[PAT-甲级]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

解题思路：数学模拟题，详见代码及注释

代码：

#include<stdio.h>#include<algorithm>using namespace std;// 自定义排序：从大到小排序 bool cmp(int a, int b){return a > b;}// 将n中每一位保存到num数组中 void toArray(int n, int num[]){for(int i = 0; i < 4; i ++){num[i] = n % 10;n /= 10;}}// 将数组num转换成数字 int toNumber(int num[]){int sum = 0;for(int i = 0; i < 4; i ++){sum = sum*10 + num[i]; }return sum;}int main(){// MAX保存num数组转换过来的最大值;MIN同理 int n, MIN, MAX;scanf("%d", &n);int num[5];while(1){toArray(n, num);sort(num, num + 4);MIN = toNumber(num);sort(num, num + 4, cmp);MAX = toNumber(num);n = MAX - MIN;printf("%04d - %04d = %04d\n", MAX, MIN, n);// 出口 if(n == 0 || n == 6174)break;}return 0;}