读书-算法《程序设计导引及在线实践》-简单计算题2：棋盘上的距离

#include <stdio.h>
#include <math.h>

void main()
{
    int ncase, i;
    scanf("%d",&ncase);

    for(i = 0; i < ncase; i++)
    {
        char begin[3], end[3];/*这里定义为3,考虑到了还要存入字符串结束符‘\0‘*/
        int x, y;
        scanf("%s %s",begin, end);
        
        x = abs(begin[0] - end[0]);
        y = abs(begin[1] - end[1]);

        if (0 == x && 0 == y)
        {
            printf("0 0 0 0\n");
        }
        else
        {
            /*steps of king*/
            if (x < y)
            {
                printf("%d ", y);
            }
            else
            {
                printf("%d ", x);
            }

            /*steps of queen*/
            if (x == y || 0 == x || 0 == y)
            {
                printf("1 ");
            }
            else
            {
                printf("2 ");
            }

            /*steps of rook*/
            if (0 == x || 0 == y)
            {
                printf("1 ");
            }
            else
            {
                printf("2 ");
            }
            
            /*steps of bishop*/
            if (abs(x-y)%2 != 0)
            {
                printf("Inf\n");
            }
            else if(x == y)
            {
                printf("1\n");
            }
            else 
            {
                printf("2\n");
            }
        }
    }
}