pat 1069 The Black Hole of Numbers（20 分）

Posted 2021-01-04 getcharzp

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

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <string>
 6 #include <map>
 7 #include <stack>
 8 #include <vector>
 9 #include <queue>
10 #include <set>
11 #define LL long long
12 #define INF 0x3f3f3f3f
13 using namespace std;
14 const int MAX = 1e5 + 10;
15 
16 int n, A[10], ans_min, ans_max, cnt;
17 
18 int my_pow(int x, int n)
19 {
20     int ans = 1;
21     while (n)
22     {
23         if (n & 1) ans *= x;
24         x *= x;
25         n >>= 1;
26     }
27     return ans;
28 }
29 
30 int main()
31 {
32 //    freopen("Date1.txt", "r", stdin);
33     scanf("%d", &n);
34     if (n == 6174)
35     {
36         printf("7641 - 1467 = 6174
");
37         return 0;
38     }
39     while (1)
40     {
41         if (n == 6174) return 0;
42         ans_max = ans_min = cnt = 0;
43         while (n)
44         {
45             A[cnt ++] = n % 10;
46             n /= 10;
47         }
48         sort(A, A + 4, less<int>());
49         for (int i = 0; i < 4; ++ i)
50             ans_max += A[i] * my_pow(10, i);
51         sort(A, A + 4, greater<int>());
52         for (int i = 0; i < 4; ++ i)
53             ans_min += A[i] * my_pow(10, i);
54 
55         if (ans_min == ans_max)
56         {
57             printf("%04d - %04d = 0000
", ans_max, ans_min);
58             return 0;
59         }
60         n = ans_max - ans_min;
61         printf("%04d - %04d = %04d
", ans_max, ans_min, n);
62     }
63     return 0;
64 }