# PAT(A) 1069. The Black Hole of Numbers (20)

### 1069. The Black Hole of Numbers (20)

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, 10000).

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

### 代码

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 #include "stdio.h" #include #include using namespace std; int big(int num){ vector arr; arr.push_back(num / 1000); arr.push_back(num / 100 % 10); arr.push_back(num / 10 % 10); arr.push_back(num % 10); sort(arr.begin(),arr.end()); return arr * 1000 + arr * 100 + arr * 10 + arr; } int small(int num){ vector arr; arr.push_back(num / 1000); arr.push_back(num / 100 % 10); arr.push_back(num / 10 % 10); arr.push_back(num % 10); sort(arr.begin(),arr.end()); return arr * 1000 + arr * 100 + arr * 10 + arr; } int main(){ int num = 0; scanf("%d",&num); int b = big(num); int s = small(num); int newnum = b - s; if (num == newnum) printf("%04d - %04d = %04d\n",b,s,newnum); while(num != newnum){ printf("%04d - %04d = %04d\n",b,s,newnum); num = newnum; b = big(num); s = small(num); newnum = b - s; } //system("pause"); }