Lightoj1010——Knights in Chessboard（找规律）

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

题意是m*n的棋盘上最多能放多少个马，可以分一下三种情况讨论：

1、如果m或n为1，那么很明显能都放马

2、如果m和n都大于2，的情况。如果m*n是基数，那么一定有一种颜色的方格比另一种颜色多一个，在这个颜色的方格上可以都放马，为偶数的话很明显能放一半

3、如果m或n有一个为2，则又可以分三种情况（假设m=2）

     一、n==4时，则这是两个田字格，可以放4个马

     二、n==1时，也都可以放马

     三、n==2、3时，最多放4个马

     以上面三种情况对n进行贪心算法，先找出4的倍数个格子，再在剩下的格子里判断条件二和条件三

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<cmath>
#define MAXN 10000
using namespace std;
int main()
{
int ans;
int t,cnt=1,i,j,m,n;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&m,&n);
if(m==1||n==1)
ans=max(m,n);
else if(m==2||n==2)
{
int temp=max(m,n)%4;
ans=max(m,n);
if(temp)
{
ans=max(m,n)/4*4;
if(temp==1)
ans+=2;
else
ans+=4;
}
}
else if(m*n%2==0)
ans=m*n/2;
else
ans=m*n/2+1;
printf("Case %d: %d\n",cnt++,ans);
}
return 0;
}