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过了,帖代码,有注释#include <stdio.h>
#include <string.h>
#include <math.h>
int main()
{
int cases=0;
int n1=0;
char inf[4]={'\0'};
int king=0;//王所需要走的步数
int queen=0;
int car=0;
int elephant=0;
int dx=0,dy=0;
scanf("%d",&cases);
getchar();
for(;n1<cases;n1++)
{
scanf("%c%c %c%c",inf,inf+1,inf+2,inf+3);
getchar();
/*
putchar(inf[0]);
putchar(inf[1]);
putchar(inf[2]);
putchar(inf[3]);
*/
dx=abs(inf[2]-inf[0]);
dy=abs(inf[3]-inf[1]);
if((dx==0)&&(dy==0))
{
printf("0 0 0 0\n");
continue;
}//起始位置跟终止位置相同的情况
//王
king= (dx>dy)?dx:dy;
//王后
if((dx==dy)||(dx==0)||(dy==0))
{
queen=1;
}
else queen=2;
//车
if((dx==0)||(dy==0))
{
car=1;
}
else
{
car=2;
}
printf("%d %d %d ",king,queen,car);
//象
if(abs(dx-dy)%2!=0)
{
printf("Inf\n");
}
else if(dx==dy)
{
printf("1\n");
}
else
{
printf("2\n");
}
}
return 0;
}
注意两点 首先是象能走全部的黑格 白格不能走
然后 在相同的位置的时候 应该输出0 0 0 0
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