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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 22:17что за тест?
Edited by author 16.10.2011 17:12
Edited by author 16.10.2011 17:12 Check your formulas for overflows а у меня ва31...
Dmitriy, разобрались что не так?
Разобрался... я магическим образом степень тоже по модулю 10^9+7 брал)
Edited by author 31.10.2011 12:42 |
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