1069. The Black Hole of Numbers (20)

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本文介绍了一种独特的数学现象：对于任意一个四位数（除了所有数字都相同的特殊情况），通过一种特定的操作流程，可以最终得到一个固定不变的数6174，即所谓的“黑洞”数。文章详细解释了这一过程，并通过实例演示如何计算得到6174。

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, 10000).

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

注意本身有可能就是6174

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
int tomax(int a[]){
	int sum=0;
	for(int i=3;i>=0;i--){
		sum=sum*10+a[i];
	}
	return sum;
}
int tomin(int a[]){
	int sum=0;
	for(int i=0;i<=3;i++){
		sum=sum*10+a[i];
	}
	return sum;
}
int main() {
	int n;
	cin>>n;
	do{
		int fz=n;
		int a[10]={0},cnt=0;
		while(fz){
			a[cnt++]=fz%10;
			fz/=10;
		}
		sort(a,a+4);
		int maxn=tomax(a);
		int minn=tomin(a);
		n=maxn-minn;
		printf("%04d - %04d = %04d\n",maxn,minn,n);
	}while(n!=6174&&n!=0);
    return 0;
}