pat 1069 The Black Hole of Numbers（20 分）

1069 The Black Hole of Numbers（20 分）

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 (0,10​4​​).

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>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <queue>
#include <set>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int MAX = 1e5 + 10;

int n, A[10], ans_min, ans_max, cnt;

int my_pow(int x, int n)
{
    int ans = 1;
    while (n)
    {
        if (n & 1) ans *= x;
        x *= x;
        n >>= 1;
    }
    return ans;
}

int main()
{
//    freopen("Date1.txt", "r", stdin);
    scanf("%d", &n);
    if (n == 6174)
    {
        printf("7641 - 1467 = 6174\n");
        return 0;
    }
    while (1)
    {
        if (n == 6174) return 0;
        ans_max = ans_min = cnt = 0;
        while (n)
        {
            A[cnt ++] = n % 10;
            n /= 10;
        }
        sort(A, A + 4, less<int>());
        for (int i = 0; i < 4; ++ i)
            ans_max += A[i] * my_pow(10, i);
        sort(A, A + 4, greater<int>());
        for (int i = 0; i < 4; ++ i)
            ans_min += A[i] * my_pow(10, i);

        if (ans_min == ans_max)
        {
            printf("%04d - %04d = 0000\n", ans_max, ans_min);
            return 0;
        }
        n = ans_max - ans_min;
        printf("%04d - %04d = %04d\n", ans_max, ans_min, n);
    }
    return 0;
}