The problem is very easy to solve once you have understood it. The test cases aren’t very clear in explaining the problem. In this problem you have to find the distance of the latest ridge formed after n folds(order of folding is right to left then left to right) from the initial left side. Perform this on a paper. The solution becomes pretty clear.
Problem was super fine…and it’s good that setter has provided only 2 test cases…otherwise , problem could not make any sense. it would be easier to solve if n=3 also get provided.
I have also tried many times this question…but at last i got the pattern and simply solve it…( but i also spoiled 6-7 pages to understand the question ).
Must watch my solutions , i have literally solved for all values of n…
I too had a hard time figuring out, just started by folding paper and marked the distance on paper it was 1/2, 1/4, 3/8, 5/16, 11/32 now You have to just find the pattern
arr = 1, 1, 3, 5, 11
ans = (2*arr[n-2] + arr[n-1]) / 2^n
Just providing an alternate solution.
Simply the sum of the GP series 1/2 - 1/4 + 1/8 - 1/16… needs to be calculated.
The sum of GP series can be manually calculated in terms of n as sum = a(1 - r^n)/(1-r), where a = 1/2 and r = -1/2.
Since r is negative there will be two different cases for n = odd and n = even.
From the formula you can get the numerator and denominator, reduce them to their simplest form (using gcd) and that will be your answer.
In case you need a reference
nth term of the series was simply the average of (n-1)th and (n-2)th term. Since denominator was always even and numerator always come out to be odd giving us irreducible fraction everytime.
Here’s my solution: CodeChef: Practical coding for everyone
For any n, the values are (2^n+1)//3 and 2^n (where // rounds the division down to integer).
I worked this out by converting the entire Amazon rainforest into paper. Hope no-one noticed.
it is not that difficult also.
T(1)=1/2,
T(2)=1/2-1/4,
T(3)=1/2-1/4+1/8,
.
.
.
, T(n)=1/2-1/4+1/8-1/16+1/32…+(-1^(n+1))*(1/2)^n.
Just solve this G.P and you will get the answer
took some time… but was exciting to find values for n=4,5 and based on the sequence obtained …drawing conclusions…
Please go through my code… check at bottom for explanation… i used “-” dash rather than paper to solve this stuff
I found a cool pattern too.
I compute numerator and denominator separately. Kind of DP style, where numerator value for n foldings depends on the numerator value for n-1 foldings.
Two arrays num[] and den[] are used for numerator and denominator.
Set base case num[1], den[1], num[2], den[2] as 1, 2, 1, 4 respectively.(for 1/2, 1/4)
for i = 3 to i = N (here N = 25 max constraint):
den[i] = 2^i
if i is odd: num[i] = 2 * num[i-1] + 1
if i is even: num[i] = 2 * num[i-1] - 1
So we have precomputed till N.
We just need to display num[n] and den[n] for any given value of n (no. of foldings) now.
My Solution: CodeChef: Practical coding for everyone
I posted it in the PRDRG - Editorial topic but I’ll post it here too cuz why not?
The first ridge is at distance \frac12, the second is at \frac{1}{2}-\frac{1}{4}, the third ridge at \frac{1}{2}-\frac{1}{4}+\frac{1}{8}, the fourth ridge is at \frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}…
Notice a pattern?
The n^{\text{th}} ridge is at a distance of
Now 2 clearly does not divide 2^n-(-1)^n and 3 does.
So, the numerator is \frac{2^n-(-1)^n}{3} while the denominator is 2^n.
Time complexity: O(1)
I had the same problem when one of my senior told me that I interpreted it wrong. Only the top layer needs to be folded each time that makes it 1/2 1/4 3/8 and so on. I have done it by adding the fractions. Have attached the link to the code. For any further queries, feel free to ask any question.
I surely won’t leave this unsolved.
Feedback Forwarded to setting panel XD
C’mon… find another wife who will help you to solve this problem!! 18 hrs left…
Since MOSS penalty guarantees nonnegative rating. So technically speaking MOSS penalty should not have any effect on unrated users.
Hopefully, you leave enough trees for coming generations.
As for the problem, I sincerely wish you manage to solve this problem during the contest (with or without wife), and if not, I have written Editorial for this problem which may help you.
Lastly, for god, I hope God will find in his heart to accept you back.
Anything else?
@taran_1407 - Statistics feel theres at least one more person eagerly awaiting your editorials. :p. Or…perhaps not XD