首页 > 试题广场 >

The Black Hole of Numbers (20)

[编程题]The Black Hole of Numbers (20)
  • 热度指数:2092 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 32M,其他语言64M
  • 算法知识视频讲解
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.

输入描述:
Each input file contains one test case which gives a positive integer N in the range (0, 10000).


输出描述:
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.
示例1

输入

6767

输出

7766 - 6677 = 1089<br/>9810 - 0189 = 9621<br/>9621 - 1269 = 8352<br/>8532 - 2358 = 6174

这道题你会答吗?花几分钟告诉大家答案吧!