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Expert_Choice[1]
- Expert Choice represents a significant contribution to the decision making process 工t assists a decision maker in solving complex problems involving many criteria and several courses of action . An Expert Choice solution to a problem reflects the exp
牛顿法解方程源代码
- 牛顿法解方程之混沌情况，帮助你对牛顿法解方程的领悟，具有非常强的视觉冲击力，美仑美奂，不看将后悔一辈子的！-Newton equation solution chaotic situation and help you right Solving equations of Newton's grasp, with very strong visual impact, attractive, and will not look at a lifetime of regret!
hyplas
- ************************************************************************ * * * * * THIS IS THE H Y P L A S 2.0 README FILE * * ----------------- * * * * HYPLAS is a finite element program for implicit small and large * * strain analisys of hyperelast
fortran-code
- 徐世良第二版FORTRAN源码，很有用 徐世良第二版FORTRAN源码，很有用-fortran code,It s useful help for us! fortran code,It s useful help for us!
p
- 拟合曲线，最小二乘法和牛顿法等，有助大家基础学习。-Curve fitting, least squares method and Newton s law, the foundation will help you learn.
football
- Descr iption 2008六月激动人心的欧洲杯就要开始了，来直世界各地的人们都会涌向主办国瑞士和奥地利。由于比赛是在不同的城市进行，为了使球迷们能够尽快的到达各个城市，主办方决定在城市之间修建铁路来满足需要，铁路线当然越短越好喽。现在你的任务是帮助主办方选择最优的修建方案。 Input 输入的第一行是一个数n(2 <= n <= 50 )，表示城市的数目。 后面是一个n * n的矩阵A，A[i][j]表示第i个城市和第j个城市的路径(0 <= A[
fuhexinpusen
- 复合辛普森算积分C++实现，相信对大家学习有一定帮助-Composite Simpson' s integral operator C++ implementation, I believe this will definitely help them to learn
FORTRAN
- Fortran 77 的用户指南 希望对需要的人有帮助-Fortran 77 User' s Guide hope to those who need help
floyd
- 计算赋权图中各对顶点之间最短路径有两种方法，其一是调用 Dijkstra 算法，另一方法就是被称为Floyd 的算法，利用LINGO9.0编写了通用的FLOYD算法如下，希望对大家有所帮助,带有例题哦！-Calculated on the weighted graph in the shortest path between vertices in two ways, one is called Dijkstra algorithm, the other method is known as F
GAWZM
- 用matlab编写的遗传算法程序，挺好用，大家能参考学习交流就最好了-based on Matlab ,gennetic alg. It s very usefull.Hope it can help your need.
Gauss-Seidel
- 数值线性代数（徐方树）第四章上机题 G-S迭代法 希望有帮助-Numerical linear algebra (Xu Fangshu) the fourth chapter on the topic G-S iterative method I hope there is help
file
- Joe and his entire College, total 1337 of them, went for a picnic. They were bored and they planned a game. Everybody stand in a circle, each of them identifiers starting 1 to 1337 consecutively in a circle, clockwise. The rule is that “a numbering C
sgu250
- sgu250：Constructive Plan 题目大意： 给出一个n∗ m的01矩阵，0表示不能放，1表示能放，在其中放入三个矩形，要求满足如下条件： 1.每个矩形面积大于0。 2.这些矩形必须是一个联通块，矩形之间不能重叠。 3.矩形的左边界在同一条线上。 4.中间矩形的横向长度小于两边矩形的横向长度。 求出最大的三个矩形的总面积，无解输出− 1。-250. Constructive Plan time li
factory
- 本代码为工厂模式学习举例，有普通工厂模式，抽象工厂模式，对学习设计模式的童鞋来书大有帮助-The code for the factory pattern study, for example, the ordinary factory pattern, abstract factory pattern, design pattern of children s shoes in Hebrews of learning help