1069. The Black Hole of Numbers

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

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000

#include<stdio.h>#include<algorithm>using namespace std;int max_value = 0, min_value = 0;int swap(int x){int a[4];a[0] = x / 1000;a[1] = x % 1000 / 100;a[2] = x % 100 / 10;a[3] = x % 10;sort(a, a+4);max_value = a[3]*1000 + a[2]*100 + a[1]*10 + a[0];min_value = a[0]*1000 + a[1]*100 + a[2]*10 + a[3];return max_value - min_value;}void print(int x){int res = swap(x);printf("%04d - %04d = %04d\n", max_value, min_value, res);}int main(){int n;scanf("%d", &n);int pre_res = swap(n);while(1){print(n);n = swap(n);if(pre_res == 0)break;int next_res = swap(n);if(pre_res == next_res)break;pre_res = next_res;}return 0;}