pat乙级1019

求四位数的每一位：

1 int a[4];
for (int i = 0; i < 4; i++)
{
   a[i] = n % 10;
   n /= 10;
}

输出4位整数，少于4位，高位补0：

printf("%04d", n);

6174是一个测试点，所以用do....while语句。

#include <iostream>
#include <string>

using namespace std;

int up, down;
void getNumber(int n)
{
    int a[4];
    for (int i = 0; i < 4; i++)
    {
        a[i] = n % 10;
        n /= 10;
    }
    for (int i = 0; i < 4; i++)
    {
        for (int j = i; j > 0 && a[j] > a[j - 1]; j--)
        {
            int temp = a[j];
            a[j] = a[j - 1];
            a[j - 1] = temp;
        }
    }
    
    down = a[0] * 1000 + a[1] * 100 + a[2] * 10 + a[3];
    up = a[3] * 1000 + a[2] * 100 + a[1] * 10 + a[0];
}

int main()
{
    
    int n;
    cin >> n;

    do {
        getNumber(n);
        n = down - up;
        printf("%04d - %04d = %04d
", down, up, n);
    } while (n != 6174 && n != 0000);

    return 0;
}