Codeforces Round #817 (Div. 4)

2 года назад, # ^ |

How tf did you do this

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vitosevskich

2 года назад, # |

← Rev. 3 →

-106

Why did you put a Div.4 contest on Tuesday? It would be better if it was on let's say, Sunday because Div.4 contests always have a ton of participants. This way a lot of people won't be able to participate because of their obligations.

And virtual participation is just not the same as live participation because you can see how many people solved each problem till the end of the contest, solutions etc.

I am not writing this because I am mad because I won't be able to participate in the contest (in fact, I will be able to participate) but because a lot of people won't participate because the contest is on Tuesday.

MikeMirzayanov or whoever decides the day of the contest

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andrei_boaca

2 года назад, # ^ |

← Rev. 2 →

+15

This is the same as saying "Why is there a div 2 on Thursday?". However, I haven't seen any contest being downvoted just for the day on which it will take place and also a div 4 round may be really good for low rated people to boost their CP confidence through educational problems.

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Tom66

2 года назад, # |

← Rev. 2 →

Strange. Why all of the commments are downvoted???

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YariKartoshe4ka

2 года назад, # ^ |

-57

The comment is hidden because of too negative feedback, click here to view it

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BallBreaker

2 года назад, # ^ |

-16

look at my icon

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2 года назад, # ^ |

-24

You almost got me, unfortunately my wifi sucks so the video didn't load

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mohamedhany1024

2 года назад, # ^ |

Ha! YouTube ads saved me from that clever trick! Thx youtube!

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d_sqaure

2 года назад, # |

← Rev. 4 →

-29

Expecting some intresting problems

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SlavicG

2 года назад, # |

As a tester, I don't like sausages.

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qwexd

2 года назад, # ^ |

-53

As a tester, I disrespectfully disagree with your opinion. Sausages could have increased round quality so much.

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chromate00

2 года назад, # ^ |

-22

As a participant (out-of-competition), why?

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qwexd

2 года назад, # |

-30

Why Top G wasn't mentioned in the blog? I thought he tested the round

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actual_truthseeker

2 года назад, # ^ |

they hate a user with the name andrewtate for no reason just because of the name. god these people

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MrSavageVS

2 года назад, # |

-11

Div 4 arrives... Every Specialist be like: First unrated contest for me!

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Pew_Pew.

2 года назад, # |

-22

My first unrated contest

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Tkm_algo

2 года назад, # |

-12

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TimDee

2 года назад, # |

-13

Spoiler

a tester, mesanu should add me in this blogpost.

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bycicle

2 года назад, # ^ |

when will ratings change

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TimDee

2 года назад, # ^ |

When I will decide.

Now I don't want.

TimDee

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Miyano_Shiho

2 года назад, # |

← Rev. 2 →

What's wrong with the people who are downvoting every single comment? Is everyone supposed to not like these rounds just because you don't? This community is weird

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DonSilvio

2 года назад, # |

-6

Particularly liked the part about urging people not to register new accounts solely for narcissistic purposes =)

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Marinush

2 года назад, # |

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Cipher100

2 года назад, # |

+43

Meme

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coderdhanraj

2 года назад, # |

+10

Damnn! 5000+ unrated accounts are participating in this contest, isn't a huge number? what do you say?

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Another_Alt_123

2 года назад, # ^ |

how did you count?

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coderdhanraj

2 года назад, # ^ |

link

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Miyano_Shiho

2 года назад, # ^ |

← Rev. 2 →

The standing will probably look pretty weird Edit: Forgot about the trusted thing. Sorry..

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flakes24

2 года назад, # ^ |

Only trusted participants are shown in standings.

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flakes24

2 года назад, # |

← Rev. 2 →

+12

Is the starting time changed?

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canonica1

2 года назад, # ^ |

Yes

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Tkm_algo

2 года назад, # ^ |

Yes~10 minutes

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Yagnik007

2 года назад, # |

Any reason for delay of 15 minutes?

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MARSHALMATHERS

2 года назад, # ^ |

This delay is a complete waste of time coz can't do anything good in this delayed time

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Nero_noob

2 года назад, # ^ |

← Rev. 3 →

-15

You can downvote my comment meanwhile

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beethoven97

2 года назад, # ^ |

Now most people don't waste their time and started downvote your comment ^^

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sahilanand

2 года назад, # |

Why contest postponed?

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Tkm_algo

2 года назад, # ^ |

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bagadbilla

2 года назад, # |

Hope at least today I won't see the grid.

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HappyIvan

2 года назад, # ^ |

Just look at problem F...

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puzzlesad

2 года назад, # |

+26

My first coding competition ever! Kinda excited

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https://mirror.codeforces.com/contest/1722/submission/170265006

2 года назад, # |

+71

As a tester (I really tested it tonight!) I updated the post with the following text:

The opinions of testers about the order of problems are very contradictory. Please do not rely heavily on the order of problems in the round and read all the problems. Good luck!

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BallBreaker

2 года назад, # |

Thank you for this contest!

Favourite problem is E it is simple to understand, but it isn't easy.

Nice Contest!

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AlexMari

2 года назад, # |

-34

Wtf problem F sick implementation?!Stop proposing this kind of problems!

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Suri429

2 года назад, # ^ |

+33

it's not at all a bad problem for Div4. If you find there are lots of if-else, maybe you need to look for better observations.

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SourabhCh

2 года назад, # |

how to solve E?

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orange_trie

2 года назад, # ^ |

I don't want to spoil the whole solution ... so I'll give hints since height and width of rectangles are under 1000,

hint: use a matrix such that matrix[l][r] = Area of all rectangles with height <= l && width <= r.

Basically a prefix sum matrix.

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Hyrulia

2 года назад, # |

+12

Contest was fun and tasks were great. Thank you for this Div4!

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snowysecret

2 года назад, # |

+18

if anyone is interested, I recorded myself doing all problems

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Saucemaster102

2 года назад, # |

if only for E it was hs<=hi<=hsb and ws<=wi<=wb, it would have been so much easier, well anyway can someone give me a hint, i stored areas of all rectangles , sorted them , stored prefix sum array, then searched idx_left=upper_bound(hs*ws) and idx_right=lower_bound(hb*wb)-1 on sorted areas, then got the sum with the perfix sum array, can i tweak this to fit the hs<hi<hsb and ws<wi<wb constraints? also brute force gave me TLE X_X

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dimasavva17

2 года назад, # ^ |

I just did hs--, hb++

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Khedr

2 года назад, # |

Why this code got runtime error at test 8?

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Kneee

2 года назад, # ^ |

← Rev. 3 →

for each h, can be more than 1000 width

Runtime Error: This line is not true (line 26)

vector<vector> pre(1001, vector(1001));

Solution: You can resize each i in for loop of prefix sum

for(int i=1;i<=1000;i++) {
    pre[i].resize(v[i].size());
    for(int j=0;j<v[i].size();j++) {
        if(j == 0) pre[i][j] = i*v[i][j];
        else pre[i][j] = pre[i][j-1] + (i*v[i][j]);
    }
}

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osource

2 года назад, # |

I found F easier than E. The time limit of 6 seconds on E allowed my O(q*1000*logn) solution to pass.

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shaheeneallamaiqbal

2 года назад, # ^ |

← Rev. 2 →

can u please share your approach for E?

i used binary search four time for solving each query. my code link:https://mirror.codeforces.com/contest/1722/submission/170286434

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Kneee

2 года назад, # ^ |

-11

that is not true, because when you search for a range assume the height you got a range from L to R satisfying the condition hs < hi < hb inside this range, you can't search for another range for width because it is not sorted (monotonic function) so binary search fails here.

Think about the constraint of h and w

Code: 170265292

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SilverWing05

2 года назад, # ^ |

Yea F was actually easy, I'm suprised so less people solved/attempted it

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Cuellius

2 года назад, # ^ |

Btw, $$$O(q n)$$$ solution passes too.

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onebit1024

2 года назад, # ^ |

no?

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secret-void

2 года назад, # ^ |

Seems like you got hacked!

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Cuellius

2 года назад, # |

What is the point of $$$\mathrm{TL} = 6 \,\mathrm{s}$$$ in the problem E? To make stupid brute-force solution pass?

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mesanu

2 года назад, # ^ |

not every language is as fast as C++

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vitosevskich

2 года назад, # |

← Rev. 2 →

I was so close to solving 1722G - Even-Odd XOR (or I at least think so)! Can anybody please explain to me how to solve it?

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Pi61

2 года назад, # ^ |

← Rev. 2 →

Here's my solution: Observation 1: If the xorOdd == xorEven, then xorOdd ^ xorEven == 0, which mean xor of all elements is equal to 0.

This leads to observation 2: If n%4==0 then the answer is 0 to n-1. Consequentially, answer for n%4==3 is 1 to n.

if n%4==2 then we take the answer for n-3, then we add 3 numbers of our choice (mine was 2^30, 2^29 and 2^30 + 2^29).

if n%4==1 then we take answer for n-6, then we add 6 numbers.

AC code: 170284159

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robostac

2 года назад, # ^ |

← Rev. 2 →

I did it by adding incrementing numbers (from 0) until the array was length n-3. Then add 1 << 29 and 1 << 30. The final value is then the xor of the entire array (proven to be unique as it has both 1<<30 and 1<<29)

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2 года назад, # ^ |

You can check my code if you want

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k1r1t0

2 года назад, # ^ |

-10

if a == b, then a ^ b == 0
so you just need to make array xor = 0
you can make ai = i
and an = a0^a1...^an
now the only problem is that an can be equal to other value
if n is odd you can set 30th bit to 1 in all numbers except an
if n is even you can set 30th bit to 1 in an and some other ai (if an==a0 choose a1, else a0 for example)

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lbm364dl

2 года назад, # ^ |

Consider adding pairs of numbers which in binary are like $$$(x)0$$$ and $$$(x)1$$$, where $$$(x)$$$ are any other bits. Note that the xor will always be equal ignoring the least significant bit. When $$$n \equiv 0 \mod 4$$$, every $$$4$$$ numbers we will have the xor of the least significant bit be $$$0$$$ $$$(0 \oplus 0$$$ and $$$1 \oplus 1)$$$, so we see that the xors are indeed equal.

We now only need to reduce other cases to this one. You don't even have to think about it because the sample test cases already give you answers for $$$n = 3, 5, 6$$$, so you put those numbers first and now you add the remaining numbers, starting from, say, $$$20, 21, 22, 23...$$$.

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Andalus

2 года назад, # ^ |

← Rev. 3 →

If $$$n \% 4 == 0$$$, I gave the even indices values from $$$1$$$ to $$$n/2$$$ and the odd indices the same value except by adding a fixed high power of 2. The XORs on this high bit position cancel to 0 since there are an even number of odd indices (since $$$n$$$ is a multiple of 4).

For $$$n \% 4 == 1$$$, I did the same for the first $$$n - 1$$$ elements, then added an extra 0.

For $$$n \% 4 == 3$$$, I did the same for the first $$$n - 3$$$ elements, then added three values in which two of them XOR to the third by utilizing two high powers of 2 ($$$10 \oplus 01 = 11$$$, except these are at some high bit positions)

For $$$n \% 4 == 2$$$, I did the same for the first $$$n - 6$$$ elements, then added six values on high bit positions that get canceled out. This took considerably time, but I eventually came up with $$$1000 \oplus 0100 \oplus 0010 = 1001 \oplus 0110 \oplus 0001$$$ (these are four bits at some really high bit positions). I'm quite curious to see if there is a more elegant solution for this case.

170258132

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tsuu2092

2 года назад, # ^ |

We can easily make the XOR equal in a group of 4 numbers (i, i + 1, 2^k — 1 — i, 2^k — 1 — i — 1).

For example, k = 4: 1 (0001) ^ 14 (1110) = 2 (0010) ^ 13 (1101) = 15 (1111). We match the numbers so that the XOR result has k bits set.

Let's pick k so that 2^k > 200000.

What if the number of elements are not divisible by 4? Then we just need to add some numbers at the beginning of the sequence

n mod 4 == 1: begin with 0 because x ^ 0 = x
n mod 4 == 3: begin with 1 2 3 because 1 ^ 3 = 2

How about n mod 4 == 2? Since XOR of the rest elements are equal, then these two numbers should be equal too, which means they are not distinct. So starting with 2 numbers is not possible. But we can start with a sequence of 6 numbers. In the contest I found a sequence which is 1 2 4 6 8 9 using brute force.

Finally, subtract n by the numbers we have added to the sequence, we should have n divisible by 4 to generate the group of 4s.

The initial value i should be at least 10 to be distinct with the starting numbers.

Code: 170265768

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vitosevskich

2 года назад, # ^ |

Thank you guys.

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sgmft

2 года назад, # ^ |

You can just set the first n-1 values arbitrary, then the n_th value is the xor sum of the first n-1 values. It's almost impossible to hack.

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moonshine07

2 года назад, # |

when codeforces post its contest editorial ?

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SilverWing05

2 года назад, # ^ |

After hacking phase is completed.

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py0x7

2 года назад, # |

Can anyone explain the error in my submission?170267520

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HarshtJ

2 года назад, # ^ |

← Rev. 2 →

You are using char array but you are not resetting them after every test case. So if test cases are like this —

2
10
abcdetIMUR
5
Timur

Your code will output No for 2nd test case because strchr(str,"I") will return non null value

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amanpatel23

2 года назад, # |

Can someone hack my solution for problem G. I think my answer might generate duplicate numbers. Link

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secret-void

2 года назад, # ^ |

It won't generate duplicate numbers. Observe that all the numbers except the number you inserted, at last, are <= 2^30. The last number you inserted xor ^ temp, will actually be > 2^30, so it won't be duplicated

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10zin

2 года назад, # |

← Rev. 2 →

For the problem D i thought of using prefix sum. I was not able to implement it. What you guys did for problem D?

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Kneee

2 года назад, # ^ |

you can use an array of maps 170197935

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SilverWing05

2 года назад, # ^ |

← Rev. 3 →

What made you think to use prefix sum? Anyway, I did it by just storing who wrote which word and then checked how many people wrote it. This can be simply done by using map (I used set aswell, it just made things easier for me)

Edit: I just realised each person wrote only distinct word xD (but my solution also works for duplicates) , so there is no need for set actually, just store each occurrence

170203831

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2 года назад, # |

I think D can be done with dp. Here is my solution:

Every next answer is sum of previous answer and max points we can get by changing side of looking of some person. So ans[0] is score that we have in start. Max points we can add is actually biggest difference between points we can get from one person by its current looking and other side looking. So we can just make difference and sort it.

Here is code for my submission: https://mirror.codeforces.com/contest/1722/submission/170308742

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andrejikus

2 года назад, # ^ |

What is ans[1]?

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2 года назад, # ^ |

ans[1] will be max score we can by only one change of looking side(it can be called dp[1]).Voli

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andrejikus

2 года назад, # ^ |

najace

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om1429888

2 года назад, # |

← Rev. 2 →

can someone please explain how to do E?hints please?

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2 года назад, # ^ |

Can you make use of h<=1000 and w<=1000 ?

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om1429888

2 года назад, # ^ |

the only thing that comes in my mind is brute force,i read about about people applying prefix sum but i dont have idea about prefix sum in 2d.

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2 года назад, # ^ |

You can make such that for every h calculate prefix for all possible w

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om1429888

2 года назад, # ^ |

thanks alot,learnt something new today.

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https://mirror.codeforces.com/contest/1722/submission/170276269

2 года назад, # ^ |

You're welcome

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javiergm06

2 года назад, # |

← Rev. 2 →

Why did my code gave WA test 2 on E?

Edit i saw it =(

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_icy_

2 года назад, # |

← Rev. 2 →

Can someone hack my solution of G?

I randomly fill an array with N-2 numbers and then just add two numbers to make XOR's equal. If I fail (i.e. needed number already in the array), I randomly fill the array again.

https://mirror.codeforces.com/contest/1722/submission/170260049

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dhiru_16

2 года назад, # |

← Rev. 2 →

-8

.....

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absolute-mess

2 года назад, # ^ |

Hi, no offence, but how did you solve problem G in just 3 minutes?

I guess it was not that much easy. Even the highest-rated coders took 5-6 minutes as well.

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vk20july

2 года назад, # |

← Rev. 2 →

-11

Can anyone Explain Problem D? I used two pointer and greedy approach but don't know why is it failing for some test cases?

Here is my code

Code:

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define loop(i,a,b) for(ll i=a;i<b;i++)
#define range(a)    a.begin(),a.end()
#define p(a)        push_back(a)
#define mp          make_pair 
#define pqb             priority_queue<int>
#define pqs             priority_queue<int,vi,greater<int> >
#define w(x)            ll x; cin>>x; while(x--)
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)
#define mii    map<int, int>    
#define umii   unordered_map<int, int>    
#define mll    map<ll, ll>    
#define umll   unordered_map<ll, ll>
#define pii    pair<int, int>    
#define pl     pair<ll, ll>
#define vi     vector<int>    
#define vl      vector<ll>    
#define si      set<int>    
#define sl      set<ll>    
#define vpii     vector<pii>    
#define vpl      vector<pl>    
#define vvi       vector<vi>    
#define  vvl       vector<vl>
#define fastinout ios_base::sync_with_stdio(false);cin.tie(NULL); cout.tie(NULL);    
#define MOD 1000000007
ll binExpIter(ll a,ll b,ll mod){long long ans=1;while(b>0){if(b&1){ans=(ans*a)%mod;}a=(a*a)%mod;b>>=1;}return ans;}
ll gcd(ll a, ll b) {if (b > a) {return gcd(b, a);} if (b == 0) {return a;} return gcd(b, a % b);}
ll phin(ll n) {ll number = n; if (n % 2 == 0) {number /= 2; while (n % 2 == 0) n /= 2;} for (ll i = 3; i <= sqrt(n); i += 2) {if (n % i == 0) {while (n % i == 0)n /= i; number = (number / i * (i - 1));}} if (n > 1)number = (number / n * (n - 1)) ; return number;} //O(sqrt(N))
vector<ll> sieve(int n) {int*arr = new int[n + 1](); vector<ll> vect; for (int i = 2; i <= n; i++)if (arr[i] == 0) {vect.push_back(i); for (int j = 2 * i; j <= n; j += i)arr[j] = 1;} return vect;}


//************************************//
int main(){
   w(t){
      ll n;
      cin>>n;
      string s;
      cin>>s;
      int x=0;
      int y=n-1;
      for(int i=1;i<=n;i++){
         int c=0;
         int ans=0;
         while(x<=y){
            if(s[x]=='L'){
               s[x]='R';
               c++;
            }
            if(c==i){
               break;
            }
            if(s[y]=='R'){
               s[y]='L';
               c++;
            }
            if(c==i){
               break;
            }
            x++;
            y--;
         }
         for(int k=0;k<n;k++){
            if(s[k]=='L'){
               ans+=k;
            }
            else{
               ans+=(n-1-k);
            }
         }
         cout<<ans<<" ";
         
      }
      cout<<endl;
   } 
}

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2 года назад, # ^ |

u can see my dp solution on this comment : https://mirror.codeforces.com/blog/entry/106422#comment-948238

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dhiru_16

2 года назад, # ^ |

It can be solved without dp as well

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2 года назад, # ^ |

u can see that my implementation is easier than greedy approach.

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vk20july

2 года назад, # ^ |

Actually, I didn't learn dp

Can you implement without dp

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tgp07

2 года назад, # ^ |

yeah

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mdibuhossain

2 года назад, # |

Today is Hacking day

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beethoven97

2 года назад, # |

Is there problems with hacking process, no one unable to hack this user for the same problem and it still in queue state?

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xiaxiakeji

2 года назад, # |

Too weak pretest for A. :(

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10zin

2 года назад, # |

Can D done with prefix sum?

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2 года назад, # ^ |

Different people said they solved it like that

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saurabh_175

2 года назад, # |

Hello everyone my solution for 1 problem is accepted during contest but now it's shows its hack and solution is wrong?? What is mean I didn't get. Can you please tell how I got my solution correct.

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2 года назад, # ^ |

It means a few things: 1. The round has open hacking, which continues (as far as I know) for 12 hours after the end of the contest; 2. Every participant(as far as I know) is able to hack anyone's solution. If you find a testcase that might make others' solutions fail, you can submit the testcase and hack their solution; 3. Someone saw your solution, found a testcase for it and hacked it successfully.

After the hacking phase, all of the submissions will be re-judged with updated testcases, which will include testcases of successful hacks(as far as I know).

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AntiXoR

2 года назад, # |

Mize orz!!

→ Ответить

https://www.geeksforgeeks.org/find-n-distinct-numbers-whose-bitwise-xor-is-equal-to-k/

2 года назад, # |

Can we have rating change quickly please

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AhmedEl-Gohary

2 года назад, # |

← Rev. 2 →

-8

Attention! Your solution 170296922 for the problem 1722G significantly coincides with solutions Gaurav__Arora/170273818, om_arsa_2312/170279944, nihal_gupta/170293955, AhmedEl-Gohary/170296922. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://mirror.codeforces.com/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.

So my evidence that I didn't cheat or communicate with any one in the round is that i used a pre-written blog on geeks for geeks about finding an array of size n whose xor value is some k and I adjusted that function to solve yesterday's DIV4 G, you can take a look at my submission and compare for yourself.

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https://www.geeksforgeeks.org/find-n-distinct-numbers-whose-bitwise-xor-is-equal-to-k/

2 года назад, # ^ |

Sorry, I reverted the punishment.

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helloIamCristiano

2 года назад, # |

-8

I have got a system message that my solution coincides with some other participant's solution for the question G, and as asked I am presenting here the common source published before the competition.

As I had seen this post earlier on geeks for geeks, I simply used it.

Please consider my participation for this contest.

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2 года назад, # ^ |

Sorry, I reverted the punishment.

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daijoub_dattebayo

2 года назад, # |

-8

I have picked a portion of my code from an article of gfg: (https://www.geeksforgeeks.org/find-n-distinct-numbers-whose-bitwise-xor-is-equal-to-k/) You can verify the same. Please look into this and consider my participation. MikeMirzayanov

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2 года назад, # ^ |

Sorry, I reverted the punishment.

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daijoub_dattebayo

2 года назад, # ^ |

I got my rating. Thank you.

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2 года назад, # ^ |

← Rev. 2 →

Congrats on the rank, but I got a little confused.

Your previous rating was 1289, mine — 1283; I placed 333, you placed 394; And you ended up receiving more points than me. How does that work MikeMirzayanov?

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niru0105

2 года назад, # |

Why ratings are still not updated?

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secret-void

2 года назад, # ^ |

They are usually updated around 10 PM (IST)

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2 года назад, # |

Was this round declared unrated?

I checked quite a few profiles from the comments and this round is marked as unrated for every single of those I checked. Any updates?

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_null_ptr

2 года назад, # |

← Rev. 2 →

-19

I have got a system message that my solution coincides with some other participant's solution for the question G, and as asked I am presenting here the common source published before the competition.

https://www.geeksforgeeks.org/program-binary-decimal-conversion/ Also used previosuly posted gfg function of binary to decimal conversion. As I had seen this post earlier on geeks for geeks, I simply used it.

Please consider my participation for this contest. Thank You

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2 года назад, # ^ |

I don't see in your submissions 170280792 and 170278642 any code/ideas from https://www.geeksforgeeks.org/find-n-distinct-numbers-whose-bitwise-xor-is-equal-to-k/ Can you point out where I'm wrong?

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_null_ptr

2 года назад, # ^ |

-19

I have been coincided wrongly just for using decimal to binary conversion function.Please make the contest rated for me.

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2 года назад, # ^ |

No offence, I might be wrong,

But even 90% of your comment is copy-paste of another comment..

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2 года назад, # ^ |

Stop deceive. You cheated and therefore were punished. Please don't distract me. Just don't do it again.

Other solutions: 170249794, 170266270, 170277345.

Your submission: 170280792

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CF-DIV1F

2 года назад, # ^ |

Spoiler

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nihal_gupta

2 года назад, # |

Hello I've got a system message that my submission to problem G coincides with some participants, but I've used the code present on GFG link to the GFG article Kindly consider my participation in this round. Thank You.

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https://www.geeksforgeeks.org/find-n-distinct-numbers-whose-bitwise-xor-is-equal-to-k/

2 года назад, # ^ |

Sorry, I reverted the punishment.

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pravin_as

2 года назад, # |

I have got a system message that my solution coincides with some other participant's solution for the question G, and as asked I am presenting here the common source published before the competition.

I have even put the link above in my submission. So please revert the decision.

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