Is this comment of yours an intended one to troll not-PikMike? :thinking:

Every New Year, most of the codeforces community be like. xD

All of them are from the other members of our team. I prepared lots of problems, however, I only own like 5 ideas of all those contests. The truth hurts!)

Google)

Можешь попробовать отправить решение из разбора, если хочешь.

Pretests might be weak but you are weaker. You would've found them if you solved G.

ultraStackDeveloper just comment out the fast input and submit again

I though the same but after seeing the number of submissions,figured it was probably easy.

In problem C, I had done sorting of pairs in asscending order and then checking it belongs to which group gives WA on test 3. like this:

for(int i = 1; i < n; i++){

if(v[i].fi > v[i-1].sec){

            ans[i] = 1-ans[i-1];  // swapping grp from 1 to 0 or 0 to 1.
        }
        else{
            ans[i] = ans[i-1];  //keeping it same.
        }
    }

can someone tell me the correct approach for it?

http://codeforces.me/blog/entry/62602

This is some of my ideas for problem G (I hope it wasn't totally correct because I did not finish coding it :( ) :

1, Calculate the prefix sum array f_i

2, Represent each f_i as a binary vector.

3, Calculate the dimension of the vector space spanned by the vectors that I found in Step 2. If the dimension is 0, print out -1, otherwise print that dimension.

We also need to check if f_n is an element of some basis.

i got TLE till the end. fml need to learn cpp now.

https://codeforces.me/contest/1101/submission/48253797

Yeah..What was that, i mean because of that i got 2 penalties.

F

Binary-search can't pass timelimit!

shuffle the array randomly first

https://codeforces.me/contest/1101/submission/48246817 https://codeforces.me/contest/1101/submission/48247813 Exact same code but TLE on one.

So this round should be unrated right?

4

2 3

3 4

6 7

7 8

Ans — 1 1 2 2

Think of it simply and greedy, you will figure out that we should find first "[...:" and last ":...]" of the string which satisfy two ':' are different, then just choose all '|' between two ':'.

Same happened with me.Completely forgot to check dashboard while solving C ,otherwise i could have solved E

I solved it by finding the non intersecting segments in the union of all the given segments. After that just put the first segment in one of the group and all others in the second group.

Sort segments by L. Put segment 0 in group 1 (set t[0] = 1). Also, let r = R[0] (right border of segment 0). Now, for segments 1 to n-1: if L[i] <= r, we must put it in group 1 (t[i] = 1), also r = max(r, R[i]), as we are continuing the group. Otherwise (L[i] > r), put it and all others to its right into group 2.

Note: for simplicity's sake I wrote t[0], t[1] etc, but since you sort the segments this won't work. There is a simple workaround to this, though.

binary is tle

I think you misunderstood the problem.