Happy New Year dear friends. May this year be the cause of the best success for all people! I wish you health and happiness!

Hi there, I'm new of codeforces, so to be consistent here I'm starting a 365 days of code streak. And this is also one of my New Year Resolutions. if you are a beginner, you can join also.

Rules:

1) Write a code for at least 1 problem on codeforces. You can start with the problems with minimum difficulty i.e 800. These problems take 5-10 minutes on average. Try solving one now https://mirror.codeforces.com/problemset?order=BY_RATING_ASC&tags=800-

2) After solving the problem take the screenshot of your whole activity frame like this and send in the whatsapp group. The reason for using whatsapp is to stay connected with the group daily as most people open whatsapp at least once a day. You can mute the group also. Can be moved to a website later.

3) Don't forget to have fun!

register here: https://forms.gle/X3FJ4Wug33npFmYk6

'Please print the answer multiplied by the number 2023.' never mind...

Maybe Hello 2023 will be enough, if you get a rank of 500-600.

Best of luck.

around the same

Actually it depends on your performance and mentality, I've bad contest consecutively and get something like -250 (because I was just bored from contests, I didn't wanted to solve after B or something like that I was just bored), I became a spec after that while I was stuck on upper expert for months. But I was actually better than that, I earned +479 in 3 contests and became a candidate master finally.

Waiting for the time where tester asks for the contribution after the contest or tester criticize the problem set may be the problems those were excluded.

Hoping for +82 and finally achieving Cyan for the first time.

same

I mean, tourist isn't exactly undefeated. He has been passed a couple of times, including by Benq.

He always comes back to the top tho

The round is combined Div 1 and Div 2. That means that problems A-F will be like div 2 A-F and problems C-H will be like div 1 A-F.

This is amazing

Didn't ask.

$$$\left ( 2 + 0 + 2 + 3 \right ) ^{\left ( 2 + 0 + 2 + 3 \right )} \equiv 2023 $$$ $$$mod$$$ $$$ (2 + 0 + 2 + 3)!$$$

look Score distribution, and rate of problems u solved, the most of global round is good for everyone.

jiangly

good luck

Still I got wa on tc 67 https://mirror.codeforces.com/contest/1779/submission/187858821 Got ac later though.

Yes, and the tourist is Messi, he will take first place in the next championship.

bro in 2nd problem I'm printing 1 and -1 sequence for even and Not possible for odd

still showing WA why ?

priority_queue

Hint: priority queue

D I got TLE with Segement Tree for some reasonmy sub

Actually D <<<<<<<<<<<<<<< C if you know

There are two corner cases, which i found at the end of the contest. First m=1 (we don't need to check a1 <= 0). Second: (because of this wa2) the sum of prefix a1...am may be positive, neutral and negative(not only non-positive). The solution goes from transformation of inequality(destroy identical terms from left and right and you will get necessary and sufficient condition), given in the task for k>m and k<=m and considering corner cases...

I have done like this. Let the entire sum be s. So A[0] + (A[1] + A[2]) + (A[3] + A[4]) ......(A[n-2] + A[n-1]) = s => A[0] + (n/2)*s = s; => A[0] = s(1-n/2)

now A[1] = s-A[0] and similarly you can calculate for A[i]'s.

for n=3 . its possible A[0] = 1(1-3/2) = 0.

You can list n equations. And then, the odd position number is all equal, and the even position number is all equal as well.

Let $$$x = \lfloor \frac{n}{2} \rfloor$$$, then $$$a_{odd} = x, a_{even} = 1 - x$$$. The sum of this array is 1.