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

分析：

簡(jiǎn)單題
主要考的是閱讀理解（笑），輸入一個(gè)四位數(shù)字，讓它的數(shù)字從大到小排列-從小到大排，知道得到一個(gè)神奇的數(shù)6174為止
主要是兩個(gè)函數(shù)to_array和to_num，排序用sort就行了。

code:

#include<cstdio> 
#include<algorithm>
using namespace std;
void to_array(int n,int a[]){
    for(int i=0;i<4;i++){
        a[i]=n%10;
        n=n/10;
    }
}

int to_number(int a[]){
    int sum=0;
    for(int i=0;i<4;i++){
        sum=sum*10+a[i];
    }
    return sum;
}

bool cmp(int a,int b){
    return a>b;
}
int main(){
    int n=0;
    int max,min=0;
    int a[10];
    scanf("%d",&n);
    while(1){
        to_array(n,a);
        sort(a,a+4);
        min=to_number(a);
        sort(a,a+4,cmp);
        max=to_number(a);
        n=max-min;
        printf("%04d - %04d = %04d\n",max,min,n);
        if(n==6174||n==0)
            break;
    }
    system("pause");
    return 0;
}