1069. The Black Hole of Numbers (20)

1069. The Black Hole of Numbers (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

#include <iostream>
#include <vector>
#include <stdio.h>
#include <string.h>
#include <map>
#include <algorithm>
using namespace std;
int main(){
int num;
string temp;
char p[5]="";
scanf("%d",&num);
sprintf(p,"%04d",num);
temp=p;
do{
int numMin,numMax;
sort(temp.begin(), temp.end(),[](char a,char b){return a>b;});
sscanf(temp.c_str(),"%d",&numMax);
sort(temp.begin(), temp.end(),[](char a,char b){return a<b;});
sscanf(temp.c_str(), "%d",&numMin);
num = numMax-numMin;
printf("%04d - %04d = %04d\n",numMax,numMin,num);
if(!num)
break;
sprintf(p,"%04d",num);
temp=p;
}while (num!=6174);
return 0;
}