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  1. Prepare
  2. Algorithms
  3. Bit Manipulation
  4. AND Product
  5. Discussions

AND Product

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205 Discussions

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  • nadiacherneshev1
     + 0 comments
    # greedy, needs to be compiled to pass the time limit with PyPy3
def andProduct(a, b):
    return reduce(lambda x, count: count & x, range(a, b+1, (a%2==0)+1))

    def andProduct(a, b):
    differing_position = (a ^ b).bit_length()
    
    mask = ~((1 << differing_position) - 1)
    
    return a & mask

    The same idea as the above but handling the binary representation as a string throws a runtime error on test-case 3, however, I can't understand what is wrong

        position = b.bit_length() - (a^b).bit_length()

    return int(bin(a)[2:][:position] + '0'*(a.bit_length() - position), 2)

  • rnolas13
     + 0 comments

    The question is missing edge cases where a=b and a

  • rnolas13
     + 0 comments

    The question is missing edge cases where a=b and a

  • jmcparland
     + 0 comments

    Haskell

    I'm kind of surprised to see implementations that AND over the list in the discussion; a Haskell foldl1 (.&.) [a..b] (basically the same thing) would often timeout.

    module Main where

import Data.Bits (shiftL, shiftR, (.&.))
import Control.Monad (replicateM_)

solve :: [Integer] -> Integer
solve [a, b] = commonPrefix a b
  where
    commonPrefix x y
        | x == y = x
        | otherwise = commonPrefix (x `shiftR` 1) (y `shiftR` 1) `shiftL` 1

main :: IO ()
main = do
    cases <- readLn :: IO Int
    replicateM_ cases $ do
        getLine >>= print . solve . map read . words

  • runeforce1
     + 0 comments

    simplicity at its finest in Go

    func andProduct(a int64, b int64) int64 {
    res := a
    for i := a+1;i <= b;i++ {
        res = res & i
    }
    
    return res
}