2.从键盘上输入一个字符,判断该字符是数字.字母.空格或其它

#includeintGetNumber(intn)//用递归来实现很简单{intsum=0;if(n/10!=0){\x09sum+=GetNumber(n/10);}sum+=n%10;retur

intmain(){intn,n1,n2,n3,n4;cin>>n;n1=n/1000;n2=(n%1000)/100;n3=(n%100)/10;n4=n%10;n=n1+10*n2+100*n3+

我给你写了,挺简单的.窗体上添加两个label控件一个command控件Dima(1To10)DimbDimcAsIntegerDimdAsIntegerPrivateSubCommand1_Clic

阶乘intmain(){inti,n;doublesum=1;/*定义3个变量,一个自增,一个范围,一个结果*/printf("enteronenumber:");scanf("%d",&n);/*输

#includeintmain(){chara,b;a=getchar();scanf("%c",&b);printf("a=%c\n",a);printf("b=%c\n",b);return0;}

你所写的代码是正确的,想要查看,请用这个：#includeintmain(){charch;scanf("%c",&ch);if(ch>='A'&&ch

importjava.util.Scanner;publicclassJiecheng{Scannerin=newScanner(System.in);inta=in.nextInt();int

import java.util.Scanner;public class $ {    public sta

#includemain(){inta;printf("ÊäÈëÒ»¸öÕý&

inti;floatf;doubled;charc;scanf("%d,%f,%f,%c",&i,&f,&d,&c);printf("%d,%f,%f,%c\n",i,f,d,c);doublesum

记住一个定理：三角形两边之和大于第三边,两边之差小于第三边.这就足够了!

n=1sum=0dowhiln

#include<stdio.h>void main(){    int num;    

clearf=1input"输入一个大于1的自然数:"tonfori=2ton-1ifmod(n,i)=0f=0exitendifendforiff=1n,"是素数"elsen,"不是素数"endif

Scannerin=newScanner(System.in);intnum=in.nextInt();if(num%3==0)System.out.println("这个数可以被3整除");else

#include#includevoidmain(void){inti,j,k,f,z;scanf("%d",&i);if(i

1、clears=0m=1fori=1to10input"请输入一个数："toaifa/2=int(a/2)s=s+aelsem=m*aendifendfor?"奇数的积是：",m?"偶数的和是：",

intmain(){charstr1[50],str2[50];printf("请输入字符串:|n");gets(str1);for(inti=0;str1[1]='\0';i++){if(str1[

　　如图望采纳!

importjava.util.Scanner;classtest7{publicstaticvoidmain(Stringargs[]){Scannerscan=newScanner(System.