Puzzle 20: Do a linear scan for the num we're looking for

When the number of iterations was small, looking near the original
position of the num was a good idea. As the number of iterations
increases, the positions are fairly random, so speed of lookup is mostly
a factor of how quickly we can scan the whole buffer.

Speeds up part 2 of the problem from 2.2s to 0.12s.

1 files changed, 3 insertions, 4 deletions

diff --git a/20.c b/20.c
index e427217..677c9df 100644
--- a/20.c
+++ b/20.c
@@ -101,10 +101,9 @@ main(int argc, char ** argv)
 			int newpos;
 			struct num tmp;
 
-			// Look for the num whose ordinal is i, starting from the
-			// position of the last num we looked at.
-			while (nums[pos].ord != i)
-				pos = modulo(pos + 1, nnums);
+			// Look for the num whose ordinal is i
+			for (pos = 0; nums[pos].ord != i; ++pos)
+				;
 
 			if (nums[pos].val == 0)
 				continue;