lightoj 1010 - Knights in Chessboard (找规律思维)

1010 - Knights in Chessboard

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Time Limit: 1 second(s)

Memory Limit: 32 MB

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 attack 8 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	Output for Sample Input
3 8 8 3 7 4 10	Case 1: 32 Case 2: 11 Case 3: 20

 

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PROBLEM SETTER: JANE ALAM JAN

图中的八个方向会攻击，是日字形;

全在黑格方或者全在白格方;

特判：

只有一行,全放;

有两行，每相隔一个田字放四个棋子。

#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
	int t,n,m;
	int k=1;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		printf("Case %d: ",k++);
		if(n>m)
		{
			int temp=n;n=m;m=temp;
		}
		if(n==1)
			printf("%d\n",m);
		else if(n==2)
		{
			if(m%4==3)
				printf("%d\n",(m/4)*4+4);
			else
				printf("%d\n",(m/4)*4+(m%4)*2);	
		}
		else
			printf("%d\n",(n*m+1)/2);
	}
	return 0;
 }