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, 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 = 10899810 - 0189 = 96219621 - 1269 = 83528532 - 2358 = 6174

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <deque>
#include <queue>
#include <cstring>
#include <vector>
#include <string>
#include <iomanip>
#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <cmath>
#include <algorithm>
using namespace std;
#define max1 10010
#define RH 1
#define LH -1
#define EH 0
int main() {
int n,max2,min2;
cin>>n;
do
{
int a[4]={0};
a[0]=n/1000;a[1]=n/100%10;a[2]=n/10%10;a[3]=n%10;
if(a[0]==a[1]&&a[2]==a[1]&&a[2]==a[3])
{
printf("%04d - %04d = 0000",n,n);
break;
}
sort(a,a+4);
max2=a[3]*1000+a[2]*100+a[1]*10+a[0];
min2=a[0]*1000+a[1]*100+a[2]*10+a[3];
n=max2-min2;
printf("%04d - %04d = %04d\n",max2,min2,n);
}
while(n>0&&n!=6174&&n<10000);
return 0;
}