1069. The Black Hole of Numbers (20)-PAT甲级真题

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

#include <iostream>#include <algorithm>using namespace std;int cmp1(int a, int b) {    return a > b;}int cmp2(int a, int b) {    return a < b;}int main() {    int N;    cin >> N;    int a[4] = {0};    int t = 3;    while (N != 0 && t != -1) {        a[t--] = N % 10;        N = N / 10;    }    t = 3;    if (a[0] == a[1] && a[1] == a[2] && a[2] == a[3]) {        cout << a[0] << a[0] << a[0] << a[0] << " - " << a[0] << a[0]             << a[0] << a[0] << " = " << "0000";        return 0;    }    int result = 0;    int b[4];    int temp;    while (result != 6174) {        for (int i = 0; i < 4; i++) {            b[i] = a[i];        }        sort(a, a + 4, cmp1);        sort(b, b + 4, cmp2);        result = a[0] * 1000 + a[1] * 100 + a[2] * 10 + a[3] -        b[0] * 1000 - b[1] * 100 - b[2] * 10 - b[3];        cout << a[0] << a[1] << a[2] << a[3] << " - " <<                b[0] << b[1] << b[2] << b[3] << " = ";        temp = result;        for (int i = 0; i < 4; i++) {            a[i] = 0;        }        while (temp != 0 && t != -1) {            a[t--] = temp % 10;            temp = temp / 10;        }        t = 3;        cout << a[0] << a[1] << a[2] << a[3] << endl;    }    return 0;}