Light oj 1010 - Knights in Chessboard

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探讨在给定尺寸的棋盘上放置骑士的最大数量，确保它们之间不会相互攻击。

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 placechess knights. You have to find the number of maximum knights that can beplaced in the chessboard such that no two knights attack each other.

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

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 numberof columns of the board respectively.

Output

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

Sample Input

Output for Sample Input

8 8

3 7

4 10

Case 1: 32

Case 2: 11

Case 3: 20

题目大意：在N*M的棋盘上走"日"字，问最多能放多少批马，无冲突

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int n,m;
int main()
{
    int i,j,k,s;
    int cla;
    scanf("%d",&cla);
    for(int gr=1;gr<=cla;gr++)
    {
        scanf("%d%d",&n,&m);
        printf("Case %d: ",gr);
        if(n==1||m==1)//情况的特判
        {
           printf("%d\n",max(n,m));
           continue;
        }
        if(n==2||m==2)//情况特判
        {
            s=m*n/8*4;//<span id="transmark"></span>当 n==2&&m==2时可以全部放满，否则必须留下左右两边或者左边或者右边一个田字格。来进行和其他的格子相隔开
            if((m*n)%8==2)//如果%8为2不够一个田字，则+2
                s+=2;
            else if((m*n)%8==6||(m*n)%8==4)//如果够4个则+4,只有取余为2,4,6而无其他情况因为*了4
                s+=4;
            printf("%d\n",s);
        }
        else
            printf("%d\n",(m*n+1)/2);//+1则适用于所有奇偶
    }
    return 0;
}