They will live stream stuff, the schedule is here: https://tco19.topcoder.com/onsite-events/watch

The first one starts in half an hour.

Direct twitch link: https://www.twitch.tv/topcoder_official

Sadly all the algorithm rounds are at late at night.

They should have had them in the first half, when Eurasia is awake. Anyway.

Link: https://www.facebook.com/hackercup

Quoting:

Tune in here to see this years winners LIVE reveal at 10am PDT / 1pm EDT / 6pm GMT / 10.30pm IST.

It starts in an hour.

Edit: Submissions so far: https://www.facebook.com/hackercup/scoreboard/327966064573355/?filter=everyone

Edit: Live stream link https://www.facebook.com/pg/hackercup/videos/?ref=page_internal

Congratulations to winners:

1. tourist
2. LHiC
3. Petr

Petr was first one to solve all problems though.

Here: https://docs.python.org/3/whatsnew/3.8.html

For competitive programming this seems useful:

Single Round Match 757

Topcoder SRM 757 is scheduled to start at 21:00 UTC-4 on April 29, 2019.

Registration is now open for the SRM in the Arena or Applet and will close 5 minutes before the match begins, so make sure that you are all ready to go. Click here to what time it starts in your area.

379F - New Year Tree

Video Link

I wanted to go into details — but it would have been a full 30 minutes. Any suggestion/query is welcome.

Bug in search box. Reproduction: 0. Visit : http://codeforces.me/contest/351/standings 1. Click on search arrow, the search box now shows up. 2. Type 'a'. Then press 'backspace'. 3. The page should have all the 100 results, but it is empty.

Did anyone notice this: http://finance.yahoo.com/news/biggest-cyber-attack-history-taking-141400441.html

When I run "Run systest", it gives me segmentation fault. When I try the same test case in the applet's editor, it gives correct result.

Weird Compiler!

class TeamContest
        { 
        public: 
        int worstRank(vector <int> strength) 
            { 
            	if((int)strength.size()  == 3) {
            		return 1;
            	}
            	vector<int> remaining;
            	for(int i = 3; i < strength.size(); i++){
            		remaining.push_back(strength[i]);
            	}
            	sort(remaining.begin(), remaining.end());
            	
            	int mn = min(strength[0], min(strength[1], strength[2]));
            	int mx = max(strength[0], max(strength[1], strength[2]));
            	int sum = mn + mx;
            	int end = (int)remaining.size() - 1;
            	int required = sum + 1 - remaining[end];
//            	cerr << required << endl;
            	for(int i = 0; i < remaining.size(); i++){
            		cout << remaining[i] << "  ";
            	}
   //         	cerr << "-----------------------\n";            	
            	int ind = 0;
            	int cnt = 0;
            	while(((ind < end &mdash; 1))){
            		while((ind < end &mdash; 1) && (required > remaining[ind])){
            			ind++;	
            		}
            		if(ind >= end -1) break;
 //           		cerr << "[" << ind << "," << end << "]" << endl;
            		ind+=2;
            		end--;
            		cnt++;
            		if(end >= 0 && end < remaining.size())
            		required = sum+1-remaining[end];
            	}
            	return 1 + cnt;
            
            } 
        
        };
[/code]

See next SRM is on 13 dec 2012 — http://community.topcoder.com/tc

Contest @ https://summergames.interviewstreet.com

You will need an account though. (or facebook login is also there)

http://community.topcoder.com/tco12/algorithm-schedule/

See You there.

Here is the problem statement- if u can login — https://bytecode.interviewstreet.com/challenges/dashboard/#problem/4f497b0018bad

otherwise — http://pastebin.ca/2121976

My approach,

Make a cumulative sum array.

Now for every index starting from 0, do binary search for k, k+mod, k + 2*mod, ..., k + 76*mod on range limited by previous max-length found, i.e. if we already have found an array of length L, which sum upto k (after doing mod) , then accordingly we have to modify the binary-search space, because we do not require a shorter-length answer.

Will this approach time-out ? As I can't submit there anymore, I need your help to see whether it is correct approach or not.

This is how we can use hashing to solve #Petr :

First how to hash all sub-strings of S of length n in O(n)  :

Let the string be  "abcdefgh"

We will do cumulative hashing as following :
Lets take a prime number p;
(here a , b , c , ... , h are ascii values of 'a' , 'b' ,'c' , .. ,'h' )
 hash[0] = a;
 hash[1] = a*p + b;
 hash[2] = a*p^2 + b*p + c;
 hash[3] = a*p^3 + b*p^2 + c*p + d;
like-wise up-to hash[7] = a*p^7 + b*p^6 + ... + g*p + h;

So, you must have figured out the recursive relation for finding hash[x].
hash[x] = hash[x-1]*p + s[i].
(base case: hash[0] = s[0] )

Now as we are done with finding cumulative hash value. We need to find the hash value of any substring.
Lets see this thing:
Let there is a string "dabctabc" ,
now "abc" at 1 and "abc" at 5 are same strings , and we must wish to have same hash value for both of them . We propose that every hashed value should be of form :
(for a substring of s from i to j inclusive.)
n = length of the whole string .
s[i]*p^n + s[i+1]*p^(n-1) + .. + s[j] *p^(n-length_substring);

Can you feel/see that if we maintain this proposition above , "abc" at 1 and "abc" at 5 will have same hash values .
(make sure we understand this before moving further)

For maintaining above proposition  , we maintain an array: step[]
step[x] = step[x-1]*p ;
(base case: step[0] = 1)

Now from hash[] and step[] we can calculate hash values of each of the substrings.
Lets remind us we are talking about the string "dabctabc" :
For any substring starting at i and ending at j (inclusive) :
for "abc" at 1:
substring_length = 3.
so, hash_value should according to our proposition should be:
a*p^7 + b*p^6 + c*p^5 .
See,
hash[3]  = d*p^3 + a*p^2 + b*p + c;
hash[0]  = d;
so our required value will be : (hash[3] - hash[0]*p^3)*(p^(8-3))
i.e (hash[3] - hash[0]*step[substring_length])*step[n - length_of_substring]

So, generalised formula for any substring starting at i and ending at j (inclusive) :
(hash[j] - hash[i]*step[substring_length])*step[n - substring_length] )

Now use the above formula for calculating hash value of "abc" at 5.
We will see that this "abc" also hash the same value that is "a*p^7 + b*p^6 + c*p^5".

Our job is almost done . :)






           

Code:

#include
int main()
{
    char *s[]={"codeforces","russia","contest"};
    char **p;
    p = s;
    printf("%s ",++*p);
    printf("%s ",*p++);
    printf("%s ",++*p);
    scanf("%*d");
    return 0;
}

I think it has happened a while ago on codeforces itself, but I forgot the reason why this happens:
See this code and code below.
http://ideone.com/N7vzQ

and added 1 "cout"  to the code at 119 and result changes:
http://ideone.com/OXHpO

By adding cout <<" "; should not change 39 to 2. :|

My guess is , its compilers problem.
Thanks in Advance.