If you are getting WA you have probably made a logical error somewhere. You
might have to post your code to know where.

On 5 Jun 2017 2:35 p.m., "Reynaldo Gil-Pons" <gil...@gmail.com> wrote:

There are 100 computers, 1 microsecond = 1e-6 seconds, so time for reading
each number 30 times is: 10 ^ 8 * (0.2 * 10^ -6) / 100 * 30 = 6 seconds, so
a little bit above, but I have tested and it passes the time limit,
although giving WA, as if the probability reasoning were incorrect...

On Monday, June 5, 2017 at 12:59:48 AM UTC-4, Xiongqi ZHANG wrote:
> 10^8 * 0.2 microsecond = 20 seconds while Time Limit is 2 second.
>
>
> Reynaldo Gil-Pons <gil...@gmail.com>于2017年6月4日周日 下午7:39写道:
> 0.2 millisecons to read a single bit? They say 0.2 microseconds. I get WA
when doing what I explained, in 1.7 seconds...
>
>
>
> On Friday, June 2, 2017 at 12:09:47 PM UTC-4, Luke Pebody wrote:
>
> > I would think it can't work on the large case because it takes 0.2
milliseconds to read a single bit once and therefore 20 seconds to read all
100M bits once and 10 minutes to read them 30 times.
>
> >
>
> >
>
> > On 2 Jun 2017 4:56 p.m., "Reynaldo Gil-Pons" <gil...@gmail.com> wrote:
>
> > In the analisis they explain a solution for the Small case, and point
out it cannot work for the Large one. When I calculate the probabilities of
getting WA using 30 checks for each position, and assuming 1000000
testcases (cant be more than this right?) I get less than 1 / 1000. But I
still get WA. I calculate an upper bound on the probability of failure as 1
- (1 - 1 / (2 ** 30)) ** number_test_cases. Is there anything wrong?
>
> >
>
> >
>
> >
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