1069 The Black Hole of Numbers (20point(s)) - C語言 PAT 甲級

1069 The Black Hole of Numbers (20point(s))

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 (0,10​4​​).

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:

6767

Sample Output:

7766 - 6677 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174

Sample Input:

2222

Sample Output:

2222 - 2222 = 0000

題目大意：

1019 數字黑洞 (20point(s))

設計思路：

1019 數字黑洞 (C語言)

• 爲四位數排序，然後輸出

編譯器：C (gcc)

#include <stdio.h>
#include <stdlib.h>

int sortnumber(int *n, int *m);
int compnumber(const void *a, const void *b);

int main()
{
    int n = 0, m = 0;
    scanf("%d", &n);
    do{
        sortnumber(&n, &m);
        printf("%04d - %04d = %04d\n", n, m, n - m);
        n = n - m;
    }while(n != 0 && n != 6174);
    return 0;
}

int sortnumber(int *n, int *m)
{
    int number[4] = {*n/1000, *n%1000/100, *n%100/10, *n%10};
    qsort(number, 4, sizeof(int), compnumber);
    *n = number[0]*1000 + number[1]*100 + number[2]*10 + number[3];
    *m = *n/1000 + *n%1000/100 *10 + *n%100/10 *100 + *n%10 *1000;
    return 0;
}

int compnumber(const void *a, const void *b)
{
    return *(int *)b - *(int *)a;
}