PAT 1069. The Black Hole of Numbers (20)

http://pat.zju.edu.cn/contests/pat-a-practise/1069

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, 10000).

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 <cstdio>
#include <algorithm>
using namespace std;
int x1, x2;
void gao(int x){
	int a[4];
	for (int i = 0; i < 4; i++){
		a[i] = x % 10;
		x /= 10;
	}
	sort(a, a + 4);
	for (int i = 0; i < 4; ++i){
		x1 = x1 * 10 + a[3 - i];
		x2 = x2 * 10 + a[i];
	}
}

int main(){
	int n;
	x1 = x2 = 0;
	scanf("%d", &n);
	gao(n);
	while (1){
		printf("%04d - %04d = %04d\n", x1, x2, x1 - x2);
		if (x1 - x2 == 0 || x1 - x2 == 6174)
			break;
		int k = x1 - x2;
		x1 = x2 = 0;
		gao(k);
	}
	return 0;
}