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code : array [1..50] of longint;
zeros : longint;

when n = 1
there are only two possibilities '1' and '0'
for n = 2 we try to add '0' and '1' after them
(code[1] = 2 , zeros = 1(one number end with zero))

for number end with '1' , we can add '0' only
for '0' , we can add both '0' and '1'

so ...

as topic

I have read the pascal program,
for each set , he has performed BFS for 8 times.
My method need only 5 times, I think It can reduce the run time.