# UVA 696 – How Many Knights

This is extremely similar to UVA 278 – Chess but the tricky case is when min = 2. If you write it out on paper you will figure out the optimal placing for the knights.

```import java.io.PrintWriter;
import java.util.Scanner;

/**
*
* @author Sanchit M. Bhatnagar
* @see http://uhunt.felix-halim.net/id/74004
*
*/
public class P696 {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);

while (sc.hasNext()) {
int R = sc.nextInt();
int C = sc.nextInt();
if (R == 0 && C == 0) {
break;
}
int max = Math.max(R, C);
int min = Math.min(R, C);
if (min == 1) {
out.print(max);
} else if (min == 2) {
if (max < 3) {
out.print(min * max);
} else if (max == 3) {
out.print(4);
} else {
int tmp = max / 4;
int rem = max % 4;
out.print(tmp * min * 2 + Math.min(rem, 2) * 2);
}
} else {
int a = max / 2;
int b = (max - a);
int c = min / 2;
int d = (min - c);
out.print(Math.max(a * c + b * d, a * d + b * c));
}
out.println(" knights may be placed on a " + R + " row " + C + " column board.");
}

out.close();
sc.close();
}
}
```

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