# 【PAT甲级】The Black Hole of Numbers

### 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;
}```

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