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ACM contest basic Exercises
ACM大赛基本练习题-ACM contest basic Exercises
- 2022-02-14 06:11:36下载
- 积分:1
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数值处理算法源代码
数值处理算法源代码-numerical algorithm source code
- 2022-07-03 13:56:44下载
- 积分:1
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复合材料渐进损伤分析子程序
复合材料损伤准则多种多样,目前所用大多基于直接损伤退化,利用abaqus子程序USDFLD来实现复合材料的渐进失效。
- 2022-07-19 16:46:38下载
- 积分:1
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《数字滤波与卡尔曼滤波》本人感觉写的比较好的卡尔曼滤波图书,...
《数字滤波与卡尔曼滤波》本人感觉写的比较好的卡尔曼滤波图书,- Digital Filter and Kalman Filter I feel better to write the Kalman filter books,
- 2022-02-14 11:37:33下载
- 积分:1
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《模式分类》第二版的配套的Matlab源代码
《模式分类》第二版的配套的Matlab源代码-"pattern classification," the second version of the Matlab supporting source code
- 2022-11-26 06:55:02下载
- 积分:1
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C实例实用算法包括:枚举、递推、背、矩阵操作…
C++Example实用的算法:包括枚举,递归,回溯,矩阵运算等-C Example practical algorithm include : Enumeration, recursive, back, matrix operations, etc.
- 2023-04-03 09:00:04下载
- 积分:1
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深度搜索 dfs
深度优先遍历图的方法是,从图中某顶点v出发:
(1)访问顶点v;
(2)依次从v的未被访问的邻接点出发,对图进行深度优先遍历;直至图中和v有路径相通的顶点都被访问;
(3)若此时图中尚有顶点未被访问,则从一个未被访问的顶点出发,重新进行深度优先遍历,直到图中所有顶点均被访问过为止。 当然,当人们刚刚掌握深度优先搜索的时候常常用它来走迷宫.
- 2022-10-19 18:20:03下载
- 积分:1
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KNN分类算法
用KNN算法实现数据分类,算法简单易懂,适合初学者!简单来说,K-NN可以看成:有那么一堆你已经知道分类的数据,然后当一个新数据进入的时候,就开始跟训练数据里的每个点求距离,然后挑离这个训练数据最近的K个点看看这几个点属于什么类型,然后用少数服从多数的原则,给新数据归类。
- 2022-09-30 19:55:03下载
- 积分:1
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telnet,common telnet
import java.io.InputStream;
import java.io.PrintStream;
import org.apache.commons.net.telnet.TelnetClient;
public class TelnetTest
{
private TelnetClient telnet = new TelnetClient();
private InputStream in;
private PrintStream out;
- 2022-03-16 19:54:43下载
- 积分:1
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issue a travel home to travel home to drive a car with the minimum of cost yi us...
旅行家问题 一个旅行家想驾驶汽车以最少的费yi 用从一个城市到另一个城市(假设出发时油箱是空的)。给定两个城市之间的距离为D1、汽车油箱的容量为C(以升为单位),每升汽油能行驶的距离为 D2,出发点每升汽油价格P和沿途油站数N(N可以为零),油站i离出发点距离Di,每升汽油价格Pi(i=1,2...N)。计算结果四舍五入至小数点后两位。 如果无法到达目的地,则输出“No Solution"。-issue a travel home to travel home to drive a car with the minimum of cost yi used from one city to another city (assuming starting at the fuel tank was empty). Given the two cities for the distance between D1, car fuel tank capacity of the C (in liters) per liter petrol traveling distance to the D2, the starting point liter gasoline prices P and several petrol stations along the N (N can be zero), PFS i distance from the starting point Di per liter Steam oil prices Pi (i = 1,2 ... N). Calculation results rounded to two decimal places. If unable to reach their destination, the export of "No Solution."
- 2023-09-07 00:25:03下载
- 积分:1