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  1. Prepare
  2. Algorithms
  3. Bit Manipulation
  4. Xor-sequence
  5. Discussions

Xor-sequence

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150 Discussions

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  • jmcparland
     + 0 comments

    Haskell

    So, if you look at the sequence of running XORs over the XORs, there's a pattern. I didn't give it much thought after that -- just implement the lookup function.

    module Main where

import Control.Monad (replicateM_)
import Data.Bits (xor)

lu :: Int -> Integer
lu 0 = 1
lu 1 = 2
lu 2 = 2
lu x = do
    let (q, r) = (x - 3) `quotRem` 8
        b = 8 * (fromIntegral q) + 6 :: Integer
     in [b, b + 1, 0, 0, b + 2, b + 3, 2, 2] !! r

solve :: Int -> Int -> Integer
solve 1 b = lu (b - 1)
solve a b = xor (lu (a - 2)) (lu (b - 1))

main :: IO ()
main = do
    cases <- readLn
    replicateM_ cases $ do
        [a, b] <- map read . words <$> getLine
        print $ solve a b

  • eduardocintrama1
     + 0 comments

    Hi there! Here is my Java solution for the challenge. If you'd like to discuss programming or any other topics, don't hesitate to reach out!

    https://github.com/eduardocintra/hacker-rank-solutions/blob/master/src/br/com/eduardocintra/medium/xorsequence/XorSequence.java

  • kumar_cse_576
     + 0 comments

    c++ optimal one

    long getXor(long x){
    int n=x%8;
    if(n==0||n==1)return x;
    if(n==2||n==3)return 2;
    if(n==4||n==5)return x+2;
    else return 0;
}

long xorSequence(long l, long r) {
return getXor(r)^getXor(l-1);
}

  • krishnamadhumit1
     + 0 comments

    where is an array to search or to perform a function

  • claudia_yao2012
     + 1 comment

    like zaber04 said, we need to set up an array starting from 0 to a certain number to see the pattern. The following is the code I use to set up the array and observe the pattern:

    result = 0
further_xor = 0
print(0, "\t", result)
for i in range(1, 40):
  result = result ^ i
  further_xor = further_xor ^ result

  print(i, "\t", result, "\t", further_xor)