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 : https://www.codechef.com/ICL2019
Start Time : 20:00 IST, 28th March 2019
Prizes (tentatively) worth Rs 30000 for top Indian winners.
Registration for prizes: https://bits-apogee.org
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
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.
Winners are requested to drop a mail at email@example.com for further details.
You can find the overall ranklist here. Kudos to the winners and all the awesome participants!