Hackerrank 'Army game problem'

This post is dedicated to Hackerrank 'Army game problem'.

Here is the question:

Luke is daydreaming in Math class. He has a sheet of graph paper with  rows and  columns, and he imagines that there is an army base in each cell for a total of  bases. He wants to drop supplies at strategic points on the sheet, marking each drop point with a red dot. If a base contains at least one package inside or on top of its border fence, then it's considered to be supplied. For example:

Given  and , what's the minimum number of packages that Luke must drop to supply all of his bases?

Input Format

Two space-separated integers describing the respective values of  and .

Constraints

Output Format

Print a single integer denoting the minimum number of supply packages Luke must drop.

Sample Input 0

2 2

Sample Output 0

1

Explanation 0

Luke has four bases in a  grid. If he drops a single package where the walls of all four bases intersect, then those four cells can access the package:

Because he managed to supply all four bases with a single supply drop, we print  as our answer.

Answer:

There are multiple approaches to solve this but I would like you guys to try it for some time then look below. 

I know you guys did not try , just googled it :D.

You can consider this as denomination problem where you need to split the matrix in amounts of 2X2 , 1X2 and 1X1. This is a broad solution and a bit tricky as well ,since you need to keep a count each and every row. 

For the next bit think a bit out of the box. I mean literally out of the box .

Now try and cover this whole by just 2X2 even if you go out of the box.

Give it some time use a paper and pen you will find the pattern hidden there only.

Let me break it down for you.

Consider a  10X6 matrix 

Now for a 10X5 matrix :

Total Number of supply drops are same in both cases.

Since 10X5 and 10X6 could be fit in 10X6 matrix.

Now test for odd row and column. It has a bit of tweak in it.

I have pasted the code in C++ for your reference. I know you will copy it . So go ahead.

I am not saying it is most optimized code since there is always a room for optimization. Wrote it as I thought about it. Hope you find it useful. 

CODE:

int main(){   
     int n;    
     int m;
    cin >> n >> m;
    int twobytwo  = (m / 2) * (n / 2);
    int factor =0;
    if( (m%2 ==1) && (n%2 ==1) ){ 
       factor = ( (m/2 +1) + ((n-1) /2) );
     }else if(m%2 == 1){
        factor = n/2;
    }else if(n%2 == 1){
        factor = m/2;
    } 
   cout <<(twobytwo+factor);
    return 0;
}