lightoj 1010-Knights in Chessboard （规律）

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

3

8 8

3 7

4 10

Case 1: 32

Case 2: 11

Case 3: 20

 题意

再一个Ｎ×Ｍ的棋盘上可以放多少只马，使得他们不能互相伤害。

思路

　分三种情况：

①放满每个白格子或黑格子。

②当n=1||ｍ=1时全放。

③当n=2||m=2时每个田字格放满，隔一个田字格再放一个田。

代码

    #include<cstdio>    #include<cstdlib>    #include<cstring>    #include<iostream>#include<cmath>    using namespace std;    int main(){        int i,j,k,m,n,T,x,y;long long ans;        scanf("%d",&T);        for(int u=1;u<=T;u++){            ans=0;int flag=0;            scanf("%d%d",&x,&y);if(x==1||y==1)printf("Case %d: %d\n",u,max(x,y));else if(x==2||y==2){flag++;if(x>y)swap(x,y);for(i=1;i<=y;i++)if(flag%3!=0){ans+=2;flag=0;}else flag++;printf("Case %d: %lld\n",u,max(ans,k-ans));}else{            k=x*y; for(i=1;i<=y;i++){                ans=ans+x/2;                if(flag){flag=0;x--;}                else{flag=1;x++;}            }            printf("Case %d: %lld\n",u,max(ans,k-ans));        }}                   return 0;    }

made by 罗旅洲