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Swayam78's blog

1 line soln for Today Div 2 C
By Swayam78, history, 119 minutes ago, In English

Here , u just have to take a = c^d ,Then check if it do satisy the condition or not , if it doesn't then cout -1 , else cout a. I would appreciate if you can add proof for same.

283646208

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116 minutes ago, # |
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With the help of truth table of all possible combinations for the i pos bit.

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    113 minutes ago, # ^ |
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    Yep.

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110 minutes ago, # |
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.

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110 minutes ago, # |
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I also want proof of it as I went with most common solution of constructing a bit by bit

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108 minutes ago, # |
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Dayyum :D I went too brute i guess haha

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    105 minutes ago, # ^ |
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    Almost Everybody have done the same.

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      104 minutes ago, # ^ |
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      Yeah and I even considered carry and now i feel like carry part was totally wrong there but it was not even needed so no effect on solution :P

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        52 minutes ago, # ^ |
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        did your carry approach get ac? if yes can u explain it pls

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          46 minutes ago, # ^ |
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          Yes, it was accepted but I guess i did a bit more than what was required.

          Actually I ran it in such a way that i was considering making a binary no. that gives carry and one which doesn't. And most probably it would have worked even if i was only considering to make a binary string which would just not give carry.

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            41 minute(s) ago, # ^ |
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            ik its just overkill but i am interested to know how it works. but arent there many binary no possible solutions if u consider carry part

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              33 minutes ago, # ^ |
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              exactly there are but i was taking just any 2 of previous step. I used just two string there and at any point i could get at most 4 strings and i took just any 2 out of them and go to next point. So simply i had to work on only 62*4 strings. That ain't much ig

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67 minutes ago, # |
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wait damn how does this work? also how did you get this idea? just looking at samples or something

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10 minutes ago, # |
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Cool!

Proof: In my solution (283677605) you only ever add a bit to $$$a$$$ if exactly one of $$$c$$$ and $$$d$$$ contains that bit, otherwise you either get a contradiction on the spot, or you don't add the bit and move on

why is that

. You can skip the latter cases and check if that value of $$$a$$$ works, at the end. This is exactly equivalent to setting $$$a=c$$$ ^ $$$d$$$ and checking if it works.

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