资源列表
gpcm05a22
- 用于求解广义预测控制器，控制律为非增量式的-used for generalized predictive controller, the control laws of a non-incremental
anglcau
- 在点云处理中，我们通常需要知道两个面的夹角，这个小程序通过最小二乘拟合。求的两个面的夹角，供参考。-in point cloud processing, we often need to know the angle between two faces, the small least-squares fitting procedure. For the angle between the two face for reference.
ROCandAUC
- 计算ROC曲线和AUC的Matlab程序
meller
- 之前做的meller-幂法，数值分析算法
zizuzhi
- 使用som神经网络对数据进行分类，并可视化显示。-The use of som neural network for data classification, and visual display.
rotationmat3D
- 输入一个旋转角度，以及一个旋转轴，可以求出空间中任意点的旋转矩阵 注意角度用弧度单位-Given input: r: angle to rotate, in rads Axis: axis of rotation, defined as a 3point which connects to the origin. Output: 3x3 Matrix R such that for an arbitrary point v, Rv is the poi
KnapsackProblemCLanguagesSourceCode)
- 背包问题（贪婪算法）c语言源代码。希望有帮助！-Knapsack problem (greedy algorithms) c language source code. Hope that helps!
startCA0
- 最简单的CA（元胞自动机），一点不同是这里的元胞取值并不是离散的，而是连续的。-the easiest CA algorithm
cPP-Bomb-flat-parabolic-trajectory
- 可以计算物体做平抛运动时的轨迹，包含问题介绍和程序-Can calculate the movement of the object to make flat parabolic trajectory, contains the problem descr iption and procedures
dichotomy
- 编制了求解非线性方程的二分法，是一种数值计算方法-Compiled the dichotomy of solving nonlinear equation is a kind of numerical method
1014
- 浙大 编程能力测试 的甲级题目 第 1014. Waiting in Line (30) 又是一道服务队列问题-uppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are:
Fastest_Dirction
- 最速下降法（黄金分割法确定步长）分、解决无约束问题-Steepest descent method (golden section method to determine the steps) points, unconstrained problem solving