1069 The Black Hole of Numbers (20)

1069 The Black Hole of Numbers (20)（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 (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

注意点：在求每一位数的时候，用来保存数字的数组一定要是4位的，并且赋初值， 否者会因为减法出现的三位数，而导致死循环

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main(){
  int n, i;
  cin>>n;
  while(true){
    int  cnt=0, mmax=0, mmin=0;
    vector<int> v(4,0);
    while(n){
      v[cnt++]=n%10;
      n/=10;
    }
    sort(v.begin(), v.end());
    for(i=0; i<v.size(); i++){
      mmin = mmin*10 + v[i];
      mmax = mmax*10 + v[3-i];
    }
    printf("%04d - %04d = %04d
", mmax, mmin, mmax-mmin);
    n=mmax-mmin;
    if(mmax-mmin==6174||n==0) break;
  }
  return 0;
}