LightOJ 1010 Knights in Chessboard <贪心思维>

Knights in Chessboard

Time Limit: 1000MSMemory Limit: 32768KB64bit IO Format: %lld & %llu

Submit Status uDebug

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

令行<=列...

当行<3时单独考虑----

其他的把棋盘分成连续黑白两色---黑最多比白多1.

代码：

#include<cstdio>void s(int ca){    int m,n;    scanf("%d%d",&m,&n);    int s;    printf("Case %d: ",ca);    if (m>n)    {        s=m;m=n;n=s;    }    if (m==1)    {        printf("%d\n",n);        return ;    }    else        if (m==2)    {        int lp=n/4*2;        if (n%4==1)            lp++;        if (n%4==2||n%4==3)            lp+=2;        printf("%d\n",lp*2);    }    else    {        printf("%d\n",(m*n+1)/2);    }}int main(){    int t;scanf("%d",&t);    for (int i=1;i<=t;i++)        s(i);    return 0;}