给定一个正整数,个位数字逆序输出.例如:输入:16948 输出:84961.

#includevoidmain(){inti;printf("pleaseinputanumberbetween0and1000:");scanf("%d",&i);if(i>1000||i=100

两个地方错了.第一,scanf()读入的应为变量的地址,所以scanf("&d",x);应该是scanf("%d",&x)  

n=val(inputbox("请输入N"))t=1fori=1tonjc=1forj=1toijc=jc*jnexts=s+t*jct=-tnextprints第二题：DimxPicture1.Au

intmain(void){\x05intn;\x05intcount=0;\x05inta,b,c,d,e;\x05printf("请输入一个不低於5位的正整数：");\x05scanf("%d",

#include <stdio.h>void main(){    long int a, b,&

#include#includeintDigit(intn){returnn>0?1+Digit(n/10):0;}voidrprint(intn){n>0?printf("%d",n%10),rpr

#include#includevoidmian(){intnum,i,temp[5];printf("请输入一个低于5位的整数\n");scanf("%d",&num);for(i=0;i

#includeintmain(void){longintx;inty,i=1;printf("请输入一个不超过100000的整数:");//99999||x

#include"stdio.h"main(){chara[10];inti,j;h:printf("请输入数字,注意不能超过5位:");gets(a);for(i=0;a[i];i++);if(i>

#includemain(){longm,a,b,c,d,e;printf("请输入一个整数:(0~99999):\n");scanf("%ld",&m);e=m%10;d=m/10%10;c=m/1

#includevoidmain(){inta,b,c,d,e;unsignedinti;//定义i位无符号的整型变量printf("pleaseenterthenumber:\n");scanf("

/*从键盘上输入一个任意位数的正整数,判断它是几位数,并逆序输出该数*/#include"stdio.h"intmain(){intnum,temp,i=0;printf("请输入一个数字\n");s

你把邮箱或QQ号发来,我做完告诉你写好的：importjava.io.*;publicclasswl{publicstaticvoidmain(String[]args)throwsIOExcepti

packagecomm.db2;/***@authorjifei**/publicclassOrder{Stringstr1="";inttempi=0;publicOrder(){}publicOr

可以是可以,不过太麻烦了,见下面的代码#includeintmain(){inta,b[5],s=0;scanf("%d",&a);while(a){b[s++]=a%10;a/=10;}printf

#include"stdio.h"voidmain(){intx,a,b,c,d,e,n,j;printf("请输入一个不多于5位的正整数\n");scanf("%d",&x);a=x/10000;b

设十位数为Y,个位数为X,则有Y=3X-1,原数是10Y+X=10（3X+1）+X逆序数是10X+Y=10X+(3X+1)所以【10（3X+1）+X】-【10X+(3X+1)】=45解得X=2,Y=5

#include#includevoidmain(){inti;charstr[100];scanf("%s",&str);printf("长度为:%d\n逆序输出:",strlen(str));fo

#include#defineMAXN20intmain(){intn,t,k=0;intia[MAXN];printf("请输入一个整数:");scanf("%d",&n);while(t=n%10

#include#includeintmain(void){\x09intnum;\x09do\x09{\x09\x09printf("请输入一个不大于四位的数据:\n");\x09\x09scanf