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[4];
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=3; i>=0; i--)
        {
            dight[i]=n%10;
            n/=10;
        }
        for(i=0; i<4; i++)//由大到小
        {
            maxnum=i;
            for(j=i+1; j<4; j++)
            {
                if(dight[j]>dight[maxnum])
                {
                    maxnum=j;
                }
            }
            int t=dight[maxnum];
            dight[maxnum]=dight[i];
            dight[i]=t;
        }
        minnum=0;
        for(i=3; i>=0; i--)
        {
            minnum*=10;
            minnum+=dight[i];
        }
        maxnum=0;
        for(i=0; i<4; i++)
        {
            maxnum*=10;
            maxnum+=dight[i];
        }
        n=maxnum-minnum;
        printf("%04d - %04d = %04d\n",maxnum,minnum,n);
        if(!n)
            break;
    }while(n!=6174);
    return 0;
}