Editorial for DMOPC '18 Contest 5 P1 - A Painting Problem

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Author: Kirito

We can recursively generate the binary representation of a number $N$ ~N~ by appending the remainder of $N$ ~N~ divided by 2 to the binary representation of $N$ ~N~ divided by 2.

If we let $B(N)$ ~B(N)~ represent the binary representation of $N$ ~N~, then we have:

$\displaystyle \begin{align*} B(30) &= B(15) + \mathtt{0} \\ &= B(7) + \mathtt{10} \\ &= B(3) + \mathtt{110} \\ &= B(1) + \mathtt{1110} \\ &= \mathtt{11110} \end{align*}$

After obtaining the binary representations, we can loop through the last $K$ ~K~ characters, and count the number of positions with the same character, and print the answer.

Time Complexity: $\mathcal O(\log N + \log M + K)$ ~\mathcal O(\log N + \log M + K)~

Alternatively, if one is familiar with bitwise operations, this problem is the same as counting the number of bits set in $(N \otimes M) \mathbin{\&} (2^K - 1)$ ~(N \otimes M) \mathbin{\&} (2^K - 1)~.

Time Complexity: $\mathcal O(K)$ ~\mathcal O(K)~

Comments

-1

ross_cleary commented on Dec. 2, 2020, 9:44 p.m. ← edited →

For the bitwise operations solution, why isn't the time complexity the same as the recursive solution? Doesn't bitwise XOR have a time complexity of $\mathcal{O}(\log N)$ ~\mathcal{O}(\log N)~?

1

Genius1506 commented on Dec. 2, 2020, 9:59 p.m. ← edited →

Bitwise XOR is $\mathcal{O}(1)$ ~\mathcal{O}(1)~.