EI_Centro30

说明：  live555的mp4播放,这个是用live555 做的通过rtsp协议，传输MP4(this is a project of live video play ,and it use the live555 for mp4 by rtsp ,welcom to download the file live555-mp4.rar)

dolphin模拟器的文件，重要文件，缺失则无法运行(dolphin,Important files, missing can not run)

pcmcia source code

说明：  自适应波束形成算法，包含MVDR、lms，rls等常用算法(adaptive beamforming)

说明：  pid implementation in c++ by backward euler approcimation

Read this to understand basics about software testing.

CE修改器，最新过梦幻西游保护版！！！复活CE 。内存直读！！！(CE modifier, the latest Fantasy Westward Journey Protection Edition!!! Resurrection of CE. Memory Direct Reading!!!)

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.

说明：  贪吃蛇游戏的Java文件，class文件和讲解，他能增长，吃食物，碰到石头会死亡。(The Code and Explanation of Snake Eating)

易语言开源恶搞软件，出现两选框，但让你的朋友只能点选喜欢你按钮-Yi language恶搞open source software, there two marquee, but let your friends like you can click the button