### Problem Description：

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 (0,10​4​​).

### 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.

6767

### Sample Output 1:

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

2222

### Sample Output 2:

2222 – 2222 = 0000

### AC代码：

``````#include <bits/stdc++.h>
using namespace std;
int arr2num(int a[])   //把数组里的数按照下标次序来组成一个新的数字
{
int n = 0;
for(int i = 0; i < 4; i++)
{
n = n*10 + a[i];
}
return n;
}
int main()
{
int n;
cin >> n;
int a[4];
do{
for(int i = 0; i < 4; i++)   //把各位数存入数组里面
{
a[i] = n%10;
n /= 10;
}
sort(a,a+4);    //升序排列
int x = arr2num(a);     //得到由数组a中元素组成的最大数字
sort(a,a+4,greater<int>());   //降序排列
int y = arr2num(a);     //得到由数组a中元素组成的最小数字
n = y-x;    //作差得到一个新的四位数
printf("%04d - %04d = %04d\n",y,x,n);
}while(n!=0 && n!=6174);
return 0;
}``````