C语言程序设计教程（第三版）课后习题5.7 （C语言代码）

#include<stdio.h>
int a;
void num_bit(int);
void prireverse_order(int);
int main()
{
 int num;
 scanf("%d",&num);
 num_bit(num);
 prireverse_order(num);
 return 0;
}
void num_bit(int num)
{
 if(num>10000){
  a=5;
  printf("%d\n",a);
 }
 else if(num>1000){
  a=4;
  printf("%d\n",a);
 }
 else if(num>100){
  a=3;
  printf("%d\n",a);
 }
 else if(num>10){
  a=2;
  printf("%d\n",a);
 }
 else{
  a=1;
  printf("%d\n",a);
 }
}
void prireverse_order(int num)
{
 int ge,shi,bai,qian,wan;
 switch(a){
  case 1:printf("%d",num);break;
  case 2:shi=num/10;
   ge=num%10;
   printf("%d %d\n",shi,ge);
   printf("%d%d",ge,shi);
   break;
  case 3:bai=num/100;
   shi=(num-bai*100)/10;
   ge=num-100*bai-shi*10;
   printf("%d %d %d\n",bai,shi,ge);
   printf("%d%d%d",ge,shi,bai);
   break;
  case 4:qian=num/1000;
   bai=(num-qian*1000)/100;
   shi=(num-1000*qian-100*bai)/10;
   ge=num-qian*1000-bai*100-shi*10;
   printf("%d %d %d %d\n",qian,bai,shi,ge);
   printf("%d%d%%d%d",ge,shi,bai,qian);
   break;
  case 5:wan=num/10000;
   qian=(num-wan*10000)/1000;
   bai=(num-wan*10000-qian*1000)/100;
   shi=(num-wan*10000-qian*1000-bai*100)/10;
   ge=(num-wan*10000-qian*1000-bai*100-shi*10);
   printf("%d %d %d %d %d\n",wan,qian,bai,shi,ge);
   printf("%d%d%d%d%d",ge,shi,bai,qian,wan);
   break;
   default:puts("You are wrong.");break;
 }
}