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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 limit per test: 0.25 sec. memory limit per test: 65536 KB input: standard output: standard Oh, no! - Petya said, walking around his recently bought ground plot. Petya wants to build a new house on it. According to Petya s building project the house should look above like C character. There are many trees growing on Petya s plot. But everyone who cuts down a tree in Petya s country is sent to cut down trees for the rest of his life. So first of all Petya has to choose a place for building the house without cutting any tree. He is feeling that he is not able to find the solution on his own, so he decided to ask you to help him. The task is simplified a little by the fact that Petya s plot has a rectangular shape of size N*M, divided into 1*1 square cells. For each cell it is known whether there are any trees growing there. House can t occupy cells where trees grow. Fortunately Petya could explain how his house must look above. 1) H
Update : 2024-05-19 Size : 1024 Publisher : owaski

this m.file can read data excel directly and then test the data with Alpha cronbach. After that, rank the filed by friedman test-this m.file can read data excel directly and then test the data with Alpha cronbach. After that, rank the filed by friedman test
Update : 2024-05-19 Size : 1024 Publisher : ali

相干性分析m文件,适合于车辆试验频域分析-Coherence analysis m file, suitable for vehicle test frequency domain analysis
Update : 2024-05-19 Size : 2048 Publisher : Guicun Guo

CCS3.3环境下,TMS320F28335的EQep模块的测试。程序实现M法和T法测速。-TMS320F28335 EQep module test in CCS3.3 development environment,mainly fulfilling M and T speed measurement algorithm.
Update : 2024-05-19 Size : 601088 Publisher : ds

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定义法求解lyapunov指数测试函数,以lorenz为对象dingyi_lya.m-Solving lyapunov index test defined function to the object dingyi_lya.m lorenz
Update : 2024-05-19 Size : 23552 Publisher : 青风

倒立摆系统的辩识和控制 (1) 程序Pmodel.m 描述问题 建立倒立摆系统的数学模型 (2) 程序N-Model.m 取得系统的行为状态样本 (3) 程序chp14_3.m 根据以上所得输入值Pm和目标值Tm构造一个神经网络,并根据所描述问题的行为状态样本对网络进行训练,调整网络的权值和阀值,使网络具有与系统同样的行为特性。 (4) 程序chp14_4.m 对系统与网络的行为特性进行测试与比较。-Identification and control of inverted pendulum system Mathematical models (1) procedures Pmodel.m description of the problem to establish an inverted pendulum system (2) program N-Model.m obtain sample behavior of the system status (3) obtained according to the above procedures chp14_3.m the input value and the target value Tm Pm construct a neural network, and according to the description of the problem behavior state of samples to train the network, adjust the network weights and thresholds, the network has the system the same behavioral characteristics. (4) Program chp14_4.m behavior characteristics of the system and the network to test and compare.
Update : 2024-05-19 Size : 27648 Publisher : m

ZSY集团在这里向您介绍一款最新的激光模块,其激光束犹如铅笔一样。小型激光测距模块(ZDM/M)是为那些对于轻重量、小尺寸和低功耗的应用而设计的。小型激光测距模块ZDM/M采用USB为其本身供电和数据通信。这个OEM(开放式)型号有一个用于导航的红点指示器(通过USB开启和关闭),是2012年推出的一个完全封装的微型型号。小型激光测距模块在手持GPS定位方面应用的很不错的。-LASER TEST
Update : 2024-05-19 Size : 407552 Publisher : tonyzhang

bp神经网络分两类.首先在savedata.m里将两类数据分别标记为1和2,然后在nncw.m里将load数据训练,最后在test里load新数据测试。-bp neural network divided into two categories. First, there will be two types of data in savedata.m are labeled 1 and 2, and then load the data in nncw.m in the training, the last load in the test in a new test data.
Update : 2024-05-19 Size : 503808 Publisher : 山青

matlabmk
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M-K突变检验的算法,《现代气候统计诊断与预测技术》书本里M-K突变检验例题可用-MK mutation test algorithms, Modern climate statistics diagnosis and prediction technology books MK mutation test examples available
Update : 2024-05-19 Size : 1024 Publisher : 博士

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利用C\C++实现RSA算法的加、解密运算。 具体包括: 1)利用扩展的Euclid计算 a mod n 的乘法逆元; 2)Miller-Rabin素性测试算法对一个给定的大数进行测试; 3)实现的运算,并计算; 4)利用Euler定理手工计算,并与3)计算的结果对比; 5)实现RSA算法。并对 I LOVE NANJING UNIVERSITY OF AERONAUTICS AND ASTRONAUTICS 加解密。说明:为了方便实现,分组可以小一点,比如两个字母一组。 字母及其数字编码 字母及其数字编码 空格 00 N 14 A 01 O 15 B 02 P 16 C 03 Q 17 D 04 R 18 E 05 S 19 F 06 T 20 G 07 U 21 H 08 V 22 I 09 W 23 J 10 X 24 K 11 Y 25 L 12 Z 26 M 13 -Use of C \ C++ implements the RSA algorithm encryption and decryption operations. These include: 1) using the extended Euclid calculate a mod n multiplicative inverse 2) Miller-Rabin primality testing algorithm for a given test large numbers 3) to achieve the operation, and the calculation 4) the use of Euler Theorem manual calculation, and compared with the results of the calculation 3) 5) implement the RSA algorithm. And I LOVE NANJING UNIVERSITY OF AERONAUTICS AND ASTRONAUTICS encryption and decryption. Description: In order to facilitate the achievement of the packet may be smaller, for example, a group of two letters. Alphabet letters and their digital encoding and digital encoding Spaces 00 N 14 A 01 O 15 B 02 P 16 C 03 Q 17 D 04 R 18 E 05 S 19 F 06 T 20 G 07 U 21 H 08 V 22 I 09 W 23 J 10 X 24 K 11 Y 25
Update : 2024-05-19 Size : 1024 Publisher : 刘洋

可靠性测试系统RTS79的原始数据代码,M文件-Reliability Test System RTS79 raw data codes
Update : 2024-05-19 Size : 2048 Publisher : 荀卿染

M Eagle: The Development and Flight Testing of an Autonomous Robotic Lander Test  Bed (Research)
Update : 2024-05-19 Size : 4935680 Publisher : akhtar

local Name04 extends D O M Test Case Source Code for Andriod.
Update : 2024-05-19 Size : 1024 Publisher : fb01598275cana

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A RUBY class for generating n-element combinatatorials of any m elements (indexed by [0,1,...m-1]): C(m, n)To test, first uncomment the test block. Then run:> ruby cmbn.rb 7 4 ==> C(7, 4)or> ruby cmbn.rb 7 ==> {C(7,0), C(7,1), ..., C(7,7)}
Update : 2024-05-19 Size : 1024 Publisher : zhuang2840628

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用单链表编一个约瑟夫环问题。设编号为1,2,• • • ,n的n个人围坐一圈,约定编号为k(1≤k≤n)的人从1开始报数,数到m的那个人出列,他的下一位又从1开始报数,数到m的那个人又出列,依次类推,直到所有人出列为止,由此产生一个出队编号的序列。压缩文件里面包括了代码,运行文件,还有实验报告。-Joseph compiled a list with a single ring problem. Let numbered 1,2, n individual • • • , n of sitting around, agreed number k (1≤k≤n) who reported that the number starting with 1, the number m to the man out of the line, his the next and 1 countin, count to m the man was out of the line, and so on, until everyone out of the line up, thereby generating a sequence number of the team. Compressed file which contains the code, run the file, and test reports.
Update : 2024-05-19 Size : 88064 Publisher :

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支持向量机,实现2或多分类,基于matlab仿真,内有说明-ex6.m- Octave script for the rst half of the exercise ex6data1.mat- Example Dataset 1 ex6data2.mat- Example Dataset 2 ex6data3.mat- Example Dataset 3 svmTrain.m- SVM rraining function svmPredict.m- SVM prediction function plotData.m- Plot 2D data visualizeBoundaryLinear.m- Plot linear boundary visualizeBoundary.m- Plot non-linear boundary linearKernel.m- Linear kernel for SVM [?] gaussianKernel.m- Gaussian kernel for SVM [?] dataset3Params.m- Parameters to use for Dataset 3 ex6 spam.m- Octave script for the second half of the exercise spamTrain.mat- Spam training set 1 spamTest.mat- Spam test set emailSample1.txt- Sample email 1 emailSample2.txt- Sample email 2 spamSample1.txt- Sample spam 1 spamSample2.txt- Sample spam 2 vocab.txt- Vocabulary list getVocabList.m- Load vocabulary list porterStemmer.m- Stemming function readFile.m- Reads a le into a character string submit.m- Submission script that sends your solutions to our servers submitWeb.m- Alternative s
Update : 2024-05-19 Size : 583680 Publisher : 张伟强

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带约束的随机序列 使用模拟的方法求解投资组合的有效前沿面,m文件生成随机数,测试文件是通过计算风险与收益,将生成的五维随机序列映射到二维空间中-Random sequence with constraints Using the simulation method to solve the portfolio effective frontier. M file to generate a random number, test file is by calculating the risk and return, will generate the five dimensional random sequence is mapped into a two-dimensional space
Update : 2024-05-19 Size : 1024 Publisher : LindaPan

用类描述计算机中CPU的速度和硬盘的容量。要求Java应用程序有4个类,名字分别是PC、CPU、HardDisk和Test,其中Test是主类。  PC类与CPU和HardDisk类关联的UML图如下: 其中,CPU类要求getSpeed()返回speed的值,要求setSpeed(int m)方法将参数m的值赋值给speed。HardDisk类要求getAmount()返回amount的值,要求setAmount(int m)方法将参数m的值赋值给amount。PC类要求setCPU(CPU c)将参数c的值赋值给cpu,要求setHardDisk(HardDisk h)方法将参数h的值赋值给HD,要求show()方法能显示cpu的速度和硬盘的容量。  主类Test的要求 (1) main方法中创建一个CPU对象cpu,cpu将自己的speed设置为2200; (2) main方法中创建一个HardDisk对象disk,disk将自己的amount设置为200; (3) main方法中创建一个PC对象pc; (4) pc调用setCPU(CPU c)方法,调用时实参是cpu; (5) pc调用setHardDisk(HardDisk h)方法,调用时实参是disk (6) pc调用show()方法。-With class describes the computer CPU speed and hard drive capacity. Requirements Java application has four categories, names are PC, CPU, HardDisk and Test, where Test is the main class.  PC with UML class diagram CPU and HardDisk class association as follows: where, CPU type requires getSpeed ​ ​ () Returns the value of the speed, requiring setSpeed ​ ​ (int m) method of the value of the parameter m is assigned to speed. HardDisk class requires getAmount () Returns the value of the amount of required setAmount (int m) method parameter m is assigned to amount. PC class requires setCPU (CPU c) the value of the parameter c is assigned to cpu, requires setHardDisk (HardDisk h) method parameter values ​ ​ h is assigned to HD, requirements show () method can show cpu speed and hard drive capacity. Requirements  main class Test of (1) main method to create a CPU target cpu, cpu own speed is set to 2200 (2) main method to create a HardDisk targe
Update : 2024-05-19 Size : 31744 Publisher : 张文蔷

基于matlab的交互式医学图像分割,操作非常类似于photoshop的磁性套索工具。主要用于ct/mri图像实质区域如肿块的分割提取。test为主m文件。在Matlab R2010a以上证实可用。-An interactive segmentation method based on matlab for medical images.
Update : 2024-05-19 Size : 16384 Publisher : Zhang Yuwei

Otherhere
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最大报销额问题 问题描述 实验室经常要报销一定额度的发票,Sao 负责这项工作。允许报销的发票类型包括买图 书(A 类)、文具(B 类)、差旅(C 类),要求每张发票的总额不得超过1000 元,每张发票 上,单项物品的价值不得超过600 元。 数据输入 测试输入包含若干测试用例。每个测试用例的第1行包含两个正数Q 和N,其中Q 是 给定的报销额度,N(<=30)是发票张数。随后是N 行输入,每行的格式为: m Type_1:price_1 Type_2:price_2 ... Type_m:price_m 其中正整数m 是这张发票上所开物品的件数,Type_i 和price_i 是第i 项物品的种类和 价值。物品种类用一个大写英文字母表示。当N 为0时,全部输入结束,相应的结果不要输 出。-The maximum reimbursement issues Problem Description Laboratories often reimbursed a certain amount of the invoice, Sao responsible for this work. Allow reimbursement invoice type including buying FIG. Book (A class), stationery (Class B), the total amount of travel (Class C), the requirements of each invoice shall not be more than 1000 yuan, each invoice The value of the individual items may not be more than 600 yuan. data input Test input contains several test cases. Each row of the first test case contains two positive numbers Q and N, wherein Q is Given the reimbursement amount, N (<= 30) is the invoice number of sheets. Followed by N-line input, the format of each line is: m Type_1: price_1 Type_2: price_2 ... Type_m: price_m Wherein m is a positive integer on this invoice open items in quantity, Type_i and price_i is kind items and item i value. Categories of items with a capital letters. When N is 0, all the input end of the corresponding results do not lose Out.
Update : 2024-05-19 Size : 1024 Publisher : 宋家
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