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 (0,104).
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 + ; int n, A[], ans_min, ans_max, cnt; int my_pow(int x, int n)
{
int ans = ;
while (n)
{
if (n & ) ans *= x;
x *= x;
n >>= ;
}
return ans;
} int main()
{
// freopen("Date1.txt", "r", stdin);
scanf("%d", &n);
if (n == )
{
printf("7641 - 1467 = 6174\n");
return ;
}
while ()
{
if (n == ) return ;
ans_max = ans_min = cnt = ;
while (n)
{
A[cnt ++] = n % ;
n /= ;
}
sort(A, A + , less<int>());
for (int i = ; i < ; ++ i)
ans_max += A[i] * my_pow(, i);
sort(A, A + , greater<int>());
for (int i = ; i < ; ++ i)
ans_min += A[i] * my_pow(, i); if (ans_min == ans_max)
{
printf("%04d - %04d = 0000\n", ans_max, ans_min);
return ;
}
n = ans_max - ans_min;
printf("%04d - %04d = %04d\n", ans_max, ans_min, n);
}
return ;
}