资源列表
java
- 创建一个point,用随机函数定义5对坐标（x,y）并储存在一个该类的数组中，安于原点的距离从远到近输出各点的坐标和原点的距离-Create a point, with random function definitions 5 on coordinates (x, y) and stored in a such an array, the distance from the origin of the content with far to nearly output of different
vff
- 使用matlab打开扩展名为vff文件的程序-Matlab extension used to open documents vff
liantong
- 在某一网络拓扑结构中,检验是否使每一个节点都保持连通状态,在电网规划中有重要作用-In a network topology, the test is to enable each node to maintain connectivity of a state, in the grid plays an important role in planning
MJCZ
- 面积重载，可以减少程序员的工作强度，提高工作效率。-An area of heavy-duty, can reduce the intensity of the work of programmers to improve the work efficiency.
find
- 查找-find
osaUtils
- PyVISA code for the Anritsu MS9710C Optical Spectrum Analyser.
Q12
- 语音信号经过一个FIR滤波器，包括前后的频谱图，光谱图，滤波器的零值点，以及频谱-Voice signal through a FIR filter, including the front and rear of the frequency spectrum, spectral map, the filter zero points, and the spectrum
wave
- TESTBENCH OF CARDIO SYS-TESTBENCH
BA_quant
- This fuction performs backward adaptive quantisation of the input vector x It returns the quantised signal yb, the quantised index ind for each sample of yb and the quantiser step size ss used for each sample of yb The algorithm commences wi
XOR
- 有道难题：另类异或 描述： 对于普通的异或，其实是二进制的无进位的加法。 这里我们定义一种另类的异或A op B, op是一个仅由^组成的字符串，如果op中包含n个^，那么A op B表示A和B之间进行n+1进制的无进位的加法。 下图展示了3 ^ 5 和 4 ^^ 5的计算过程： 0 1 1 (3) ^ 1 0 1 (5) --------- 1 1 0 (6) 0 1 1 (4) ^^ 0 1 2 (5) --------- 0
967sou
- 输入两个数据，经过判断计算，最后输出m,n之间的所有符合要求的自守数。-Two data input through a computation, the final output m, n between the number of all meet the requirements of the automorphic.
emitator
- Sourcecode in BASCOM AVR