Arduino 试验

中国人运用和算盘，运行后可进行运算。 王建民：13071396160-use of the Chinese people and thinking, after running for computing. Reelection : 13071396160

倒立摆-采用LQR（线性二次型最优控制）方法实现倒立摆控制-inverted pendulum-use LQR (linear quadratic optimal control) method inverted pendulum control

vc的一些原码,对于帮助学习和提高很有帮助-vc some of the original code, for help to learn and improve very helpful

解压后为多媒体电子琴源代码,希望大家喜欢！-unpacked for multimedia flower source code, hope you like them!

介绍如何用单片机配置FPGA，仅供参考，如有疑问请联系我。-how to use microcontroller FPGA configuration, for reference purposes only, if any questions, please contact me.

用户登陆This function is leting user load,but it is not very useful and very common-This function is leting user load,but it is not very useful and very common

编写筛选查找素数函数： void sieve(bool isPrime[], int n) 其中isPrime[ ]为一个布尔型数组，n为数组大小。由于2是第一个素数， 所以设置isPrime[0]和isPrime[1]的值为false，并设置其余的元素初值为true。 然后对从4到n－1的每一个i，判断i是否能够被2整除，如果i能够被2整除，则设置isPrime[i]为false。 对从2到n/2的每一个可能的数值重复以上处理，当操作结束时，所有值为true的isPrime[i]所对应的i就是素数，它们将是从筛子中掉下来的。-Find prime numbers prepared by screening the function: void sieve (bool isPrime [], int n) which isPrime [] as a Boolean array, n is the size of the array. Because 2 is the first prime number, so set the isPrime [0], and isPrime [1] a value of false, and set the initial value of the remaining elements of true. And then from 4 to n-1 for each i, to determine whether i can be divisible by 2, if i can be divisible by 2, then set the isPrime [i] is false. From 2 to n/2 possible values for each one to repeat the above processing, when the operation ended, all the true value of the isPrime [i] corresponding to i is a prime number, they will be falling from the sieve in the past.

VANET是一个模拟器VANET它可用于学生和其他人

一些常用的经典例程，可以用来学习的软件，例如：控件、表格、结构控制、声音控制和文件I/O等。-Some commonly used classic routines, can be used to study the software, such as: control, form, structure, control, voice control and file I/O and so on.

DVD软件开机画面更换工具，适用于凌阳DVD-DVD Software boot screen replacement tool for Sunplus DVD