题目描述

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.

输入描述

Each input file contains one test case which gives a positive integer N in the range (0,10​4​​).

输出描述

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.

输入例子1

6767

输出例子1

7766 - 6677 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174

输入例子2

2222

输出例子2

2222 - 2222 = 0000

我的代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
bool cmp(int a,int b){
    return a<b;
}
int main(){
    int num,a[4];
    scanf("%d",&num);
    a[0]=num/1000;
    a[1]=num%1000/100;
    a[2]=num%1000%100/10;
    a[3]=num%1000%100%10;
    if(a[0]==a[1]&&a[0]==a[2]&&a[0]==a[3]){
        printf("%d%d%d%d - %d%d%d%d = 0000",a[0],a[0],a[0],a[0],a[0],a[0],a[0],a[0]);
        return 0;
    }
    int max,min;
    do{
        a[0]=num/1000;
        a[1]=num%1000/100;
        a[2]=num%1000%100/10;
        a[3]=num%1000%100%10;
        sort(a,a+4,cmp);
        min=a[0]*1000+a[1]*100+a[2]*10+a[3];
        max=a[3]*1000+a[2]*100+a[1]*10+a[0];
        num=max-min;
        printf("%d%d%d%d - %d%d%d%d = %04d\n",a[3],a[2],a[1],a[0],a[0],a[1],a[2],a[3],num); 
    }while(num!=6174 && num!=0);
    return 0;
}