Where is the ranklist of CCWI2019

Hey, @onkar2000 where is the rank list, in whole contest, rank list is missing and after contest still missing, please update ASAP.

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Add link to the contest as well :slight_smile:

https://www.codechef.com/CCWI2019

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There was technical issue from Codechef side…We are in touch with the support team…Ranklist will be updated soon…Stay Tuned. Sorry for the inconvenience caused…

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Bro can you please provide me the editorial or approach for Problem: CCWI1908

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haan my approach was
if there are k envelops
put 1 crorepati in 1 envelop as long as possible
and put the remaining all in last enevelop
so if n>k-1:
ans=1/k+1/k+1/k+…(k-1) times
because we have put 1 crorepati in 1 enevelope and the probability of choosing each envelope is 1/k
if there are still some crorepati are left
then rem crorepati=n-(k-1)
and total number of cards in last coupon =n-(k-1)+m
winning probabilty is (n-k+1)/(n-k+1+m)
and the probability of choosing this coupon is 1/k
so ans=k-1/k + 1/k *(n-k+1/n-k+1+m)
if n<=k-1 then
we can keep n crorepatis in n enevelops and leave the remaining ones
ans=n/k

Example 1 :
n=5 m=2 k=3
put 2 creorepatis in 2 envelops
probability to choose these two enevelops=2/3
now put 3 crorepatis and 2 roadpatis in last enevelop
the winning probability =1/3 *(3/5)=1/5
ans=2/3+1/5
here find lcm=15
numerator=10+3
denominator=15
prob=13/15

Example 2:
n=2 m=2 k=3
put 2 crorepatis in 2 envelops and 2 roadpatis in 3rd enevelop
probability of choosing first 2 enevelops=2/3

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Thanks bro.
Really nice approach.

@admin please look into this matter.As the organizers of the contest have to choose the winners.

Bro can you please make the image of last question public.
Problem: CCWI1902

My rank was 29

.Are we eligible for finals?

Was the expected time complexity for the last problem : - O(V^3 . K) . where, V=600 and K=500 , this shouldn’t have passed the 1 second time limit, but many solutions (almost all) had this time complexity and they got AC.

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Graph-K?

Summary

Expected TC should be O(M*K)
CodeChef: Practical coding for everyone 2nd Subtask.

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Here the graph is a very bad graph with multiple edges and self-loops, so I stored a[i][j]=(number of direct-edges from ‘i’ to ‘j’).
Now, for k=2, answer= (Matrix)^2
For k=3, answer=(Matrix)^3

Also, they are asking for all k's from 1 to 500, not just for a particular ‘K’ , so can’t use that method…(Maybe I am wrong?)

And, to my surprise this gave AC,(matrix-multiplication, 'k'-times)

I did not even do matrix exponentiation because they ask for all k’s :frowning:

Thanks for the link !

Ok. Got it. Yup, if I knew the tcs would be weak I would submit it in the contest for sure.

I don’t see any solution with O(V^3 * K). Almost solutions have time complexity of O(M * K).

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@anon55659401 No you are not eligible for the finals