ZCMU--1379: The Black Hole of Numbers(C语言)

Description

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

Each input file contains one test case which gives a positive integer N in the range (0, 10000).

Output

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

6767

2222

Sample Output

7766 - 6677 = 1089

9810 - 0189 = 9621

9621 - 1269 = 8352

8532 - 2358 = 6174

2222 - 2222 = 0000

解析：n化为4位数后，n数字降序 — n数字升序，如果结果是0或者是6174，结束。

注意点：输入n可能小于4位数，如果直接数值型计算需要进行补零，不然会死循环一直输出，导致Output Limit Exceed，不过可以用数组来存一下，排个序，相减的数即x=a[3]*1000+a[2]*100+a[1]*10+a[0];//数字降序
y=a[0]*1000+a[1]*100+a[2]*10+a[3];//数字升序

#include 
int a[4];
int main()
{
	int n,i,j,t,x,y;
	while(~scanf("%d",&n)){
		while(1){
		  for(i=0;i<4;i++) a[i]=n%10,n/=10;//将数字保存到数组
		  for(i=0;i<3;i++){	//冒泡从小到大排序(就4个数，当然sort没问题👍)
			for(j=i+1;j<4;j++){
				if(a[i]>a[j]) t=a[i],a[i]=a[j],a[j]=t;
				}
			}
			x=a[3]*1000+a[2]*100+a[1]*10+a[0];//数字降序
			y=a[0]*1000+a[1]*100+a[2]*10+a[3];//数字升序
			printf("%04d - %04d = %04d\n",x,y,x-y);
			n=x-y;//更新n
			if(n==0||n==6174) break;//满足，退出即可
		}
	}
	return 0;
}