java从键盘输入一个数,是3位数就加上100,是4位数就除以100

importjava.io.*;importjava.util.*;classMain{publicstaticintgcd(intm,intn){while(true){if((m=m%n)==0)

packagetest1;importjava.util.Scanner;publicclassTest6{publicstaticvoidmain(String[]args){Scannerinpu

importjava.util.Scanner;publicclassSum{publicstaticvoidmain(String[]args){Scanners=newScanner(System

publicstaticvoidprintNum(intn){for(inti=2;ij){if(i%j==0){m++;}}else{break;}}if(m<1){System.out.print

importjava.util.Arrays;importjava.util.Scanner;publicclassDataCompare{\x09publicstaticvoidmain(Strin

import java.util.Scanner;  public class TestJT {  public&nbs

a*a=b假设你先循环2到a,发现b都不能整除,这时你在循环a+1到a*a就没有意义了,因为b=a*a,所以b/(a+x)是肯定小于a的,而2到a已经循环过了不是吗?不过一般代码里面都是循环2到b/2

input "输入一个数：" to srsrc=ALLTRIM(STR(sr))lensr=LEN(src)yxsr=""FOR&

Scannerscanner=newScanner（system.in)intx,y,z,temp;System.out.println("请输入x:");x=scanner.nextInt();Sy

importjava.util.Scanner;publicclassTest{publicstaticvoidmain(String[]args){System.out.print("请输入第一个整

//写好了#includeintmain(){boolishave=false;inta,b,num1,num2,temp;intc,GCD,LCM;//最大公约数GCD,最小公倍数LCMprintf

#include"stdafx.h"voidmain(){intnum;printf("输入一个数:");scanf("%d",&num);if(num%3==0){if(num%5==0){prin

判断从键盘输入的三个数能否构成三角形.如果是特殊的三角形（如等边、等腰、直角,等腰直角）需要能判断出来.

oolcalc(unsignedintn){if(n

clearinput"请输入一个数："toxifx%2=1?"是奇数"else?"是偶数"endif第2题是什么?

publicclassJs{privatedoublea;privatedoubleb;privatedoublec;Js(doublea,doubleb,doublec){this.a=a;this

其实这种逻辑运算的,提现不错Java的面向对象的概念的,你直接写逻辑语句就好,首先判断等边,即if(a==b==c)System.out.print("该三角形是等边三角形");这样写下去就好了.希望

#includevoidmain(){intn;sacnf("%d",&n);if(n%27==0)printf("Yes\n");elseprintf("No\n");}

classOddSum//定义类{publicstaticvoidmain(Stringargs[])//main方法{intsum=0;//定义一个整形变量for(inti=1;i

哥们怎么又是你来?再问：您会不会的啊再答：packagetestt;importjava.util.Scanner;publicclassTest{\x09publicstaticvoidmain(S