APT甲级——A1069 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

Output Specification:

If all the 4 digits of

Sample Input 1:

Sample Output 1:

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

Sample Input 2:

Sample Output 2:

2222 - 2222 = 0000

注意点：
（1）题目并未保证输入的n一定是在[1000,10000)之间的数，所以对于输入为53这样的数，第一行输出应为5300 - 0035 = 5265

（2）当输入的数为0或者6174时要特殊判断，输入0输出应为0000 - 0000 = 0000；输入6174输出应为7641 - 1467= 6174，而不能没有输出

（3）输出的数必须为4位，不够4位要在高位补0

 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 using namespace std;
 5 int main()
 6 {
 7     int A, B, preAns = -1, Ans = 0, flag = 0;
 8     string Na, Nb;
 9     cin >> Ans;    
10     while (preAns != Ans)
11     {
12         preAns = Ans;
13         Na = to_string(Ans);
14         while (Na.length() < 4)
15             Na += "0";
16         sort(Na.begin(), Na.end(), [](char a, char b) {return a > b; });
17         A = stoi(Na.c_str());
18         sort(Na.begin(), Na.end(), [](char a, char b) {return a < b; });
19         B = stoi(Na.c_str());
20         Ans = A - B;
21         if (flag == 1 && preAns == Ans)
22             break;
23         printf("%04d - %04d = %04d
", A, B, Ans);
24         flag = 1;//怎么都得有一行输出
25     }
26     return 0;
27 }