资源列表
burgers
- 大名鼎鼎的burges 方程的Fortran语言程序。Burges方程式源项为零的NS方程，在某些情况下市很有用的。-Famous burges equation Fortran language program. Burges equation source term to zero NS equations, in some cases, city, very useful.
FDTD-used-for-electromagnetic-field
- FDTD(时域有限差分法)计算电磁场分布，可以在visual c++环境中运行。其中采用的格式是用C语言写的，适用于初学者。程序中ex与ey分别代表电场分量与磁场分量。-FDTD (finite difference time domain method) calculation of the electromagnetic field distribution in the visual c++ environment. Which format is used to write in C l
inversion_matrix
- this file calculaate inversion_matrix
Spiral-Matrix
- 用c语言实现螺旋矩阵顺序输出和逆序输出！-C language spiral sequence output and reverse output of the matrix!
0.618
- 学会用MATLAB编写程序； 学会一维搜索法； 精确的一维线性搜索法：0.618法； 非精确的一维线性搜索法：goldstein法；-Learn how to use MATLAB programming Society of one-dimensional search method precise linear one-dimensional search method: 0.618 law non-accurate one-dimensional linear search method:
Square2
- 简单的二位有限元程序设计，有助于学生对有限元的理解-Simple two finite element program design, help student understanding of finite element
lorenz
- lorenz系统 绘制Lorenz方程，并研究相关特性，进一步理解非线性系统。 1、 用计算机绘制Lorenz方程 2、 研究Lorenz方程的相关特性-Analyzed using Matlab evolution of Lorenz system familyDraw Lorenz equations, and research-related characteristics, and further understanding of nonlinear systems. 1, with t
lglr
- 拉格朗日插值算法的实现，主要参照数值计算方法上的流程图实现-a example about lagelangri
exchange-issues
- 贪心算法求解货币兑换问题。算法里的经典问题，用C实现的，开发环境是vs2010-Greedy algorithm currency exchange issues
CCF
- 已知两个时间序列数据，求两者之间的互相关系数-two time series data, solve their mutual cross-correlation
hust20151
- 输入一串字符，并验证其是否为正确的电话号码。长度必须为11位第一位必须为1，第二位为3、5、6、8中得一个，其余位必须为0-9中的一个数-nter a string of characters, and verify that the correct phone number. Length must be 11, the first must be 1, the second was a bit of 3,5,6,8 in the rest position must be a number
Shortest-Palindrome
- 给定一个字符串,你可以将它转换成一个回文通过添加字符在它前面。找到并返回最短的回文可以发现通过执行这个转换。 例如: 鉴于“aacecaaa”,返回“aaacecaaa”。 鉴于“abcd”,返回“dcbabcd”。-Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest p