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最新推荐文章于 2021-03-24 08:55:47 发布

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本文探讨了一个经典的棋盘问题——如何在m*n的棋盘上放置最多的马，使得任意两个马不会互相攻击。通过分析发现当m和n均大于2时，答案为棋盘上黑色格子和白色格子的最大数量；当m或n为1或2时，则需采用特定的放置策略来获得最大数量。

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

8 8

3 7

4 10

Sample Output

Case 1: 32

Case 2: 11

Case 3: 20

题目大意：求m*n的棋盘中最多放几个马

思路：T<=41000，搜索肯定超时，只能找规律。

m，n都大于2时ans=max（黑块数，白块数）

其中一个等于1或2时要特殊考虑，是2时要遵循四个一块放，隔四个的规律才能放最多。模拟出这个过程即可

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
    int t,i,m,n,ans;
    cin>>t;
    for(i=1; i<=t; i++)
    {
        scanf("%d%d",&m,&n);
        if(m>n) swap(m,n);
        if(m==1) ans=n;
        else if(m==2)
        {
            int x=(m*n)%8;
            if(x>4) x=4;
            int y=(m*n)/8;
            ans=x+y*4;
        }
        else
        {
            if((m*n)&1) ans=(m*n+1)/2;
            else ans=(m*n)/2;
        }
        printf("Case %d: %d\n",i,ans);
    }
    return 0;
}