#include <stdio.h>
#include <string.h>
#define min(a,b) (a<b?a:b)
char inp[100];
int main()
{
int test, x, y;
char a;
scanf("%d",&test);
getchar();
while (test--)
{
gets(inp);
sscanf(inp,"%c %d %d",&a,&x,&y);
if (a=='k')
{
printf("%d\n",(x*y+1)/2);
} else if (a=='K')
{
printf("%d\n",((x+1)/2) * ((y+1)/2));
} else if (a=='Q'||a=='r')
{
printf("%d\n",min(x,y));
}
}
return 0;
}
Monday, October 17, 2011
[UVa] 278 - Chess
Analyze all 4 cases and soon you'll equations for all of them. The given advantage is the lower limit of 4. Below that the derived theorems fail.
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