java语言从键盘输入输入1个四位数,将其加密后输出.

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

#includevoidmain(){\x09intn,i,j,t,a[10000];\x09printf("请输入正整数N:");\x09scanf("%d",&n);\x09printf("请依次

2L你好夸张!把下面代码放到main()方法中Scanners=newScanner(System.in);System.out.println("请输入a的值：");inta=s.nextInt()

运行通过,有问题HI：#includeintmain(){intnum,n=0,m=0;do{scanf("%d",&num);if(num>0)m++;elseif(num

voidmain(){intn=0;inti=0;intsum=0;printf("pleaseinputn:\n");scanf("%d",&n);if(n

#includevoidperm(int*data,intn,intcurr){if(curr==n-1){for(inti=0;iprintf("%d",data[i]);printf("\n");

VC6.0调试通过了,while循环结构#include"stdio.h"voidmain(){inti,sum=0,n;i=1;printf("Inputn:");scanf("%d",&n);wh

#include "stdio.h"int main(){   int count1=0,count2=0,i;&nbs

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

#include"stdio.h"intmain(intargc,char*argv[]){\x09intarr[10];\x09for(inti=0;i

#includeintmain(void){intcount=0,n,sum=0;whlie(scanf("%d",&n)!=EOF){if(n==-1)break;elsesum+=n;count+

inta,sum,count;doubleavg;sum=0;count=0;while(1){printf("请输入正整数（输入0结束输入）:");scanf("%d",&a);if(a==0){b

#include<stdio.h>void main(){ int a[5]={0}; int i,j,t; printf(&q

BufferedReaderbuf=newBufferedReader(newInputStreamReader(System.in));System.out.print("请输入时间(注意格式必须为

#include<stdio.h>#include<stdlib.h>#include<string.h>typedef struct Link

importjava.util.Scanner;publicclassTest01{publicstaticvoidmain(String[]args){Scannerscan=newScanner(

main(){inta,b,c;scanf("%d%d%d",&a,&b,&c);if(a+b==c)printf("\nRIGHT");elseprintf("\nERROR");}

有清晰的二维数组最好了?这是不可能实现的,只能用我的方法模拟出一个二维数组.我的代码现在行和列就是键盘输入的了呀scanf("%d%d",&m,&n);m和n就是行和列至于键盘输入数组数据,把*(p+

while循环for循环太多了也太乱了.我给你修改了一些,也调试成功了,你看一下吧#include#includeintmain(void){inta,b,n,count,i,x,y;scanf("%

用Scannerscan=newScanner(System.in);Strings=scan.next();if(s!="y"||s!="n"){System.out.println(