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Dear all,

I would use your help for a very similar challenge :

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3191

I read a few posts before :

You are counting some occurences of 13 more than once
if you say, AB contains 13, and CD are any digit ...

Dear all,

It would be great if someone helps me with this challenge :
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3191

When the number M cannot "overlap" itself, it is not a big deal to compute it recursively.
By "overlapping itself", here is an ...

Finally got AC !!!

The trick is that there are more conditions on L_1 and L_2 (the lengths of first and second string resp.) :

L_1 <= L_2 <= 2*L_1

Dear reader,

If you solved or tried this challenge, can you please tell me if my reasoning is correct ? If so, do you think there is tricky test cases ?

Here is an URL to the statement :
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3196

Thank ...

Dear all,

I keep on having wrong answer for this one. The basic idea behind this challenge is simple : the length a string is the sum of the lengths of the 2 previous strings. Hence if we find the lengths of strings 1 and 2, by using Fibonacci numbers, we can compute the length of every arbitrary ...