1.编写一个方法,方法实现求2个整数的最大值

第一题：两个TextBox,一个Command.OptionExplicitPrivateFunctionMaxModNum_(ByValaAsLong,ByValbAsLong)_AsLongDim

importjava.util.Scanner;classRound{\x09publicdoubleperimeter(doubleradius){\x09\x09returnradius*2;\x

#include "stdio.h"void main(){  int n,sum=0;  printf(&q

#include"stdio.h"#include"stdlib.h"#includeintmain(){intx=0,y=0,z=0,count=0;for(x=0;x

#include#includeintmain(){inti=2;intn=0;ints=0;for(;s+i再问：你好。问下system("pause");这句什么意思？再答：暂停控制台。在

c:intfib(intn){return(n

publicclassTest{publicstaticvoidmain(String[]args){System.out.println("圆的面积："+area(2));System.out.pr

publicstaticdoublemultiple(doublea,doubleb){returna*b;}publicstaticdoubledvd(doublea,doubleb){return

public class Test {\x09public static double area(double length,do

08电商的同学,报上姓名,把分给我吧,不然.罚你把这题目抄100遍.chai

public　interface　Area　｛public　double　calculateArea（）；｝public　class　MyRectangle　｛　　　　　　　　　double　leng

usingSystem;classRectangle{privateintwidth;privateintheight;publicRectangle(intw,inth){width=w;heigh

#includeintmain(){inta[4][4]={{7,4,3,2},{8,6,5,7},{3,7,2,5},{4,8,6,5}};inti,sum=0;for(i=0;i{sum+=a[i

1.#include"stdio.h"//#defineRECURSION1#ifdefRECURSIONlongfact(intn){if(n

FunctionS(AAsInteger,BAsInteger,CAsInteger)AsIntegerDimxAsInteger,yAsInteger,zAsIntegerx=1y=1z=1Fori

方法一：filter2clear all;I=imread('lena.bmp');%读入预处理图像imshow(I)%显示预处理图像K1=filter2(fspecial(

函数中选择分类---逻辑选用IF函数c2=if(b2="Y",a2,b2)

usingSystem;usingSystem.Text;namespaceMynamespace{publicclassProgram{publicstaticintmain(){inttemp=0

publicinterfaceCircle{doublearea(doubler);}publicclassCircleImplimplementsCircle{publicdoublearea(do

public class getWord{ public static void main(String[] args){&nbs