I love Codeforces.

Frfr

Just curious, are you still continuing with Project Euler until you reach 1000 problems or more?

Reading this blog also makes me want to try Project Euler, but I’m not sure how to start. Do you think solving one problem per day is doable, or are there some problems that would take multiple days to think through?

I don't know either. simasimarisa, sorry for pinging you, can you help him :>

Some (other) Asia-Pacific Teams:

• BurnedChicken, syl123456, abc864197532, Taiwan, National Taiwan University.
• kizen, windva, leinad2, Korea, Seoul National University.
• SorahISA ub33 nella17, Taiwan, National Yang Ming Chiao Tung University
• Gabp DippleThree ISPADEI, Philippines, De La Salle University
• Zanite bonk Chrisedyong, Indonesia, BINUS University

• bkdn24.leevox bkdn24.Ravenous BaoJiaoPenguin, Vietnam, University of Science and Technology — The University of Danang

• slah007 leo020630 kwoncycle, South Korea, Pohang University of Science and Technology

• cn449 sheyasutaka physics0523, Japan, Kyoto University

• simasimarisa ponjuice Series_205, Japan, Institute of Science Tokyo
• sounansya ryu150 i18hiromasa(AtCoder), Japan, National Institute of Technology, Tokuyama College
• yuto1115 penguinman Wailydest, Singapore, National University of Singapore
• middle_man Kaling mjhmjh1104, South Korea, KAIST
• marvinthang lmqzzz nguyentunglam, Vietnam, National Economics University
• yijw0930 gs25 stonejjun, South Korea, Korea University

Join OCPC to get:

gud luck to your team too :3

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Also, is this team 3 VNese?

I also added some teams on Codeforces, including:

• Kuuhaku
• N-Dimensional Sum
• DolAndDool
• Jägermeister
• Penguin Feeders
• 404 Skill Not Found
• Lebah Ganteng
• PhoKing
• Heroku
• BKDN.Arcane
• Arcane
• nameless
• 7 is our favourite number
• Codemaxxing

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Updated til this

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Damn, huge effort for this :3

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Auto comment: topic has been updated by DeMen100ns (previous revision, new revision, compare).

Congrats for #49

As a tester, meow

https://codeforces.me/profile/DeMen100ns

https://codeforces.me/profile/shironeko1702

As a tester, problems is excellent and cute :>

it's just rollback rating issue :") sr

As a do-nothing-setter, gluck everyone!

You can use Open in IDE button. It exists for every USACO problem.

Auto comment: topic has been updated by DeMen100ns (previous revision, new revision, compare).

Thanks for report, this is a mistake, i'll fix it now

Auto comment: topic has been updated by DeMen100ns (previous revision, new revision, compare).

As a stupidest author, love you SPyofgame

I think only Div.1 should be unrated because there isn't any problem with Div.2

Upd: I was wrong because E Div.2 is also well known:

As a tester, this round's problem-set interested me so much. Highly recommend reading all problems.

b-but isn't this div2 :>

As a stupidest tester, GLHF everyone :3

Veryyyyyyyyyyyy near...

f--k you SPyofgame

Nice video

And now i have delta -108 instead of -3 :(

You can read the section 10.4 of CP Handbook to see why this algorithm got accepted. In the contest, i got accepted with same strategy but used bitset instead.

If you want the solution without bit optimization then you can going to a solution with SQRT decomposition.

If $$$n \lt m$$$, you can use a 2D array $$$ans$$$ that contain the solution of queries 2 for any $$$(i,j)$$$. Then you can update (query 1) in $$$O(n)$$$ and get the answer (query 2) in $$$O(1)$$$.

If $$$n \gt m$$$, you can easily get the answer with brute force in $$$O(m)$$$ and update in $$$O(1)$$$.

Note that $$$min(n,m) \le \sqrt{nm}$$$, so the complexity of this algorithm will be $$$O(nm\sqrt{nm})$$$. As i remember, in the contest, there is a constestant got accepted with this solution

Simpler code for C with O(n) complexity

how so orz sir

I think maybe it can solve this problem : IOI 2005 — Mountain

+2, if you need an another-orange tester, ping me :)

We call 2 subsets have sum $$$S$$$ with $$${a[i1], a[i2], ..., a[ik]}$$$ and $$${a[j1], a[j2], ..., a[jr]}$$$ We know that these $$$r+k$$$ elements have distinct index in $$$a$$$ which mean $$$r+k \leq n$$$.

Let build array $$$b$$$ in this way :

• $$$b[1]=0$$$

• $$$(2 \lt =y \lt =k)$$$ $$$- \gt $$$ $$$b[u+1]=b[1]+a[i1]+...+a[iu]$$$

• $$$(k+1 \lt =u \lt =k+r)$$$ $$$- \gt $$$ $$$b[k+u+1]=b[1]+a[j1]+...+a[ju]$$$

For any element in other $$$n-(r+k)$$$ elements in $$$a$$$, we can easy set to last $$$n-(r+k)$$$ elements in $$$b$$$. Now we have array $$$b$$$ with $$$n+1$$$ elements and we also have $$$b[k+1]=b[k+r+1]=S$$$. Now just remove $$$b[k+1]$$$ to make array $$$b$$$ with $$$n$$$ elements.

The target of this problem is to build an array b that contains n numbers instead of n+1 so that if there are 2 subsets have the same sum S, you can build an array b having n+1 numbers with S appears twice, and you just delete one of it to make an array b have n numbers.

Me when in the contest :>

When hackers get hacked :>

thanks alot :))

"DELETED"

Maybe 1439A2 - Binary Table (Hard Version) :D

Hi, can you predict for us

DeMen100ns

Mondeus

:D