PAT (Advanced Level) 1069 The Black Hole of Numbers （20 分）

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最新推荐文章于 2022-02-09 20:20:02 发布

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本文介绍了一种数学现象，即通过特定操作将任意四位数最终导向固定数值6174的过程，这一过程被称为进入数字的黑洞，而6174则被称为Kaprekar常数。文章详细解释了如何从任意四位数开始，通过不断重复特定步骤，最终达到6174这一神秘数值的方法。

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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:

Sample Output 1:

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

Sample Input 2:

Sample Output 2:

2222 - 2222 = 0000

Code:

#include <iostream>
#include <algorithm>
#pragma warning(disable:4996)

using namespace std;

void n2a(int arr[], int n)
{
	arr[0] = n / 1000;
	arr[1] = n % 1000 / 100;
	arr[2] = n % 100 / 10;
	arr[3] = n % 10;
	sort(arr, arr + 4);
}

int a2n(int* a, bool flag)
{
	return (flag == 0 ? a[0] * 1000 + a[1] * 100 + a[2] * 10 + a[3] : a[3] * 1000 + a[2] * 100 + a[1] * 10 + a[0]);
}

int main()
{
	int n;
	scanf("%d", &n);
	int arr[4];
	n2a(arr, n);
	if (arr[0] == arr[1] && arr[1] == arr[2] && arr[2] == arr[3])
	{
		printf("%04d - %04d = 0000\n", n, n);
		return 0;
	}
	int temp = n, n1, n2;
	do							// 用do……while防止特例6174
	{
		n2a(arr, temp);
		n1 = a2n(arr, 1);
		n2 = a2n(arr, 0);
		temp = n1 - n2;
		printf("%04d - %04d = %04d\n", n1, n2, temp);
	} while (temp != 6174);
	return 0;
}