PROBLEM LINK:
Contest - Division 3
Contest - Division 2
Contest - Division 1
DIFFICULTY:
CAKEWALK
PROBLEM:
Find the number of black cells in a n*n chessboard (where n is even).
EXPLANATION:
Since n is even, and each row of the board has an alternating sequence of colours, the number of black cells in each row is exactly \frac{n}{2} (What would happen if n wasn’t even?). Therefore, the number of black cells over all n rows is n*\frac{n}{2}.
TIME COMPLEXITY:
For each test case, computing the answer takes O(1) time.
SOLUTIONS:
Editorialist’s solution can be found here.
Experimental: For evaluation purposes, please rate the editorial (1 being poor and 5 excellent)
- 1
- 2
- 3
- 4
- 5
0 voters