1069 The Black Hole of Numbers (20)

最新推荐文章于 2024-05-24 08:20:13 发布

原创最新推荐文章于 2024-05-24 08:20:13 发布 · 240 阅读

0 ·

CC 4.0 BY-SA版权

PAT (Advanced Level) Practice 专栏收录该内容

84 篇文章

订阅专栏

本文介绍了一个有趣的数学现象——通过特定操作使得任意一个不含完全相同数字的四位数最终都会到达一个固定的数值6174，即Kaprekar常数。并提供了计算过程示例及实现这一过程的C++代码。

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

#include<cstdio>
#include<algorithm>
using namespace std;
bool cmp(int a, int b){
	return a>b;
}
void to_array(int n, int num[]){
	for(int i=0; i<4; i++){
		num[i]=n%10;
		n/=10;
	}
}
int to_number(int num[]){
	int sum=0;
	for(int i=0; i<4; i++){
		sum=sum*10+num[i];
	}
	return sum;
}
int main(){
	int n, MIN, MAX;
	scanf("%d", &n);
	int num[5];
	while(1){
		to_array(n, num);
		sort(num, num+4);
		MIN=to_number(num);
		sort(num, num+4, cmp);
		MAX=to_number(num);
		n=MAX-MIN;
		printf("%04d - %04d = %04d\n", MAX, MIN, n);
		if(n==0 || n==6174)break;
	}
	return 0;
}