pat 1069. The Black Hole of Numbers

原题链接：https://www.patest.cn/contests/pat-a-practise/1069

      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

怎么写都有一个实例错误，求大神帮忙找找吧。。

// pat_The Black Hole of Numbers.cpp : 定义控制台应用程序的入口点。//#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;int a[5], b[5];int main(int argc, char* argv[]){int n, i, j;scanf("%d", &n);int m = n;i = 0;while (n){a[i] = n % 10;n /= 10;i++;}for (i = 0; i < 3; i++){if (a[i] != a[i + 1])break;}if (i >= 3){printf("%04d - %04d = 0000\n", m, m);return 0;}sort(a, a + 4);for (i = 0; i < 4; i++)b[i] = a[4 - i - 1];while (1){int sum1 = 0, sum2 = 0, sum;for (i = 0; i < 4; i++)sum1 = sum1 * 10 + b[i];for (i = 0; i < 4; i++)sum2 = sum2 * 10 + a[i];sum = sum1 - sum2;printf("%04d - %04d = %04d\n", sum1, sum2, sum);if (sum == 6174 || sum == 0)break;i = 0;while (sum){a[i] = sum % 10;i++;sum /= 10;}sort(a, a + 4);for (i = 0; i < 4; i++)b[i] = a[3 - i];}return 0;}