Ewwa's blog

By Ewwa, history, 14 months ago, In English,

Hello ! Can somebody explain to me how to solve this problem in O(logn) ?

Given N ( N <= 10^18 ) find the number of ordered pairs (a, b) for which a ^ b = X. ( 1 <= a, b <= N )

X is the first number >= N which has all it's bits equal to 1. For example if N = 5(101) then X = 7(111) and if N = 7 then X = 7.

Additional challenge : X is any number between 1 and 10^18, independent of N.

Problem link if somebody wants to submit : https://www.hackerearth.com/problem/algorithm/roy-and-maximum-xor/


14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Auto comment: topic has been updated by alittlezz (previous revision, new revision, compare).

14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Suppose a, b ≥ 0 instead; let 2p be the maximum digit of N.

Then X can't have any digits larger than 2p; if its maximum digit is 2p, then (w.l.o.g.) a must be  ≥ 2p and b < 2p, otherwise (if it's smaller) a, b must be both  ≥  or  < 2p.

In all cases except a, b ≥ 2p, the smaller number is uniquely given by the larger one, so that's trivial. Otherwise, we can reduce the problem to X, N - 2p and solve that recursively. We have steps in the recursion and each of them can be solved in O(1), so that gives an solution.

  • »
    14 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I dint understand why you got so many cases

    Assume 2p <=N-1<2p + 1

    So now the max possible number is 2p + 1-1. And number of ways is (N-2p)*2.

    Since only one of them must be greater than or equal to 2p.Nowing if we fix that then other one is unique. So that gives the number of ways. With a corner case of N=1.

    I am sorry if u are trying convey the same.I couldnt get it.

    PS:I am talking abt the hackerearth problem.