For each of 2^K players labeled by 1 … 2^K, how many permutations of their initial positions can allow him to advance to the final? The tournament is as following figure. The winner of a match is the one with larger label.

EXPLANATION:

For a given player X, if he advanced to the final, his label should be the greatest among the players in his semi-final part.

Assume he is the winner of semi-final one. His position has 2^(K-1) choices. And, there are X-1 players with smaller label. Therefore, the answer is 2^(K-1) * (X-1) * (X-2) * … (X - 2^(K-1) + 1) * (2^(K-1))!.

With X varying from 1 to 2^K, it is easy to maintain the product (multiply a number and divide a number), because the MOD number is a prime, according to the Euler Theorem.

If we calculate the inverse by the fast power, similar to the fast multiply I mentioned in a previous editorial, we can get a O(2^K * K) algorithm.

AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here.
Tester’s solution can be found here.

“…happens a battle between a knight on the 2.jth position and a knight on the 2.j+1th position”.
The above line copied from the problem statement is misleading. Should have been: “…happens a battle between a knight on the 2.jth position and a knight on the 2.j-1th position”. Faulty statements like this should be guarded against in future problem statements.

great question loved solving it… The Permutations n Combinations part was easy but the inverse modulo part and calculations were quite a bit. And there was a problem in the problem statement itself. as @bodmas pointed.
my solution can be viewed: http://www.codechef.com/MARCH14/status/BINTOUR,abcdexter24

it is modulo multiplicative inverse property that
a^(-1)%mod=a^(mod-2)%mod when a and mod are coprime…bigmod is just calculating a^(mod-2) in log(mod) time…