pat1069. The Black Hole of Numbers (20)

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

---

提交代码

 #include<cstdio>
 #include<stack>
 #include<cstring>
 #include<iostream>
 #include<stack>
 #include<set>
 #include<map>
 using namespace std;
 int dight[];
 int main()
 {
     //freopen("D:\\INPUT.txt","r",stdin);
     int n,maxnum,minnum;
     scanf("%d",&n);
     int i,j;
     do{//有可能n一开始就是6174
         for(i=; i>=; i--)
         {
             dight[i]=n%;
             n/=;
         }
         for(i=; i<; i++)//由大到小
         {
             maxnum=i;
             for(j=i+; j<; j++)
             {
                 if(dight[j]>dight[maxnum])
                 {
                     maxnum=j;
                 }
             }
             int t=dight[maxnum];
             dight[maxnum]=dight[i];
             dight[i]=t;
         }
         minnum=;
         for(i=; i>=; i--)
         {
             minnum*=;
             minnum+=dight[i];
         }
         maxnum=;
         for(i=; i<; i++)
         {
             maxnum*=;
             maxnum+=dight[i];
         }
         n=maxnum-minnum;
         printf("%04d - %04d = %04d\n",maxnum,minnum,n);
         if(!n)
             break;
     }while(n!=);
     return ;
 }