LIGHT OJ 1010 Knights in Chessboard

Given an m x n chessboard where you want to place chess knights. You have to find the number of maximum knights that can be placed in the chessboard such that no two knights attack each other.

Those who are not familiar with chess knights, note that a chess knight can attack8 positions in the board as shown in the picture below.

Input

Input starts with an integer T (≤ 41000), denoting the number of test cases.

Each case contains two integers m, n (1 ≤ m, n ≤ 200). Here m and n corresponds to the number of rows and the number of columns of the board respectively.

Output

For each case, print the case number and maximum number of knights that can be placed in the board considering the above restrictions.

Sample Input

3

8 8

3 7

4 10

Sample Output

Case 1: 32

Case 2: 11

Case 3: 20

题意： 有一个n*m大的棋盘，问你最多能放几个入如上图的棋子，如图所示，棋子周围带点的地方不能放别的棋子；

思路：纯粹公式总结，分为以下几种情况：

           当n或m为偶时 并且n,m不为2时（为2要特判） 棋盘填的形式应该为每个两行填满一行， 

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           中间空格代表两行；

            当为奇数时    应该是填   （每个空代表空一格） 

=     =    

    =                                                

=     =                                         

          接下来就是特判了，显而易见，当n,m为1 时 最多肯定是填满所有的格子 即n*m;]

          当n,m为2时   填的方式变了，即填田字格,并且每个田字格周围应空一个田字格，并且当剩余2时应该加上2，剩余4,6时应加上4，画画图就能知道 ；

        下面附上代码：

         

#include<cstdio>
int main()
{
	int t,m,n,k=0,s;
	scanf("%d",&t);
	while(t--){
		scanf("%d %d",&m,&n);
		if(m==1||n==1) printf("Case %d: %d\n",++k,m*n);
		else if(m==2||n==2) {
			s=(n*m)/8*4;
			if(n*m%8==2) printf("Case %d: %d\n",++k,s+2);
			else if((n*m)%8==4||(n*m)%8==6) printf("Case %d: %d\n",++k,s+4);
			else printf("Case %d: %d\n",++k,s); 
		}
		//else if(m%2==0) printf("Case %d: %d\n",++k,m*n/2);
		else printf("Case %d: %d\n",++k,(m*n+1)/2);
	}
	return 0;
}