1069. The Black Hole of Numbers (20)

题目链接：http://www.patest.cn/contests/pat-a-practise/1069

题目：

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

分析：

提取各个位数的数字

AC代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
int main(){
 freopen("F://Temp/input.txt", "r", stdin);
 int n;
 scanf("%d", &n);
 while (true){
  int num[4];
  num[0] = n % 10;
  n /= 10;
  num[1] = n % 10;
  n /= 10;
  num[2] = n % 10;
  num[3] = n / 10;
  sort(num, num + 4);
  int max = num[3] * 1000 + num[2] * 100 + num[1] * 10 + num[0];
  int min = num[0] * 1000 + num[1] * 100 + num[2] * 10 + num[3];
  int cha = max - min;
  printf("%04d - %04d = %04d\n", max, min, cha);
  if (cha == 0 || cha == 6174)
   break;
  n = cha;
 }
 return 0;
}

截图：

——Apie陈小旭