1069. The Black Hole of Numbers (20)-蒲公英云

1069. The Black Hole of Numbers (20)

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

题目大意：

代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
int main()
{
    int i,j,n,m,k,t,num[5];
    scanf("%d",&n);
    while(true)
    {
       num[0]=n%10;
       n/=10;
       num[1]=n%10;
       n/=10;
       num[2]=n%10;
       n/=10;
       num[3]=n%10;
       sort(num,num+4);
       int Max=num[0]+num[1]*10+num[2]*100+num[3]*1000;
       int Min=num[0]*1000+num[1]*100+num[2]*10+num[3];
       n=Max-Min;
       printf("%04d - %04d = %04d\n",Max,Min,n);
       if(n==0||n==6174)
        break;
    }
}