please someone explain the question bintour in march challenge 2014 or it’s test cases
Well it’s very difficult to explain the whole problem again, instead you can tell which part you couldn’t understand.
The Question is something like this. You are given that in a tournament where are 2^k players where k is given as the input you have to find out the number permutations in which player i (i varying from 1 to 2^k) gets to the final round.
k = 2
then there are 4 players 1,2,3,4.
Player 1 :
There are no permutations in which he gets to the final round.
Player 2 :
The possible permutations in which he gets to the final round are
1,2,3,4 Player 1 and 2 play where player 2 wins and then he enters the final round with player 4
2,1,3,4 Player 1 and 2 play where player 2 wins and then he enters the final round with player 4
1,2,4,3 " "
2,1,4,3 " "
4,3,2,1 " "
4,3,1,2 " "
3,4,1,2 " "
3,4,2,1 " "
So 8 cases and similarly for other
sir if j<=2^(k-i) then if i=1 and k=2 then value of j can be 1 and 2 but if j =2 then battle will be in knights standing at position 2j i.e 4 and 2j+1 i.e 5 but maximum value can be 4 so how is this poosible
thnks … i understand that sir the problem is that position are given as 2j and 2j +1 so if j=1 then the match will be in 2 and 3 and when j=2 the match will be in 4 and 5 but how is this poosible accoding to my understanding and your example i think the match is in between 1st and 2nd , 3rd and 4th knight
Actually I was doing the same mistake ,just start the j from 0 not 1
it really solves the problem