题目

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:

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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:

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6767

Sample Output 1:

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7766 - 6677 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174

Sample Input 2:

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2222

Sample Output 2:

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2222 - 2222 = 0000

题解

思路

  • 题目不是很难，主要是字符串的处理不要忘了0
  • 就是当我们计算出两数只差时，要对不足四位数的前面补足0,再参与下一轮的排序并做差。
  • 这点使用Python很方便
  • c = "0" * (4 - len(str(a - b))) + str(a - b)即可
  • 同时a和b的更新，是对c进行排序。Python排序后会返回一个列表，要拼接转为字符串。并且做差之前要再转为整数。
  • b = int("".join(sorted(c))) 即可

数据结构

  • a 是被减数
  • b 是减数
  • c 是差

算法

  • 减法

代码

  • 由于使用Python能AC，因此只放了Python的代码。
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temp = input()
c = "0" * (4 - len(temp)) + temp
a, b = int("".join(sorted(c, reverse=True))), int("".join(sorted(c)))
while True:
    a, b = int("".join(sorted(c, reverse=True))), int("".join(sorted(c)))
    c = "0" * (4 - len(str(a - b))) + str(a - b)
    print("%04d - %04d = %s" % (a, b, c))
    if c == "6174" or c == "0000":
        break