PAT 甲级 1069  The Black Hole of Numbers

1069 The Black Hole of Numbers （20 point(s)）

For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in non-increasing order first, and then in non-decreasing order, a new number can be obtained by taking the second number from the first one. Repeat in this manner we will soon end up at the number ​​6174​​ -- the black hole of 4-digit numbers. This number is named Kaprekar Constant.

For example, start from ​​6767​​, we'll get:

7766 - 6677 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174
7641 - 1467 = 6174
... ...

Given any 4-digit number, you are supposed to illustrate the way it gets into the black hole.

Input Specification:

Each input file contains one test case which gives a positive integer N in the range (0,104).

Output Specification:

If all the 4 digits of N are the same, print in one line the equation ​​N - N = 0000​​​. Else print each step of calculation in a line until ​​6174​​ comes out as the difference. All the numbers must be printed as 4-digit numbers.

Sample Input 1:

6767

Sample Output 1:

7766 - 6677 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000

经验总结：

这一题，如果可以熟练的使用sscanf与sprintf函数，解决起来就十分简单啦~

AC代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
bool judge(char s[])
{
  char a=s[0];
  for(int i=1;i<4;++i)
    if(s[i]!=a)
      return false;
  return true;
}
bool cmp(int a,int b)
{
  return a>b;
}
int main()
{
  int n,c,d;
  char s[5];
  scanf("%d",&n);
  sprintf(s,"%04d",n);
  if(judge(s))
    printf("%s - %s = 0000",s,s);
  else
    do
    {
      sort(s,s+4,cmp);
      printf("%s - ",s);
      sscanf(s,"%d",&c);
      sort(s,s+4);
      printf("%s = ",s);
      sscanf(s,"%d",&d);
      printf("%04d\n",c-d);
      sprintf(s,"%04d",c-d);
    }while(c-d!=6174);  
  return 0;
}