1019 digital black hole

Given any number that is not exactly the same 4 Positive integer , If we put 4 Numbers are sorted in non incremental order , And then in non decreasing order , Then use the 1 A number minus 2 A digital , Will get a new number . Keep doing this again and again , We will soon stop having “ Digital black hole ” It's called.  6174, This magic number is also called Kaprekar constant .

for example , We from 6767 Start , Will get

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

Any 4 Positive integer , Please write a program to demonstrate the process of reaching the black hole .

Input format ：

Type to give a  (0,104)  A positive integer in an interval  N.

Output format ：

If  N  Of 4 The digits are all equal , Then output... In one line  N - N = 0000; Otherwise, each step of the calculation will be output in one line , until  6174  Appear as a difference , See the example for the output format . Pay attention to each number press  4  Bit format output .

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

The code is as follows ：

#include<stdio.h>
int shen(int ar[5])
{
	int i, j, k;
	for (i = 0; i < 4 - 1; i++)
	{
		for (j = 0; j < 4 - i - 1; j++)
		{
			if (ar[j] < ar[j + 1])
			{
				k = ar[j];
				ar[j] = ar[j + 1];
				ar[j + 1] = k;
			}// Descending order 
		}
	}
	int zh = 0;
	for (i = 0; i < 4; i++)
		zh = zh * 10 + ar[i];
	return zh;
}
int jian(int ar[5])
{
	int i, j, k;
	for (i = 0; i < 4 - 1; i++)
	{
		for (j = 0; j < 4 - i - 1; j++)
		{
			if (ar[j] > ar[j + 1])
			{
				k = ar[j];
				ar[j] = ar[j + 1];
				ar[j + 1] = k;
			}
		}
	}
	int zh = 0;
	for (i = 0; i < 4; i++)
		zh = zh * 10 + ar[i];
	return zh;
}
int fack(int ar[5])
{
	int i;
	for (i = 0; i < 4 - 1; i++)
		if (ar[i] != ar[i + 1])
			return 0;
	return 1;
}
int pan(int n)
{
	if (n >= 1000)
		return 0;
	else if (n >= 100 && n < 1000)
		return 1;
	else if (n > 10 && n < 100)
		return 2;
	else
		return 3;
}
void print(int shi,int ji, int f2)
{
	printf("%d - ", shi);
	switch (f2)
	{
	case 0:printf("%d", ji); break;
	case 1:printf("0%d", ji); break;
	case 2:printf("00%d", ji); break;
	case 3:printf("000%d", ji); break;
	}
	printf(" = ");

	int f = pan(shi - ji);
	switch (f)
	{
	case 0:printf("%d", shi-ji); break;
	case 1:printf("0%d", shi-ji); break;
	case 2:printf("00%d", shi-ji); break;
	case 3:printf("000%d", shi-ji); break;
	}
	printf("\n");
}
int main()
{
	int a, i, j;
	int sh, ji,f2;
	int ar[5] = { 0 };
	scanf("%d", &a);
	//zh = a;
	for (i = 0; i < 4; i++,a/=10)
		ar[i] = a % 10;
	sh=shen(ar);
	ji = jian(ar);
	j = fack(ar);
//	printf("%d - %d = %d\n", sh - ji);
	if (j == 1)
		//printf("%d - %d = %d\n", sh,ji,sh - ji);
	{
		f2 = pan(ji);
		print(sh, ji, f2);
	}
	else
	{
		while (sh - ji != 6174 && j != 1)
		{
		//	int f1 = pan(sh);
			 f2 = pan(ji);
			//printf("%d - %d = %d\n",sh,ji, sh - ji);
			print(sh, ji, f2);
			a = sh - ji;
			for (i = 0; i < 4; i++, a /= 10)
				ar[i] = a % 10;
			sh = shen(ar);
			ji = jian(ar);
			j = fack(ar);
		}
		//printf("%d - %d = %d\n", sh, ji, sh - ji);
		//int f3 = pan(sh - ji);
		f2 = pan(ji);
		print(sh, ji, f2);
	}
}