# Invitation to International Coding League 2018- External Rated Contest

Hi Codechef community,

Computer Science Association & Association of Computing Machinery, BITS PILANI, Pilani Campus presents you their flagship contest, “International Coding League 2018”. It is being organised as a part of our annual technical fest, Apogee.

The contest will be External rated contest and will be of 2.5 hours.

Start Time : 20:00 IST, 27th February’ 2017

Prizes: Worth Rs 20000 (10k + 6k + 4k) for top 3 Indian winners.

Also top 5 Global winners and top 5 Indian winners will get 250 laddus each.

The contest consists of 5 algorithmic problems of varying difficulty. The problems have been set such that even beginners can attempt a few problems and even the best coders find tough job ahead at the hard ones. It is advised to read every problem in the contest. The detailed editorials of all the problems with the Setter’s & Tester’s code will be uploaded immediately after the contest so that you may benefit from it.

The setting and testing panel includes me, akulsareen, ankit, divesh, priyank and vibhav. I would also like to thank Arjun arul from Codechef team for going through the problems and improving the statements as well.

To participate in the contest, you just need a Codechef handle and then just click on the link mentioned above for participating for the contest.

I hope you enjoy the contest and Happy Coding

You can find the link of last year contest here

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The ranking will be similar to ACM ICPC and the penalty time for wrong submission has been made 10 minutes for the contest.

You can find the intended solutions here. I will be uploading the short editorials in a while. Detailed ones will be put up on codechef discuss forum by tomorrow. Hope you all enjoyed the contest.

Any feedback regarding the clarity of problems, time limits, test data being weak/strong etc?

Short Editorial of first 4 problems:

Problem 1 — Matrix Game
Greedy solution applies and Matrix has no rolw in the problem. Just consider the numbers and at each moment of time every person tries to take the largest available number. Find the sum they can accumulate after this process and print answer accordingly.

Problem 2 — Stingy Strings
Again greedy solution applies in the sense that we replace all occurences of a given character by its corresponding number or replace none of them. Next part is to modify the cost function as “length — 2*(count of numbers) + (sum of numbers)/k”. Using this, we can say that we replace only those ccharacters with numbers whose value is greater than or equal to 2*k.

Problem 3 — Maze love

Let number of steps taken in each direction be N, S, E, W. So we have following equations:

N — S = m, E — W = n, Na + Sb + Ec + Wd = P

Replace S, W in last equation, we get N * A + E * B = C, where
A = a + b, B = c + d, C = P + mb + nd

The other constraints are N, S, E, W are integers and N >= m, S >= 0 and E >= n, W>= 0.

Solution exists when gcd(A, B) divides C. After this check, we simply brute force over the possible values of N and update answer (i.e. minimum value of N+S+E+W) if present.

Problem 4 — Replace and Substring queries.

Idea is to find the number of substrings with given mask in string B easily. For this, we fix a particular index and find the first point of occurrence of every character on the right and add the corresponding contribution to the mask. For example- In “aabbbbdac” for index 1, “a” occurs at “1”, “b” occurs at “3”, “d” occurs at “7” and “c” occurs at “9”. So number of substring with one endpoint at 1 and having mask 1 are 2, mask 3 are 3, mask 11 are 2 and mask 15 is 1.

One this initial computation is done, we try to handle the queries and updates. For query on string “a”, we just need to find the mask of the substring a[l:r] and this requires to compute if a particular character occurs in a range. With updates on string “a”, this can be easily handled using fenwick tree or segment tree.

To handle updates on string “b”, idea is to first remove contribution of every substring which contains the xth character and then add the contribution back. For finding number of mask of substrings with contain xth character, we see the substring can either end at x, start at x or contain x in the middle. The first 2 are easy to calculate and are similar to the construction by finding first occurrence to the left and right of index x. For the substring which contain “x” in the middle we can visualise them as 2 parts, left and right and their mask being the “OR” of the left and right mask. Thus doing this OR convolution of the left and right masks in naive manner i.e. O(ALPHA**2) we can find their contribution too.

Complexity of precomputation is O(m * ALPHA * log m) and query and update on string “a” is O(ALPHA * logn) and O(log n) while update on string “b” is O(ALPHA * log m + ALPHA**2).

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Short explanation for last problem: Raid Systems

The first idea of the problem is to see that files on xth hard disk can be retrieved using files on hard disk 1. The file indices differ by exactly (x-1). For example, if at any moment of time, hard disk 1 has files f1, f2, f5 when n = 5, then hard disk 2 has files f(1+2-1), f(3+2-1), f(5+2-1) i.e. f2, and f3 (f7 is discarding). See the table to convince regarding this observation. Thus, the queries of both type can be handled very easily using binary search only provided we are able to find the files in sorted order on hard disk 1 on any day.

Next thing is to notice is that the pattern of files being stored on hard disk 1 will repeat after some time, to be precise after 2^(ceil(log2(n)). Draw a table for n = 8 and n = 5 to get view of it. Next observation is that total number of files we require for all possible days is bounded by 3^ceil(log2(n)). For example: Consider n = 4

0: f1
1: f1, f2, f3, f4
2: f1, f3
3: f1, f2
4: same as 0 i.e. pattern repeats.

The total number of files stored i.e. slots used is (1 + 4 + 2 + 2) = 9, 3^2. You can see this thing for any general n. Complete the table for n=5 and observe the same. Now, we just need to come up with a 3^(ceil(log2(n)) generation algorithm, which gives us the files in sorted order on any day. The memory complexity will also be the same. Many constructive algorithms can exist and you can try to find your own. My solution uses just one of my finding, idea is again similar to binary search i.e. can be divide the files (in sorted order) in half, calculate the files on left side and use it to calculate the files on right side. Basically, I mean to say that if a particular bit is set in “y” and we calculate the lower half files, then the upper half files are basically the mirror image on the lower half files. You can check my “gen” function in the solution code for details.


(https://github.com/likecs/ICL-18-Solutions/blob/master/Raid%20systems.cpp)
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For each test case, output a single line containing two space seperated integers a and b such that a / b = Pmax - the maximum possible power achievable after transformation and gcd(a,b) = 1

I don’t get this part. Can anyone explain?

In problem 2

I considered lsum(count of letter) and rsum(-count of numbers + (sum of numbers/K))

So,For each i since sum += 1(count of letter) or -1(count of number) +(val(s*)/k)

int num = (int)a*-96;
if(num/k - 1 > 1)
lsum += num/k - 1;
else
rsum++;


What is wrong with this



[1]: https://www.codechef.com/viewsolution/17560153

In problem 1
Can we solve by checking the frequency of each number from 1 to 100 and if frequency of every element is even then there will be “draw” otherwise “cyborg” will win?

OR can anyone can provide testcases were this solution fails https://www.codechef.com/viewsolution/17561707

In problem2. “Geno” can never win… that’s cool observation

@vivek_1998299 ALPHA refers to alphabet size, which is 20 here. OR convolution refers to polynomial multiplication where c_0 x^{p_0} \cdot c_1 x^{p_1} = c_0c_1 x ^{p_0 | p_1}. To put it simply the coefficients at p_0 and p_1 affect p_0 | p_1 instead of p_0 + p_1.

This is relevant here in the following manner: assume you know there are c_0 substrings with mask p_0 that begin at “x”. Similarly there are c_1 substrings with mask p_1 that end at “x”. So there will be c_0c_1 substrings with mask p_0 | p_1 that contain “x” irrespective of where it lies.

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Sorry for the delays, the editorials are given in below links:

Contest starts in 20 minutes.

In problem 2
Next part is to modify the cost function as “length — 2(count of numbers) + (sum of numbers)/k”
I am not able to get this! Can you elaborate a bit more on this?

Count of numbers + count of characters = n (i.e. length of array). So we replace count of characters in the cost function using the above equation.

For example after simplifying the function, you find the value as 12/10, So you should print the answer as “6 5”.

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This problem is quite interesting and it seems the accepted solutions use various different methods. My solution uses the fact that the queries can be reduced to the form of find the position of the a^{th} odd number in the b^{th} row of Pascal’s triangle, due to the pattern that shows up. And it turns out that all the odd numbers are at the positions which are numbers obtained by removing zero or more set bits from the binary representation of b (source).

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Won’t the length of array change on replacing character to number?
Eg: Character : z
Number : 26

solution:https://www.codechef.com/viewsolution/17558988
what is wrong with solution?

I get it. Thanks

@dishant_18, don’t see numbers literally, try to understand them as single object occupying a position and what value it adds if it was present.

1 Like