Invitation to International Coding League 2019-External Rated Contest

Hi Codechef community,

Association of Computing Machinery & Computer Science Association, BITS PILANI, Pilani Campus present you their flagship contest, “International Coding League 2019”. It is being organised as a part of our annual technical fest, Apogee.

The contest will be External rated contest for div. 2 participants only.

The contest will be of 2.5 hours.

Contest Link :

Start Time : 20:00 IST, 28th March 2019

Prizes (tentatively) worth Rs 30000 for top Indian winners.

Registration for prizes:

The contest consists of 6 algorithmic problems of varying difficulty. The detailed editorials of all the problems with the Setter’s & Tester’s code will be uploaded immediately after the contest.

I would 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.

I hope you enjoy the contest and Happy Coding :slight_smile:

You can find the link of last year’s contest here
PS: Contest starts at 20:00 please ignore the typo on the poster.
PPS: The round is over. There has been no penalty for wrong submissions.

PPSS: The brief solution to the problems are as follows:-

1901: 7 *11*13=1001. Hence on performing the three operations on a 3-digit number abc, you get abcabc. Thus, your answer is the number of distinct 3-digit numbers possible from the initial 3 digits.

1902: Subtract the largest possible square of an integer from N till you get 0. The number of times you can perform the subtraction is your answer.

1903: This is a standard variation of a linear diophantine. The solution uses inverse modulo.

1904: The answer is pow(26, n - |s|) + 25 * |s| * pow(26, n - |s| - 1). Take care of taking modulo appropriately.

1905: We iterate over all possible (L, R) from 1 to N. For each interval we calculate the profit. To compute the answer for each interval in constant time, we use prefix array for sums of the cost array.
Profit[L][R] = A * (R - L + 1) - (prefix[R] - prefix[L-1]) - (Dmax - Dmin)^2

1906: Store as A,B where A>B. Double B till A>B>A/2. Then decrement both till you get B=A/2. Double B then increment all the way to zero.

Hope you enjoyed the contest. We regret the inconvenience caused due to weak test-cases in ICL1903. Official, detailed editorials will be published in couple of days.

Alright guys, the wait is over.
Top 2 Indian participants are:
1. @chant_coder
2 @hitman623

Winners are requested to drop a mail at for further details.

You can find the overall ranklist here. Kudos to the winners and all the awesome participants!


Hey, so the prize would be given considering div2 ranklist or overall ranklist?

Hey, the prizes will be given according to the overall ranklist. Note that the live ranklist will only feature div. 2 contestants. We will get the unified ranklist only after the contest ends and the prizes will be given away accordingly.

When will the winners be announced? And when will the ranklist be updated with Div 1 participants?

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  • Test cases of ICL1903 seem very weak.
  • ICL1905 is almost a copy of 1107G?!

Very poor contest and not upto the STANDARDS of codechef, can we have a dislike button feature in the new Codechef discuss forum ? :slight_smile:


Hey, We are extremely sorry for the weak test cases in ICL1903. It was unfortunately a last minute mistake. Regarding ICL1905, i think it is fairly different from the CF problem. The D array is non-sorted and the gap function is redefined. We tried to simplify the solution to match div2 levels. We initially had planned a dynamic convex hull trick solution for this problems with strict constraints but reckoned it would be unfit for this contest. Again, we are extremely sorry for the weak test cases in ICL1903. Thank you for participating though!

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Please explain ICL1901 problem given test case ?
123 5

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First round:
123*13*11*7 = 123*1001 = 123123
Number of distinct 3 digit numbers possible by using each digit of 123123 exactly once is 24.
(Other than 111, 222 and 333, any number is possible).
Second round:
By passing numbers which have 2 1s, we can get 111. Similarly for 222 and 333.
Thus, after 2 rounds, we can get all the distinct 3 digit numbers possible from the initial 3 digits.


Number of distinct 3 digit numbers possible by using each digit of 123123 exactly once is 24.

How value 24 came ?

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Still waiting for the unified ranklist. Idk why codechef takes so much time for such things. :confused:


Can someone please explain the logic of ICL1904 ? and where are the detailed editorials of this contest ??


You can use basic permutations and combination of find that out.

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It’s been a while, when will the ranklist be released? Any updates?

Loved ICL1905, though I was not able to solve it :stuck_out_tongue:

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Not yet. Will attach the list at the end of the post when it’s available with me. Also editorials coming out tonight, thanks for being patient!


yes waiting for those editorials

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I wish Codechef spent more time in doing useful things like MOSS process, ranklist update, laddus distribution etc rather than creating shitty features like “Funny Mode” :slight_smile:


“…editorials coming out tonight…”

April Fools’ prank?


Undoubtedly! I honestly don’t understand such delay in work!