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Автор Lets__go, история, 5 лет назад, По-английски

Can some one help me on how to solve this problem?

Editorial is given in Japanese, i used google translater to understand it but some sentence doesn't make any sense to me and i couldn't understand solution from editorial.

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Problem tag — Digit DP.

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Here is my submission, which you can refer to: Atcoder Submission 8567969

I don't know the solution of the editorial, just give it on my own.

Prerequesite knowledge: For every $$$a, b \in \mathbb{N}$$$ then $$$a+b = a\oplus b + 2\times (a\ and\ b) =(a\ or\ b) + (a\ and\ b)$$$. Well this is easy to prove.

Therefore let $$$x = a\ or\ b,\ y = a\ and\ b,\ u = a\oplus b,\ v = a+b$$$ then $$$x+y = v, x-y = u$$$. So for each pair $$$(u, v)$$$ in the answer maps with exactly one pair $$$(x, y)$$$, and each pair $$$(x, y)$$$ such that $$$y \subseteq x, y \le x \le N$$$ corresponds with only one pair $$$(u, v)$$$. Here $$$y \subseteq x$$$ means in binary representation, each bit of $$$y$$$ is smaller than the corresponding bit of $$$x$$$.

So we can count the number of pair $$$(a, b)$$$ such that $$$b \subseteq a, a+b \le N$$$.

Let say in binary representation, $$$a = \overline{a_{60}a_{59}...a_{1}a_{0}}$$$ and $$$b = \overline{b_{60}b_{59}...b_{1}b_{0}}$$$ then we can solve the problem using Digit DP. You can refer to my solution.

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