use the LOD terrain simulation technology.)
- 2008-12-18 22:25:14下载
- 积分:1
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cr-cz
蒸发冷凝的udf,,包括传热传质,对应于水(Evaporative condensation of UDF, including heat and mass transfer, corresponds to water)
- 2017-06-28 21:39:50下载
- 积分:1
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row socket 网络监听
row socket 网络监听
- 2014-04-15下载
- 积分:1
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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...
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
- 2022-08-18 05:53:02下载
- 积分:1
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C#程序开发范例宝典 图书光盘附带源码 第四章
C#程序开发范例宝典 图书光盘附带源码 第四章-C# Program development paradigm Baodian Book CD-ROM attached to Chapter IV source
- 2022-01-25 21:44:18下载
- 积分:1
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Downloads
借鉴别人,欧拉回路,测试输入包含若干测试用例。每个测试用例的第1行给出两个正整数,分别是节点数N ( 1 < N < 1000 )和边数M;随后的M行对应M条边,每行给出一对正整数,分别是该条边直接连通的两个节点的编号(节点从1到N编号)。当N为0时输入结束。(Referring to others, Euler circuit, test input contains several test cases. The first line of each test case gives two positive integers, namely the number of nodes N (1 < N < 1000) and the number of edges M; the subsequent M lines correspond to M edges, and each line gives a pair of positive integers, which are the number of the two nodes directly connected to the edge (node numbers from 1 to N). When N is 0, the input ends.)
- 2019-05-22 19:54:11下载
- 积分:1
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ASP.NET MVC5+EF6后台管理系统源码(含部署文档以及数据库脚本)
mvc后台,登录账户:admin 密码:zxczxc
- 2019-12-17下载
- 积分:1
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getInverse
Calculate the inverse of a mod n using Extended Euclidean s Algorithm. Source code available.
- 2015-02-21 02:19:12下载
- 积分:1