it's a cheat! =)

At least you should be able to write bruteforce in O(n^2 * 2^n) :)

For a pair of numbers the number of actions is equal to <length of the first number> + <length of the second number> - 2*<length of the longest common prefix> (numbers are treated as binary strings)

Bo`ladigan gap gapirgin-e!!!

Your must understand hint of 'aropan'. And get answer = 2^L*(L - 3*2^L + L*2^L + 3)
P.S.: Mathcad can help you to find sum of such expression as FOR(z=0,..l)z*2^z. It's only math problem.
P.P.S.: To find 2^L use http://en.wikipedia.org/wiki/Exponentiation_by_squaring

Edited by author 22.05.2012 22:17

чё это????