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

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本文详细介绍了如何通过一系列数学操作，将任意四位数除所有数字相同的情况外，逐步演算至被称为Kaprekar常数的6174这一神秘数字的过程。通过实例演示了从给定的四位数开始，通过不断重新排列数字并相减，最终达到6174这一固定点的奇妙现象。

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题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805400954585088

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

如果输入6714，则输出7641-1467=6174

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int main(){
    int a[5];
    int n;
    cin>>n;
    int minn,maxx,t;
    do{
    	for(int i=0;i<4;i++){
    		a[i]=n%10;
    		n/=10;
		}
		sort(a,a+4);
		minn=a[0]*1000+a[1]*100+a[2]*10+a[3];
		maxx=a[3]*1000+a[2]*100+a[1]*10+a[0];
		t=maxx-minn;
		printf("%04d - %04d = %04d\n",maxx,minn,t);
		n=t;
	}while(n!=6174&&n!=0);
	return 0;
}