1069. The Black Hole of Numbers (20)

客官°小女子只卖身不卖艺 2022-05-30 00:15 272阅读 0赞

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:**

6767

**Sample Output 1:**

7766 - 6677 = 1089
    9810 - 0189 = 9621
    9621 - 1269 = 8352
    8532 - 2358 = 6174

**Sample Input 2:**

2222

**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;
        }
    }

参考： [点击打开链接][Link 1]

[Link 1]: http://blog.csdn.net/apie_czx/article/details/48243753