资源列表
Knapsack_problem2
- 动态规划算法实现的另一类背包问题// 背包问题 6个物体，重量为：5， 2， 1.7， 3.5， 1， 5.1 背包最大10，,追求装下的最大重量-Another knapsack problem dynamic programming algorithm to achieve// knapsack problem six objects, weight: 5, 2, 1.7, 3.5, 1, 5.1 backpacks maximum 10, the pursuit of maximum we
Pattern3344
- A General Code for The Radiation Pattern Representation of The Array Factor of A Linear Antenna Array
6_shuzhizhuanhuan
- 程序的输入是两个数字 M 和 N( 其十进制的值不超过 1000000000) ，它们的进制在 2~36 之间。对于十以下的数字，用 0~9 表示，而十以上的数字，则使用大写的 A~Z 表示。 求出分别在 2~36 哪种进制下 M 和 N 相等。若相等则输出相应的进制，若不等则输出错误信息。-Enter the program are two numbers M and N (its decimal value is not more than 1 billion), which is b
Dijkstra
- Heaps where the parent key is greater than or equal to (≥) the child keys are called max-heaps those where it is less than or equal to (≤) are called min-heaps. Efficient (logarithmic time) algorithms are known for the two operations needed to implem
chazhi
- 用C语言编程使用拉格朗日插值法求函数的值-Interpolation of a function value
MK
- 用于非线性时间序列的滑动突变检验，非参数方法-detection abrupt MK
A.Sofien
- probleme Binpacking. Donc on va essayer de resoudre ce probleme a l aide d un programme C, qui va ^etre developer par nous m^eme. La resultat de ce programme est un chier de type LP contient la forme lineaire de notre probleme Binpac
Newton-iterative-method
- 用牛顿迭代法求解非线性方程，在程序中修改待求方程即可-With Newton iteration method to solve nonlinear equations
IDW
- idw插值算法代码，里面有纤细的注释，对AE开发者，有些帮助。-The idw interpolation algorithm code, there are slim annotation, AE developers, some help.
celluar-automata2012
- 元胞自动机 一个简单的元胞自动机 fortran-cellular automata
Dynamic-UI-Algo
- Dynamically Calculate Control Dimension at UI
den
- denominate Min-Coin Change Min-Coin Change is the problem of using the minimum number of coins to make change for a particular amount of cents, n, using a given set of denominations d_1 \ldots d_m. This is closely related to the Coin Change pr