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 (.
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
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 }