日历程序,用Basic编写的

贪婪算法解决背包问题-greedy algorithm to solve knapsack problem

Caro Game, VTM Vietnam, this is execute files

java里面实现的接口的应用，对不同的类中，如果有相同的属性，就可以创建接口。这样就便于管理-java inside the interface to achieve the application of a different category, if the same attributes, we can create interfaces. This manageable

　介绍了暖通空调系统(HVAC) 故障检测与诊断(FDD) 的概念,发生故障的原因以及主要的故障检测与诊断方 法,比较了常用的故障检测与诊断方法的优缺点,分析了国内、外暖通空调故障检测与诊断的研究现状和应用情况,对暖 通空调故障检测与诊断的发展过程与方向进行了总结和展望-Introduction of the HVAC system (HVAC) fault detection and diagnosis (FDD) concept, the reasons for failure as well as major fault detection and diagnosis methods, compared the commonly used method of fault detection and diagnosis of the strengths and weaknesses, analyzes the domestic and foreign HVAC Fault Detection and Diagnosis Research and application of HVAC Fault Detection and Diagnosis with the direction of the development process are summarized and prospects

c 语言代码， 欢迎入门单片机的参考， 温度传感器程序代码-c language code, please refer to SCM entry, the temperature sensor code

算法实现题 最长公共子序列问题 给定两个序列 X = { x1 , x2 , ... , xm } Y = { y1 , y2 , ... , yn } 求X和Y的一个最长公共子序列 举例 X = { a , b , c , b , d , a , b } Y = { b , d , c , a , b , a } 最长公共子序列为 LSC = { b , c , b , a }-Algorithm to achieve the longest common sub-title sequence of the problem given two sequences X = (x1, x2, ..., xm) Y = (y1, y2, ..., yn) for X and Y, a maximum common substructure sequence, for example X = (a, b, c, b, d, a, b) Y = (b, d, c, a, b, a) the longest common sub-sequence for LSC = (b, c, b, a )

基于FPGA的16点FFT快速傅立叶变换的Verilog源代码。-the FFT implement of Verilog based on FPGA

pid 模块化化得编程，是增量式的参数,是很基础的东西-pid

数根 数也有根吗？有！数根就是把一个数的各个数位上的值相加，直到得到一个一位数为止。请你编写一个程序来计算任意数N（0≤N≤1000000000000）的根。 输入： 输入有多个案例。每一个案例为一行，每行只有一个整数N。当输入的N是-1时，输入结束。 输出： 每一个案例输出一行。每行只有一个数字，就是N对应的数根。 Sample input 39 -1 Sample output 3-Number of root number also has roots do? There is! The number of roots is to a number of different values on the digital sum until they have been a number of up to one. Please write a program to calculate an arbitrary number N (0 ≤ N ≤ 1000000000000) roots. Input: Enter a number of cases. Each case for the line, each line is only an integer N. When the input of N is-1, the input end. Output: the output line of each case. There is only one number per line, that is, N corresponds to the number of roots. Sample input 39-1 Sample output 3