LightOJ1010---Knights in Chessboard (规律题)

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

Problem Setter: Jane Alam Jan

规律题

假设n?m是偶数且都大于2

答案是n?m/2

假设n和m都是奇数且都大于2

答案就是一半的行放m/2+1个,还有一半放m/2个

假设n m 都小于等于2

n(m)为1,答案就是m(n)

m(n)有一个为2,那么我们能够考虑能够分出多少个2*2的格子

设有x个,答案是(x/2+x % 2)?4

剩下来可能有一个2*1的,这时假设x是奇数,不能放,假设x是偶数能够放

/*************************************************************************
    > File Name: LightOJ1010.cpp
    > Author: ALex
    > Mail: [email protected]
    > Created Time: 2015年06月08日 星期一 14时24分24秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

int main() {
    int t, icase = 1;
    scanf("%d", &t);
    while (t--) {
        int n, m;
        scanf("%d%d", &n, &m);
        int ans = 0;
        if (n >= 3 && m >= 3) {
            if (n * m % 2 == 0) {
                ans = n * m / 2;
            }
            else {
                int s = m / 2 + 1;
                ans += (n / 2 + 1) * s;
                ans += (n / 2) * (s - 1);
            }
        }
        else {
            if (n == 1) {
                ans = m;
            }
            else if (m == 1) {
                ans = n;
            }
            else {
                if (m == 2) {
                    swap(n, m);
                }
                if (m <= 3) {
                    ans = 4;
                }
                else {
                    int cnt = m / 2;
                    ans = (cnt / 2 + cnt % 2) * 4;
                    if (m % 2 && cnt % 2 == 0) {
                        ans += 2;
                    }
                }
            }
        }
        printf("Case %d: %d\n", icase++, ans);
    }
    return 0;
}