LightOJ

Knights in Chessboard

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

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Description

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

Source

Problem Setter: Jane Alam Jan

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题意： 棋盘上放旗子 , 原题 给了个图 , 说明了棋子会相互攻击情况,  求最多可以放多少棋子;

   观察可得, 棋子全放白的或全放黑的满足条件,

　两种特殊情况 ;

　n==1时全放上都可;

　n==2时先放满正方形的四个格,  然后空出四个格,  最后可能会出现空1, 2, 3,格情况 , 而空三格也是最多放2个 ; 得出代码--

#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
int main()
{
    int t;
    scanf("%d", &t);
    int Q=1; 
    while(t--)
    {
        int n, m;
        scanf("%d%d", &n, &m);
        if(n>m)
            swap(n, m);
        int x, y;
        x=n*m;
        x=x/2;
        y=n*m-x;
        y=max(x, y);
        if(n==1)
            x=m;
        if(n==2)
            x =2*(2*(m/4)+min(2, m%4));
        printf("Case %d: %d
", Q++, max(x, y));  //行*2 == 两行;
    } 
    return 0;
}