1069 The Black Hole of Numbers (20)

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  • 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
     1 #include<iostream>
 2 #include<vector>
 3 #include<algorithm>
 4 using namespace std;
 5 int main(){
 6   int n, i;
 7   cin>>n;
 8   while(true){
 9     int  cnt=0, mmax=0, mmin=0;
10     vector<int> v(4,0);
11     while(n){
12       v[cnt++]=n%10;
13       n/=10;
14     }
15     sort(v.begin(), v.end());
16     for(i=0; i<v.size(); i++){
17       mmin = mmin*10 + v[i];
18       mmax = mmax*10 + v[3-i];
19     }
20     printf("%04d - %04d = %04d
", mmax, mmin, mmax-mmin);
21     n=mmax-mmin;
22     if(mmax-mmin==6174||n==0) break;
23   }
24   return 0;
25 }

     



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