1069 The Black Hole of Numbers

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

#include 
#include 
#include 
using namespace std;
int a[4], b[4];
 
int main() {
	int n, t, x = 0, y = 0;
	cin >> n;
	do {
		x = 0, y = 0;
		a[0] = n % 10;
		a[1] = n / 10 % 10;
		a[2] = n / 100 % 10;
		a[3] = n / 1000;
		t = a[0];
		if (a[0] == t && a[1] == t && a[2] == t && a[3] == t) {
			cout << n << " - " << n << " = 0000";
			return 0;
		}
		for (int i = 0; i < 4; i++) {
			b[i] = a[i];
		}
		sort(a, a + 4);
		sort(b, b + 4, greater<int>());
		for (int i = 0; i < 4; i++) {
			x = x * 10 + a[i];
			y = y * 10 + b[i];
		}
		cout << b[0] << b[1] << b[2] << b[3] << " - " << a[0] << a[1] << a[2] << a[3] << " = " << setfill('0') << setw(
		         4) << y - x << endl;
		n = y - x;
	} while (n != 6174);
	return 0;
}