UVa - 696 ( How Many Knights Solution ) C Code

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Tip : A same problem is present at LightOJ . You can find it there . To find the solution you have to consider different cases. For [ M = 1 or N = 1 ], [ M = 2 or N = 2 ], and for All others . Find out the Formula . If you Can't find it yet then Google it . Even if you won't find it the see ,

For[ M = 1 or N = 1] Ans is M*N. 

For [ M = 2 or N = 2 ] 

If M = 2 then k=N. If not k=M. Then  C = (k%4==0)?k/4:(k/4)+1 and 

D = ((k-1)%4==0)?(k-1)/4:((k-1)/4)+1.   Ans is 2*(C+D). 

For Others Ans is ((m*n)%2==0)?(m*n)/2:((m*n)/2)+1.

Code :

#include<stdio.h>

#include<math.h>

int main()

{

    int m, n, t, j=0, ans, k, c, d;

    while(scanf("%d %d", &m, &n)==2)

    {

        if(m==0 && n==0) break;

        if(m==1 || n==1)

            ans = m*n;

        else if(m==2 || n==2)

        {

            if(m==2)

                k=n;

            else

                k=m;

            c = (k%4==0)?k/4:(k/4)+1;

            d = ((k-1)%4==0)?(k-1)/4:((k-1)/4)+1;

            ans= 2*(c+d);

        }

        else

            ans = ((m*n)%2==0)?(m*n)/2:((m*n)/2)+1;

        printf("%d knights may be placed on a %d row %d column board.\n", ans, m, n);

    }

    return 0;

}

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