pat 1069 The Black Hole of Numbers（20 分）

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 <algorithm>
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <map>
 #include <stack>
 #include <vector>
 #include <queue>
 #include <set>
 #define LL long long
 #define INF 0x3f3f3f3f
 using namespace std;
 const int MAX = 1e5 + ;

 int n, A[], ans_min, ans_max, cnt;

 int my_pow(int x, int n)
 {
     int ans = ;
     while (n)
     {
         if (n & ) ans *= x;
         x *= x;
         n >>= ;
     }
     return ans;
 }

 int main()
 {
 //    freopen("Date1.txt", "r", stdin);
     scanf("%d", &n);
     if (n == )
     {
         printf("7641 - 1467 = 6174\n");
         return ;
     }
     while ()
     {
         if (n == ) return ;
         ans_max = ans_min = cnt = ;
         while (n)
         {
             A[cnt ++] = n % ;
             n /= ;
         }
         sort(A, A + , less<int>());
         for (int i = ; i < ; ++ i)
             ans_max += A[i] * my_pow(, i);
         sort(A, A + , greater<int>());
         for (int i = ; i < ; ++ i)
             ans_min += A[i] * my_pow(, i);

         if (ans_min == ans_max)
         {
             printf("%04d - %04d = 0000\n", ans_max, ans_min);
             return ;
         }
         n = ans_max - ans_min;
         printf("%04d - %04d = %04d\n", ans_max, ans_min, n);
     }
     return ;
 }