PAT(A)1069 The Black Hole of Numbers (20分)

Sample Input

6767

Sample Output

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

思路：
输入一个四位数，然后依次用最大的排序方式减最小的输出方式，如果结果时6174或者0000则输出。
我直接字符串输入，然后用高精度加法的方式去写，每次重排并赋值一下。
代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

bool cmp1(int a, int b)
{
    return a > b;
}

bool cmp2(int a, int b)
{
    return a < b;
}

int x[4];
int a[4], b[4], aa[4], bb[4];
int r[4];

int main()
{
    char n[4];

    scanf("%s", &n);

    for (int i = 4 - strlen(n); i <= 3; ++i)
        x[i] = n[i - 4 + strlen(n)] - '0';

    while (1)
    {
        sort(x, x + 4, cmp1);
        for (int i = 0; i <= 3; ++i)
            a[i] = x[i], aa[i] = x[i];


        sort(x, x + 4, cmp2);
        for (int i = 0; i <= 3; ++i)
            b[i] = x[i];

        for (int i = 3; i >= 0; --i)
        {
            if (a[i] - b[i] < 0)
            {
                a[i - 1]--;
                r[i] = a[i] - b[i] + 10;
            }
            else
                r[i] = a[i] - b[i];
        }

        for (int i = 0; i <= 3; ++i)
            printf("%d", aa[i]);
        printf(" - ");
        for (int i = 0; i <= 3; ++i)
            printf("%d", b[i]);
        printf(" = ");
        for (int i = 0; i <= 3; ++i)
            printf("%d", r[i]);
        if (r[0] == 6 && r[1] == 1 && r[2] == 7 && r[3] == 4)
            break;
        else if (r[0] == 0 && r[1] == 0 && r[2] == 0 && r[3] == 0)
            break;
        else
        {
            for (int i = 0; i <= 3; ++i)
                x[i] = r[i];
            printf("\n");
        }
    }
    // getchar(); getchar();

    return 0;
}