Hello,
>> "But, for example after 68 days, it will contribute R1*2^5 rings to the
total rings, right?"
That's wrong! It still does contribute 0.
The number of R1 contributes 0 total for ANY day larger then 62.
Why is that? :
Day 63: R1 * 2^63 ≡ 0 (mod 2^63) 2^63 is a multiple of 2^63 so
the remainder MUST be 0
Day 64: R1 * 2^64 ≡ 0 (mod 2^63) 2^63 is a multiple of 2^63 so
the remainder MUST be 0
R1 * 2^(63 + 1) ≡ 0 (mod 2^63) 2^64 is less than
2^63 and therefore a multiple 2^63 so the remainder STILL MUST be 0.
So when querying day 324, you can ONLY identify R(6) at day 0.
Am Di., 30. Apr. 2019 um 17:35 Uhr schrieb Ramesh Kumar <
[email protected]>:
> On Tuesday, April 30, 2019 at 8:18:56 PM UTC+5:30, /dev/joe wrote:
> > The 1-day rings duplicate every day, so their contribution to the sum
> moves one bit left in the binary representation every day. After 63 days,
> since we only get the value mod 2^63, the 1-day rings make no contribution
> to the value you get.
> >
> >
> > On Tue, Apr 30, 2019 at 10:05 AM Ramesh Kumar <[email protected]>
> wrote:
> > Hi All,
> >
> >
> >
> > It was mentioned we can use two chances for Code Jam 2019 Round 1B
> second problem - Draupnir. But I think we can use only one number 324
> which is sufficient and find the solution. But, I am unable to get the
> accepted answer. Is there anything wrong to use only 324 number to find all
> the required six numbers? Please help me to understand.
> >
> >
> >
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> But, for example after 68 days, it will contribute R1*2^5 rings to the
> total rings, right?
>
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