Encoder, Techspardha'18 Announcement

Hello Community,

alt text

Encoder, Techspardha’18 will take place on Friday, October 26, 2018 at 21:30 IST with a duration of 2.5 hours. The round will be conducted on Codechef and is rated for Div 2 participants. It will consist of 6 problems. You can participate here.

About Techspardha: Techspardha is NIT Kurukshetra’s annual techno-managerial fest, consisting of various events. This year, it will be conducted from 26th to 28th October.

Encoder’18 is a short programming contest where you can show off your programming skills, compete with the best minds across the planet, win exciting prizes, be recognized or simply, solve for fun. I would like to invite you to participate in Techspradha’s annual competitive programming contest ‘Encoder’.

There will be ACM Style Ranklist- Users are ranked according to the most problems solved. Ties will be broken by the total time for each user in ascending order of time. There is no partial scoring.

The round has been prepared by l0gic_b0mb, yashgulani1997, sahil070197 and me (abhineet14). We hope you’ll find the problem set engaging, interesting and will have fun while solving the problems.

Prizes

5000 INR worth prizes for the overall winner (Div. 1 + Div. 2), 3000 INR worth prizes for the winner from NIT Kurukshetra.
Though the contest is rated for Div 2, the prizes are combined for Div 1 + Div 2. Register for prizes here.

Do participate and challenge your friends as well.

Nice Contest, Where we can find the rank list of Div1??

They will publish it soon. The @admins might be asleep right now and I dont wanna disturb them :3… so need to wait until tomorrow.

When will ratings be updated ?

“Soon.” We are waiting until tomorrow to analyze how many people in div2 might be affected due to Uptree problem issue by submissions or received mails. If its too many, we have to decide what to do next.

Can someone help me find error in this code for second question “PAINT THE WALL”…or just provide some corner test cases
SOLUTION_LINK

My approach:

  1. store for each height …the color and time

  2. then sort the height and remove duplicates…

  3. then iterate from largest to smallest and increment ans if time is more and color is different …

1 Like

when you sort the heights you are missing the order.
if the order is lost you don’t know which color is on the top and which color is hidden

have they published?

@vivek_1998299 You’re probably 13th-14th considering all those penalties XP

The ratings on the graph have been updated but the current rating has not changed

I received no reply yet. I will publish their short solutions shared with me.

It will, soon.

//