#  User  Rating 

1  tourist  3757 
2  jiangly  3647 
3  Benq  3581 
4  orzdevinwang  3570 
5  Geothermal  3569 
5  cnnfls_csy  3569 
7  Radewoosh  3509 
8  ecnerwala  3486 
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#  User  Contrib. 

1  maomao90  171 
2  awoo  164 
2  adamant  164 
4  TheScrasse  159 
5  nor  155 
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8  Petr  147 
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10  pajenegod  145 
0
Try to solve problem D if we consider subsequence instead of substring! It is an interesting problem. 
0
yeah thank you ! I just realised that 
+11
Can we say goodbye for "good bye 2020"? 
+15
why the time limit for the problem E is so tight?? My solution O(n*60) gives me TLE but the solution O(n*MaxLog(Xi)) gives accepted. I don't know but i think you should increase the time limit a little bit:( 
On
Endagorion →
Codeforces Round 596 (Div.1, Div.2) and Technocup — Elimination Round 2, 5 years ago
0
I solved it like this and it worked 
0
Thanks, first i think like that but i consider that we will have inclusion exclusion principle 
0
how to solve I,i try a O(n(logn)^2) approach but it fail 
0
https://ide.geeksforgeeks.org/qorNuMmBI1 link to code for k 
0
In D , the lenght of S should be <=10^4 it will be more fair 
0

+47
Can you make the round unrated for only people affected? 
0
yes we can prove that it is optimaly to use powers of 2 as a base to generate 1 to x, consider you have n buckets of number(1,2,4,..2^(n1)) those numbers can generate 2^n ****different numbers**** and all this numbers are between 0 and 2^n1 so if we get the binary representation of x and let d is the highest significant bit so let x=1000010101 if we want to generate all number less than x and the highest significant bit is less than d we want n1 number and these number generate 2^(n1) different numbers and finaly to get x we only need one another number wish is 2^d and like this you get all number between 0 and x with least size of bucket 
0
can we solve G without dp?can we say that dp(l,q)=((l+1)^ql) 
0
ohhh you did not participate why you still wating>? 
Name 
