CSES-Two Knights problem: help needed!

can someone please help me in this problem??
https://cses.fi/problemset/task/1072
thanks!! :slight_smile:

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+1 i also want to know the same …

from all possibilities of placing two knghts , you can substract those ones in which they both attack each other…(which are 4*(n-1)*(n-2) for given nXn board)

Total possibilities are (n^2)C2

sorry, my fault.
i got ac with (n^2)C2 - 4*(n-1)(n-2)
can you please explain how did you get this 4
(n-1)*(n-2) term?
thanks!!

int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        if(i==1)
            cout<<0<<"\n";
        else if(i==2)
            cout<<6<<"\n";
        else
        {
            int p=i*i;
            int j=i-2,k=p*(p-1)/2;
            k-=8*j*(j+1)/2;
            cout<<k<<"\n";
        }
        
    }

Possible Duplicate: Number of ways two knights can be placed such that they don’t attack

@venturer have shared a link there: Explanation

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can u please explain??

https://oeis.org/A172132

Hope this will help.

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