1069 The Black Hole of Numbers (20 分)

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 (.

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

#include<bits/stdc++.h>
using namespace std;
const int maxn=1010;
#define  inf  0x3fffffff
int number[4];
bool cmp(int a,int b){
    return a>b;
}
void tran(int n){
    int i=0;
    while(n!=0){
        number[i]=n%10;
        n/=10;
        i++;
    }
}
void init(){
    for(int i=0;i<=4;i++){
        number[i]=0;
    }
}
int main(){
    int m;
    scanf("%d",&m);
    if(m==0){
        printf("0000 - 0000 = 0000
");
        return 0;
    }
    if(m==6174){
        m=7641;
    }
    int n=m;
    tran(n);
    while(true){
        if(n==6174||n==0){
            break;
        }
        init();
        tran(n);
        sort(number,number+4,cmp);
        int num1=0;
        for(int i=0;i<4;i++){
            num1=num1*10+number[i];
        }
        sort(number,number+4);
        int num2=0;
        for(int i=0;i<4;i++){
            num2=num2*10+number[i];
        }
        n=num1-num2;
        printf("%04d - %04d = %04d
",num1,num2,n);
    }
    
    
    return 0;
}
//注意：0，1，11，111，1000，6174