PAT A1069. The Black Hole of Numbers (20)

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 = 10899810 - 0189 = 96219621 - 1269 = 83528532 - 2358 = 6174

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000

输出形式是从来没有接触过的

#include <cstdio>#include <stdlib.h>#include <algorithm>#include <vector>#include <cstring>#define MAX 1001using namespace std;bool cmp(int a,int  b){ return a>b;}void breakn(int m,int m1[5]){for(int i=3;i>=0;i--){m1[i]=m%10;m/=10;}} int togethern(int m1[5]){    int m=0;for(int i=0;i<4;i++){m=m*10+m1[i];}return m;} int  main(){int n,n0[5],n1,n2;scanf("%d",&n);while(1){breakn(n,n0);sort(n0,n0+4,cmp);n1=togethern(n0);sort(n0,n0+4);n2=togethern(n0);n=n1-n2;printf("%04d - %04d = %04d\n",n1,n2,n);if(n==0||n==6174)break;}    system("pause");    return 0;}