虫虫首页| 资源下载| 资源专辑| 精品软件
登录| 注册

SUM

  • ADIAL Basis Function (RBF) networks were introduced into the neural network literature by Broomhead

    ADIAL Basis Function (RBF) networks were introduced into the neural network literature by Broomhead and Lowe [1], which are motivated by observation on the local response in biologic neurons. Due to their better approximation capabilities, simpler network structures and faster learning algorithms, RBF networks have been widely applied in many science and engineering fields. RBF network is three layers feedback network, where each hidden unit implements a radial activation function and each output unit implements a weighted SUM of hidden units’ outputs.

    标签: introduced literature Broomhead Function

    上传时间: 2017-08-08

    上传用户:lingzhichao

  • Solve the 8-puzzle problem using A * algorithme. Input: Program reads start state and goal state

    Solve the 8-puzzle problem using A * algorithme. Input: Program reads start state and goal state and heuristic (N or S) from EightPuzzle.INP file.0 representing blank. There are 2 Heuristic: 1. N: Number of misplaced tiles 2. S: SUM of Manhattan distance of current location and target location. Format: The first line write type of heuristic (N or S). Next is the status of departing and landing status. Between 2 states of 1 line blank. See examples EightPuzzle.INP

    标签: state algorithme Program problem

    上传时间: 2017-08-12

    上传用户:jjj0202

  • 第一章JSP简介 例子1(效果如图1.1所示) Example1_1.jsp: < @ page contentType="text/html charset=GB2312" >

    第一章JSP简介 例子1(效果如图1.1所示) Example1_1.jsp: < @ page contentType="text/html charset=GB2312" > <HTML> <BODY BGCOLOR=cyan> <FONT Size=1> <P>这是一个简单的JSP页面 < int i, SUM=0 for(i=1 i<=100 i++) { SUM=SUM+i } >

    标签: contentType Example charset 2312

    上传时间: 2017-08-31

    上传用户:zycidjl

  • Java的面向对象编程 掌握类的定义以及继承机制

    Java的面向对象编程 掌握类的定义以及继承机制,熟悉方法的多态性。1、 成下面父类及子类的声明:(1)声明Student类。属性包括学号、姓名、英语成绩、数学成绩、计算机成绩和总成绩。方法包括构造方法、get方法、set方法、toString方法、equals方法、compare方法(比较两个学生的总成绩,结果分大于、小于、等于)、SUM方法(计算总成绩)和testScore方法(计算评测成绩,即平均成绩

    标签: Java 对象 定义 编程

    上传时间: 2017-09-12

    上传用户:ecooo

  • 记录学生的信息 :(1)声明Student类。属性包括学号、姓名、英语成绩、数学成绩、计算机成绩和总成绩。方法包括构造方法、get方法、set方法、toString方法、equals方法、compa

    记录学生的信息 :(1)声明Student类。属性包括学号、姓名、英语成绩、数学成绩、计算机成绩和总成绩。方法包括构造方法、get方法、set方法、toString方法、equals方法、compare方法(比较两个学生的总成绩,结果分大于、小于、等于)、SUM方法(计算总成绩)和testScore方法(计算评测成绩,即平均成绩)。

    标签: toString Student equals compa

    上传时间: 2014-11-21

    上传用户:q123321

  • LatentSVM论文

    The object detector described below has been initially proposed by P.F. Felzenszwalb in [Felzenszwalb2010]. It is based on a Dalal-Triggs detector that uses a single filter on histogram of oriented gradients (HOG) features to represent an object category. This detector uses a sliding window approach, where a filter is applied at all positions and scales of an image. The first innovation is enriching the Dalal-Triggs model using a star-structured part-based model defined by a “root” filter (analogous to the Dalal-Triggs filter) plus a set of parts filters and associated deformation models. The score of one of star models at a particular position and scale within an image is the score of the root filter at the given location plus the SUM over parts of the maximum, over placements of that part, of the part filter score on its location minus a deformation cost easuring the deviation of the part from its ideal location relative to the root. Both root and part filter scores are defined by the dot product between a filter (a set of weights) and a subwindow of a feature pyramid computed from the input image. Another improvement is a representation of the class of models by a mixture of star models. The score of a mixture model at a particular position and scale is the maximum over components, of the score of that component model at the given location.

    标签: 计算机视觉

    上传时间: 2015-03-15

    上传用户:sb_zhang

  • oracle查询语句

    最近在学习Oracle,对测试人员而言必须掌握两种语言:第一种是DML,数据操纵语言 (Data Manipulation Language) 是SQL语言中,负责对数据库对象运行数据访问工作的指令集,以INSERT、UPDATE、DELETE三种指令为核心,分别代表插入、更新与删除。第二种是:DQL,数据查询语言 (Data Query Language) 是SQL语言中,负责进行数据查询而不会对数据本身进行修改的语句,这是最基本的SQL语句。核心指令为SELECT,以及一些辅助指令,如FROM、WHERE等,FROM:表示来源,可以搭配JOIN做链接查询; WHERE:过滤条件;GROUP BY:在使用聚合函数时用到,如SUM,COUNT,MAX,AVG;HAVING:对聚合结果进行筛选,这是和WHERE的不同点;ORDER BY:排序。

    标签: oracle 基础 资料

    上传时间: 2016-09-15

    上传用户:天涯云海

  • 批处理感知器算法

    批处理感知器算法的代码matlab w1=[1,0.1,1.1;1,6.8,7.1;1,-3.5,-4.1;1,2.0,2.7;1,4.1,2.8;1,3.1,5.0;1,-0.8,-1.3;     1,0.9,1.2;1,5.0,6.4;1,3.9,4.0]; w2=[1,7.1,4.2;1,-1.4,-4.3;1,4.5,0.0;1,6.3,1.6;1,4.2,1.9;1,1.4,-3.2;1,2.4,-4.0;     1,2.5,-6.1;1,8.4,3.7;1,4.1,-2.2]; w3=[1,-3.0,-2.9;1,0.5,8.7;1,2.9,2.1;1,-0.1,5.2;1,-4.0,2.2;1,-1.3,3.7;1,-3.4,6.2;     1,-4.1,3.4;1,-5.1,1.6;1,1.9,5.1]; figure; plot(w3(:,2),w3(:,3),'ro'); hold on; plot(w2(:,2),w2(:,3),'b+'); W=[w2;-w3];%增广样本规范化 a=[0,0,0]; k=0;%记录步数 n=1; y=zeros(size(W,2),1);%记录错分的样本 while any(y<=0)     k=k+1;     y=a*transpose(W);%记录错分的样本     a=a+SUM(W(find(y<=0),:));%更新a     if k >= 250         break     end end if k<250     disp(['a为:',num2str(a)])      disp(['k为:',num2str(k)]) else      disp(['在250步以内没有收敛,终止']) end %判决面:x2=-a2*x1/a3-a1/a3 xmin=min(min(w1(:,2)),min(w2(:,2))); xmax=max(max(w1(:,2)),max(w2(:,2))); x=xmin-1:xmax+1;%(xmax-xmin): y=-a(2)*x/a(3)-a(1)/a(3); plot(x,y)

    标签: 批处理 算法matlab

    上传时间: 2016-11-07

    上传用户:a1241314660

  • java学生数据库

    /*import java.util.Scanner; //主类 public class student122 {   //主方法   public static void main(String[] args){     //定义7个元素的字符数组     String[] st = new String[7];     inputSt(st);       //调用输入方法     calculateSt(st);   //调用计算方法     outputSt(st);      //调用输出方法   }   //其他方法   //输入方法 private static void inputSt(String st[]){     System.out.println("输入学生的信息:");   System.out.println("学号 姓名 成绩1,2,3");   //创建键盘输入类   Scanner ss = new Scanner(System.in);   for(int i=0; i<5; i++){     st[i] = ss.next(); //键盘输入1个字符串   } }   //计算方法 private static void calculateSt(String[] st){   int SUM = 0;         //总分赋初值 int ave = 0;         //平均分赋初值 for(int i=2;i<5;i++) {   /计总分,字符变换成整数后进行计算   SUM += Integer.parseInt(st[i]); } ave = SUM/3;         //计算平均分 //整数变换成字符后保存到数组里 st[5] = String.valueOf(SUM); st[6] = String.valueOf(ave); }   //输出方法 private static void outputSt(String[] st){     System.out.print("学号 姓名 ");   //不换行   System.out.print("成绩1 成绩2 成绩3 ");   System.out.println("总分 平均分");//换行   //输出学生信息   for(int i=0; i<7; i++){     //按格式输出,小于6个字符,补充空格     System.out.printf("%6s", st[i]);   }   System.out.println();            //输出换行 } }*/   import java.util.Scanner;   public class student122 {   public static void main(String[] args) { // TODO 自动生成的方法存根 String[][] st = new String[3][8]; inputSt(st); calculateSt(st); outputSt(st); }   //输入方法 private static void inputSt(String st[][]) { System.out.println("输入学生信息:"); System.out.println("班级 学号 姓名 成绩:数学 物理 化学"); //创建键盘输入类 Scanner ss = new Scanner(System.in); for(int j = 0; j < 3; j++) { for(int i = 0; i < 6; i++) { st[j][i] = ss.next(); } } } //输出方法 private static void outputSt(String st[][]) { System.out.println("序号 班级 学号 姓名 成绩:数学 物理 化学 总分 平均分"); //输出学生信息 for(int j = 0; j < 3; j++) { System.out.print(j+1 + ":"); for(int i = 0; i < 8; i++) { System.out.printf("%6s", st[j][i]); } System.out.println(); } }     //计算方法     private static void calculateSt(String[][] st)     {      int SUM1 = 0;      int SUM2 = 0; int SUM3 = 0;      int ave1 = 0;      int ave2 = 0;      int ave3 = 0;      for(int i = 3; i < 6; i++)      {      SUM1 += Integer.parseInt(st[0][i]);      }      ave1 = SUM1/3;           for(int i = 3; i < 6; i++)      {      SUM2 += Integer.parseInt(st[1][i]);      }      ave2 = SUM2/3;           for(int i = 3; i < 6; i++)      {      SUM3 += Integer.parseInt(st[2][i]);      }      ave3 = SUM3/3;           st[0][6] = String.valueOf(SUM1);      st[1][6] = String.valueOf(SUM2);      st[2][6] = String.valueOf(SUM3);      st[0][7] = String.valueOf(ave1);      st[1][7] = String.valueOf(ave2);      st[2][7] = String.valueOf(ave3);     } }

    标签: java 数据库

    上传时间: 2017-03-17

    上传用户:simple

  • C语言算法排序问题

    1.Describe a Θ(n lg n)-time algorithm that, given a set S of n integers and another integer x, determines whether or not there exist two elements in S whose SUM is exactly x. (Implement exercise 2.3-7.)

    标签: 算法 排序

    上传时间: 2017-04-01

    上传用户:糖儿水嘻嘻