LightOJ - 1010

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

3
8 8
3 7
4 10

Sample Output

Case 1: 32
Case 2: 11
Case 3: 20

题意

根据中国象棋“马”的棋子的跳棋规则，在给出的一个已知的棋盘大小中求出最大可以共存的“马”的数量。

思路

根据上图所示，可以发现“马”如果放在深色格子中，那么必会吃掉浅色格子中的“马”，所以，所有的“马”要有最大的共存量，那么所有的“马”都应该放在深色格子中或者浅色格子中。

AC代码

#include<stdio.h>

int main()
{
	int t;
	scanf("%d",&t);
	int c = 1;
	while(c <= t)
	{
		int m, n, p;
		int sum = 0;
		scanf("%d%d",&m, &n);
		if(m > n)//让 m 必须比 n 小
		{
			p = m;
			m = n;
			n = p;
		}
		if(m % 2 == 0 && m != 2)//m 为偶数的时候（m = 2 的时候是不一样的情况）
		{
			sum += m / 2 * n;
		}
		else if(m > 2)//m为基数的时候
		{
			sum += m / 2 * n + ((n + 1) / 2);
		}
		else if(m == 1)
		{
			sum += n;
		}
		else if(m == 2)
		{
			int q = n % 4;
			if(q == 0)
			{
				sum += n;
			}
			if(q == 1)
			{
				sum += n / 4 * 4 + 2;
			}
			if(q == 2 || q == 3)
			{
				sum += n / 4 * 4 + 4;
			}
		}
		printf("Case %d: %d\n",c, sum);
		c++;
	}
}

//这题的主要是找规律，找到规律后就是一道水题。但值得注意到是里面有一个坑，即m = 2 的时候，他就不能遵循一般规律了，得单独拿出来讨论。