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Here's another python version that is (hopefully?) reasonably well-commented to make it clear what's happening.

BTW I have to admit that I really find it astonishing how much cleaner the Python solutions to many of these problems look compared to other languages! To me, it's really much closer to the pseudocode I'd write. Or is it just me?

importosdeffind_digit(A,B,n):lengths=[len(A)]ifn<=len(A):returnA[n-1]lengths.append(len(B))ifn<=len(B):returnB[n-1]idx=1# Building up until we hit total digit length n.# Only count length though, do not build the actual string!whileTrue:idx=idx+1lengths.append(lengths[idx-2]+lengths[idx-1])iflengths[idx]>=n:break# Going back recursively until we know where exactly to # find digit n. Use the fact that # sequence[i] = [sequence[i-2], sequence[i-1]].len_diff=nwhileidx>1:iflengths[idx-2]<len_diff:# Digit n is in the second part of the sequence # section currently considered.len_diff=len_diff-lengths[idx-2]idx=idx-1else:# Digit n is in first part of the sequence section # currently considered.idx=idx-2ifidx==0:returnA[len_diff-1]else:returnB[len_diff-1]if__name__=='__main__':fptr=open(os.environ['OUTPUT_PATH'],'w')num_tests=int(input().strip())tests=[]foriinrange(num_tests):A,B,n_str=input().rstrip().split()num=find_digit(A,B,int(n_str))fptr.write(str(num)+'\n')fptr.close()

## Project Euler #230: Fibonacci Words

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Here's another python version that is (hopefully?) reasonably well-commented to make it clear what's happening.

BTW I have to admit that I really find it astonishing how much cleaner the Python solutions to many of these problems look compared to other languages! To me, it's really

muchcloser to the pseudocode I'd write. Or is it just me?