资源列表
linefuntion
- 用line画直线,用来学习和交流使用的编程
am9
- am信号的调制解调matlab的源程序 可以实现不同调制系数下的调制-am signal modulation and demodulation matlab source modulation factor can be achieved under different modulation
fsmnum
- 编译原理课程中通过有限状态机模型实现判定输入字符串是否为实数的源程序。-In compiler construction courses，use the finite state machine to implement to determine whether the input string is the real number.
fvslsgn
- 密歇根大学的ECG（心电图）检测结果数据库-University of Michigan, ECG (electrocardiogram) test results database
y520
- 实现数字发生器！类似摇奖机已经调试过，清楚明了-Digital generator!摇奖machine similar to one
Prim
- prim算法是常用的一种解决最小生成树问题的算法-There are three reasons for the changes that have taken place in our life.Firstly，people s living standard has been greatly improved.Secondly，most people are well paid， and they can afford what they need or like.Last but not
ColorBlobExtractor
- This is a rudimentary color blob extractor. You may choose ExtractBlobs(int r, int g, int b) and call it with a tolerance for each value of red, green and blue channel. Afterwards you get the rectangles containing mainly this colors. Note: The
sectorbuffer
- 先对点做指定半径的圆形缓冲区，然后根据某个方向做一扇形的缓冲区 -First generate loop buffer according to appointed point, then generate sector butter according to direction
WinformConsole
- winform example code written in c sharp good luck
zuoye3
- 码头扩建问题 某市有一码头，每次仅容一辆船停泊装卸货，由于经常有船等候进港，部分人提出要扩建码头。码头平均每月停船24艘，每艘船的停泊时间为24±20小时，相邻两艘船的到达时间间隔为30±15小时，如果一艘船因有船在港而等候1小时，其消耗成本为1000元。经预算，扩建码头大约需要1350万元，故市长决策如下：如果未来五年内停泊船只因等候的成本消耗总和超过扩建码头花费则扩建码头，否则，不予扩建。-Problem in a city with a terminal expansion dock,
zhuanbmp
- 任意文件转bmp(可改长宽比)，可以实现将文件加密成图像，源码-Any file transfer bmp (can change the aspect ratio), you can achieve the encrypted file into an image, the source
randplusminus1
- This gives the generation of random integer.