PROBLEM LINK:
Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3
Contest: Division 4
Author: Lavish Gupta
Testers: Satyam, Abhinav Sharma
Editorialist: Nishank Suresh
DIFFICULTY:
To be calculated
PREREQUISITES:
None
PROBLEM:
A queen is placed at cell (X, Y) on a N\times N chessboard. How many cells does it attack?
EXPLANATION:
Let’s separately count the squares that are under attack horizontally, vertically, and diagonally. Our answer is the sum of these squares.
 Horizontally: The queen can move any number of squares horizontally. So, every square in its row is under attack, apart from the one it’s standing on. This gives us N1 squares.
 Vertically: The exact same reasoning as the horizontal case gives us N1 more squares in this case as well.

Diagonally: There are two diagonals to consider, so once again we count them separately and add up.
 One diagonal consists of all squares (A, B) such that A+B = X+Y. A little experimenting on paper should show you that the number of squares on this diagonal is the difference between N+1 and X+Y, i.e, (N+1)  (X+Y). Subtract one from this to account for the square the queen is standing on.
 The other diagonal consists of all squares (A, B) such that A  B = X  Y. Once again, a little experimenting will show that the number of squares on this diagonal is N  X  Y, from which we subtract one.
Add up the answers to these four cases to obtain the final answer.
TIME COMPLEXITY:
\mathcal{O}(1) per test.
CODE:
Python
for _ in range(int(input())):
n, x, y = map(int, input().split())
ans = 2*n # row + column
ans += n  abs(x  y) # diagonal 1
ans += n  abs(n+1  (x + y)) # diagonal 2
print(ans  4)