#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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