java键盘输入一个4位数,用循环结构求出各个位置的数字

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.io.*;publicclasscheck{publicstaticvoidmain(String[]args)throwsIOException{System.out.prin

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

其实这是很容易的一个题的,不知道你是只要代码还是要解决过程,代码给你贴上,我测试过,能正常运行（在eclipse下,如果你要直接用dos编译运行的话,要改）packagecom;importjava.

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

packagea1;importjava.util.Scanner;publicclassTest1{publicstaticfloatzhouchang(floatx,floaty){return2

publicstaticvoidmain(String[]arg){intk=0;intm=0;System.out.println("输入一个四位数");Scannerinput=newScanne

你好像对C语言的赋值符号不理解?e=(1000*a)+(100*b)+(10*c)+(1*d);这样是把(1000*a)+(100*b)+(10*c)+(1*d)的值赋给e,而不是把e分解成a,b,c

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

假设你输入的数为x;unsignedchars[4];s[0]=x/1000;//四位数的千位s[1]=x/100%10;四位数的百位s[2]=x/10%10;四位数的十位s[3]=x%10;四位数的

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

#includeintmain(){inta[4];inti;printf("Pleaseinpuetthe4numbers:\n");for(i=0;i{printf("PleaseInpuetth

#includeusingnamespacestd;intmain(){char*str=newchar[255];cin>>str;cout再问：看不等，我刚学编程再答：把输入的数字放到一个指针数组

importjava.util.Scanner;publicclassEncpryt{publicEncpryt(){Scannerscan=newScanner(System.in);System.

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

我觉得这才是楼主的意思吧,首先判断能否同时被5和6整除,如果不能再单独判断是否能被5或者6整除.importjava.util.Scanner;publicclassTest{publicstatic

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