A1069

1069 The Black Hole of Numbers (20分)

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.

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<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
int main(){
    int n;
    cin>>n;
    while(1){
        int l=0,mul1=0,mul2=0,a[4]={0};
        while(n){
            a[l++]=n%10;
            n/=10;
        }
        sort(a,a+4);
        for(int i=0;i<4;i++){
            mul2=mul2*10+a[i];
            mul1=mul1*10+a[3-i];
        }
        printf("%04d - %04d = %04d\n",mul1,mul2,mul1-mul2);
        n=mul1-mul2;
        if(n==0||n==6174)break;
    }
    return 0;
}