SRM 707, D2, 250-Pointer (Cross)
James S. Plank
Sat Feb 11 14:26:46 EST 2017
Try this on your own. If you get stuck, return here and read what I've written below.
Given row i and column j, determining whether there is a cross
centered at row i and column j is a matter of testing to see whether
five characters equal '#'. It's one if statement, which performs the AND
of five tests.
The main challenge here is to make sure that i and j
are not in the first or last row/column. If they are in the first or last row/column,
then they cannot center a cross, and your test might segfault (or worse yet, return
a wrong answer).
So, what I did was have nested for loops, one for rows (i) and one for columns(j).
Each loop starts at one instead of zero, and the termination test for i is to make
sure that it is less than board.size()-1 rather than
board.size(), like you usually do with a loop. The termination test for j
is similar, making sure it is less than board.size()-1
rather than board.size().