搜索资源列表
SumSqDiff
- code to get optical flow using sum of square difference method
sum
- 此程序写了，任意给定两个数，互相作为底数和指数，再求和。如果不满足条件，报错。-This program was made, any given two numbers to each other as the base and index, and then sum. If does not meet the conditions, an error.
sum
- linux下的socket编程，实现服务器端接收客户端输入的两个数据，服务端将求和运算的结果返回给客户端。-under linux socket programming, client server receives two data inputs, the server will be the sum result of the operation back to the client.
pattern
- beamfoming pattern for microphones using delay and sum beamformer
clus
- 极小化误差平方和算法聚类源代码，下载后直接运行就可以了-Minimizing the error sum of squares clustering algorithm source code can be run directly after downloading the
Adaptive-Quantization-in-Min-Sum-Based-Irregular-
- Adaptive Quantization in Min-Sum based Irregular LDPC Decoder. this is a good method for LDPC decoder
Sum
- 本段程序简单的描述了一下C++教程上关于求和的算法-This section describes the procedures for a simple tutorial on what C++ algorithm on the sum
The-sum-of-Multinomial
- 数据结构中的一元多项式相加，多于考研算法实现很重要。后续部分将在后期上传！-The sum of Multinomial
Sum-score-sequence
- 本代码是LABVIEW的一个分数序列求和，简单明了-This code is a fraction of sequence LABVIEW sum, plain and simple
Sum
- 系统首先会让你输入一个3*3矩阵，然后利用数组和下标值来计算矩阵对角线上的值的和。并输出结果-first ,the program will make you input an array,then count the sum of the number.
sum
- This program return the sum of the digits of number you entered.
sason1
- the sum-rate inner bound of sason on two-user interference channel
linked-list-polynomial-sum
- 用链表表示一元多项式，写一算法完成两个一元多项式相加。-Represents a polynomial with a linked list, write an algorithm to complete two one-polynomial sum.
ssd.m
- SSD - Sum of squared Differences
image-operations-
- 3 图像基本处理操作（图像显示、读写、像素统计处理、图像文件I/O等） 4 图像质量的客观评价（峰值信噪比PSNR及其应用） 5 图像的代数运算（绝对值差函数imabsdiff，叠加函数imadd，图像的旋转imrotate） 6 视频处理（创建AVI视频） 7 图像增强（直方图、直方图均化、均值滤波、中值滤波） 8 图像形态学分析（膨胀、腐蚀、开、闭运算、轮廓提取） -3 basic image process
Sum-of-Sinusoids-method
- Sum of Sinusoids method for Rayleigh fading
SUM
- read number from keyboard and sum
sum
- 一元多项式的计算代码,是C++初学者的宝典，希望对大家有帮助。-the C++，sum of the yi yuan duoxiangshi.
sum
- 请编一个函数fun(int *a,int n,int *odd,int *even)，函数的功能是分别求出数组中所有奇数之和以及所有偶数之和。形参n给出数组中数据的个数；利用指针odd返回奇数之和，利用指针even返回偶数之和。-Please make up a function fun ( int* a, int n, int* odd, int* even ), functions are respectively derived
sum
- 一些简单数字的求和，一组不同数字按由大到小按序-Sum of simple figures, a different set of figures descending sequential