PAT (Advanced Level) Practise 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 = 10899810 - 0189 = 96219621 - 1269 = 83528532 - 2358 = 6174

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000

简单模拟操作过程即可

#include<cstdio>#include<stack>#include<cstring>#include<algorithm>#include<functional>using namespace std;const int maxn = 1e5 + 10;int n, a[maxn];int get(int x, int kind){for (int i = 0; i < 4; i++) a[i] = x % 10, x /= 10;if (kind) sort(a, a + 4); else sort(a, a + 4, greater<int>());int ans = 0;for (int i = 0; i < 4; i++) ans = ans * 10 + a[i];return ans;}int main(){scanf("%d", &n);do{int x = get(n, 0), y = get(n, 1);n = x - y;printf("%04d - %04d = %04d\n", x, y, n);} while (n != 6174 && n);return 0;}