OOP in C++ by Robert Lafore for programming concepts such as inheritance/polymorphism etc. Other than that i just think of a program and start making it in the language i am currently learning. Open up the C++ reference to find the list of functions i can use and just start making my program.
https://scontent-cdg2-1.xx.fbcdn.net/v/t1.0-9/63402_10151509531634754_1048725560_n.jpg?oh=05539b0bde4e400ffafbb7bff409f2cf&oe=57B1E8AC
I'm trying to use Eternal Blizzards approach over here. Just can't figure out how to code it up.
you could make 1 for loop that iterates over all the coordinates. Make a variable that holds the starting coordinate and a variable that hold this index. Iterate over the array finding slopes between the selected starting coordinate and others. Then make a check at the end if iteration has reached the last element. If it has do a ++ in your starting coordinate index so that it is now the 2nd element. Reset the iteration variable to the starting coordinate index + 1 ( we don't need to find slope between 2nd and 1st as it has already been found the first time). That's what i can think of without nested loops.
You're saying that the array would be inside the for loop?
Yar i still don't know how this complexity thing works. If you just have to avoid nested for loops you can reset ur iteration variables at the end of the for loop to make it serve as a nested loop. Does this reduce the complexity? Other wise i would've put that for loop i mentioned inside a while loop. It would do the same work, and save me from the work of resetting iteration variables
int rowAttack[N];
int colAttack[N];
int diaAttack[N+N-1]
for (all queens)
{
rowAttack[queen.X] = 1; //keeps a record which row has a queen when another queen comes break the loop
colAttack[queen.Y] = 1; //keeps a record which column has a queen when another queen comes break the loop
diaAttack[diagonal number] = 1; //keeps a record which diagonal has a queen when another queen comes break the loop
}This is more like n^3, definitely not o ( n ), you can see the nested loops.
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