1069. The Black Hole of Numbers (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

 1 #include<cstdio>
 2 #include<stack>
 3 #include<cstring>
 4 #include<iostream>
 5 #include<stack>
 6 #include<set>
 7 #include<map>
 8 using namespace std;
 9 int dight[4];
10 int main()
11 {
12     //freopen("D:\\INPUT.txt","r",stdin);
13     int n,maxnum,minnum;
14     scanf("%d",&n);
15     int i,j;
16     do{//有可能n一开始就是6174
17         for(i=3; i>=0; i--)
18         {
19             dight[i]=n%10;
20             n/=10;
21         }
22         for(i=0; i<4; i++)//由大到小
23         {
24             maxnum=i;
25             for(j=i+1; j<4; j++)
26             {
27                 if(dight[j]>dight[maxnum])
28                 {
29                     maxnum=j;
30                 }
31             }
32             int t=dight[maxnum];
33             dight[maxnum]=dight[i];
34             dight[i]=t;
35         }
36         minnum=0;
37         for(i=3; i>=0; i--)
38         {
39             minnum*=10;
40             minnum+=dight[i];
41         }
42         maxnum=0;
43         for(i=0; i<4; i++)
44         {
45             maxnum*=10;
46             maxnum+=dight[i];
47         }
48         n=maxnum-minnum;
49         printf("%04d - %04d = %04d\n",maxnum,minnum,n);
50         if(!n)
51             break;
52     }while(n!=6174);
53     return 0;
54 }