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 = 10899810 - 0189 = 96219621 - 1269 = 83528532 - 2358 = 6174

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000

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#include <cstdio>#include <algorithm>using namespace std;int inc, de;void gao(int x){    int a[4];    a[0] = x / 1000;    a[1] = x / 100 % 10;    a[2] = x / 10 % 10;    a[3] = x % 10;    sort(a, a + 4);    inc = a[0] * 1000 + a[1] * 100 + a[2] * 10 + a[3];    de = a[3] * 1000 + a[2] * 100 + a[1] * 10 + a[0];}int main(){    int n, s;    scanf("%d", &n);    if ((n / 1000 == n / 100 % 10) && (n / 1000 == n / 10 % 10) && (n / 1000 == n % 10)){        printf("%04d - %04d = 0000\n", n, n);        return 0;    }    do{        gao(n);        s = de - inc;        printf("%04d - %04d = %04d\n", de, inc, s);        n = s;    } while (s != 6174);    return 0;}