Codeforces Round #219

11 лет назад, # ^ |

-110

And me, please.

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forthright48

11 лет назад, # ^ |

+25

Now I know what I will do the next time one of my friends forget to logout from their CF account 3:)

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11 лет назад, # ^ |

-23

It's my account :)

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ThatMathGuy

11 лет назад, # |

← Rev. 2 →

+14

No Friday 13 special contest?

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aradhya

11 лет назад, # ^ |

← Rev. 3 →

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Nickolas

11 лет назад, # ^ |

+78

No. The previous one was a Programmer's Day special, Friday the 13th was a coincidental bonus theme. I'm playing with some ideas for the next Surprise Language Round, but this will take more time to prepare. Unless someone else volunteers...

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marat.snowbear

11 лет назад, # ^ |

+15

Thanks for making these rounds.

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the_skinny_one

11 лет назад, # ^ |

← Rev. 3 →

It was Friday 13 special test #13 in problem B (Div. 2) — many contestants failed on this test =(

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Omelianenko

11 лет назад, # |

-46

Не совсем удобное время для Украины, пары заканчиваются за пол часа до начала.

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smx

11 лет назад, # ^ |

+67

Ты смотрел фильм "Форест Гамп" ?

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ifsmirnov

11 лет назад, # ^ |

← Rev. 2 →

~~Видимо, связано с вечерним топкодером.~~ Топкодер завтра в субботу, извиняюсь за дезинформацию.

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caustique

11 лет назад, # ^ |

Так Топкодер же завтра?

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ifsmirnov

11 лет назад, # ^ |

Да, действительно. А я уж было к двум контестам подряд приготовился.

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faiyaz26

11 лет назад, # |

+20

Problems are Made in Japan!! And anyone knows Products from Japan are of good quality. :)

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khatribiru

11 лет назад, # ^ |

-6

may be

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sbasrik

11 лет назад, # |

-18

What is the specific reason of being held unusual time ?

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DEGwer

11 лет назад, # ^ |

+28

In Japan, usual time is between 0:30 AM and 2:30AM, so we want to hold the contest in earlier time.

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sbasrik

11 лет назад, # ^ |

-9

as I guess. asked it because of probability of another reason, thanks.

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Ali1377

11 лет назад, # |

+13

New authors = new personalities! Hope good problems and short statements! GL & HF!

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tsfissure

11 лет назад, # |

It's a good time for Chinese participants;thx...But some College students will take a test called "CET4/6" tomorrow,so I think the Chinese participants will reduce...For best wishes...

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Lazzaro

11 лет назад, # ^ |

Dynamics of participating CF is continuous for no reason, though tomorrow's CET6.

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MikeMirzayanov

11 лет назад, # |

← Rev. 2 →

+116

Glad to see Japanese writers! Recently, I've read a post about Japan (in Russian). Can you approve (or comment) some facts about Japan:

On Valentine's Day girls do presents, but not boys.
Hentai is extremely popular and teens under 16 can easily and legally buy it.
Snowmen are made from two but not three balls.
Colonel Sanders (KFC founder) is one of the symbols of New Year Holyday.
Many weddings (30%) are result of special meetings organized by parents.
Flats and houses in Japan do not use central heating.
Many Japanese are afraid of travels and they think that USA is very dangerous for tourists.
No tips are in restaurants.
Japanese very quick become drunk if drink alcohol.
Some subway cars are women only. Is it because of groping?

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DEGwer

11 лет назад, # ^ |

+107

1st, 3rd, 8th, and 10th are just truth. 9th one depends on individuals, but Japanese tend to become drunk very quick. 7th one is not so much, but Japan is very safe country, so some Japanese think almost all of the world is dangerous. For 2nd, 4th, 5th, and 6th, I'm not sure. But I think 4th and 6th is not truth.

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Mukit09

11 лет назад, # ^ |

+24

Thanks !!! Many interesting things are known about Japan ... :)

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DEGwer

11 лет назад, # ^ |

+103

On Valentine's Day girls do presents, and usually the present is chocolate.

And we have White Day one month later than Valentine's Day. On this day, boys do presents for girls, and it is also chocolate.

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11 лет назад, # ^ |

+27

As far as 9. goes, I heard that it's a genetic thing — a large number of Asian population is missing some pathway that helps us cope with alcohol. What a pity! (as I'm writing this, there's a bottle of beer next to me :D)

3.: A snowman from 3 balls is too unstable, not to mention harder to make. I can totally understand that one :D

should not be true, as far as "legal" goes — it's too weird, even for Japan. (although my knowledge of Japanese culture is based mostly on a certain Gintama anime...) It's not like you can't get any porn easily anywhere in the world...

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2011330300713

11 лет назад, # |

Good luck to everybody!

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firmsnail

11 лет назад, # |

It's a good opportunity to back Div.1

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EnumerativeCombinatorics

11 лет назад, # |

+10

Oh today is 13th Friday and the problem writer m.kagamiz( kagamiz )'s name is Jayson.

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_777_

11 лет назад, # |

Good oppurtunity to enter Div1

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TsunamiNoLetGo

11 лет назад, # |

-16

I am getting extra penalty on problem B Div2 because of that mistake ! I submitted two accepted codes. Is there any hope that penalty can be deleted ?

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mysterion

11 лет назад, # ^ |

Write a message to jury, via ask a question functionality. That's what I did, and my extra submissions was deleted.

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Na2a

11 лет назад, # |

Опять очередь.

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mysterion

11 лет назад, # ^ |

У меня 5 минут уже висит посылка. У кого еще дольше?:)

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11 лет назад, # ^ |

-7

у меня дольше

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saba_tavdgiridze

11 лет назад, # |

slow judge:((((((((

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ffao

11 лет назад, # ^ |

+29

Counting Submissions on Queue is Not Fun

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najim4689

11 лет назад, # |

My crucial submission on C is hung in the queue for more than 8 minutes. What should be my strategy :( Actually what should be the strategy in such situation???

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11 лет назад, # ^ |

Try hacking or thinking about other problems... or trying to check your submission on C (generate your own testcases and test it yourself, look at the code and think about it again, etc). Or, you can do it like me: just relax :D

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najim4689

11 лет назад, # ^ |

Last one seems good :p

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nurzhands

11 лет назад, # |

No hacks round :( (almost)

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11 лет назад, # |

← Rev. 5 →

+22

i submitted problem B of div-1 at 01:59:48 after a long time of being stuck at it! my hands are still shivering a bit after that last-minute panic! i just hope that it gets accepted! :)

EDIT: it (5431756) got Accepted , yay! :)

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11 лет назад, # |

The time limit on C is really close! At least if the intended solution is an O(NM) one... I hope the pretests are as strong as the lack of hacks indicated :D

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11 лет назад, # ^ |

Theoretically, 4 seconds shouldn't be even close for O(45 millions)...

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11 лет назад, # ^ |

← Rev. 2 →

+34

Hush, I'm trying to get $\text{[math]}$ accepted here :)

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tunyash

11 лет назад, # ^ |

There is accepted solution with $\text{[math]}$ time. 5429539

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11 лет назад, # ^ |

+24

And there is my failed solution with O(NM) time. Wat?

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11 лет назад, # ^ |

+16

Codeforces administrators' position is it's better to allow good written non-optimal solution to pass, than allow bad written(or on Java) optimal solution to fail. There is some logic, but i think it causes more disbalance.

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kcm1700

11 лет назад, # ^ |

← Rev. 2 →

+45

I have O(M²) solution. It just took 15 ms. 5431649

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Rubanenko

11 лет назад, # ^ |

Could you describe it briefly?

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kcm1700

11 лет назад, # ^ |

+22

We know that B_i terms are really nothing.

Let F(x, t): the minimum sum of |A_i - previous_positions| at time t with position x

If you plot this function in fixed t, you'll see that this function consists of some line segments. You get the idea here.

If there's a new firework event, one must combine |x - A_i| graph into the function.

One have to see what happens if the time passes. The shape of the graph will be expanded d × passed_time unit in both sides with lowest point as its center.

The maximum number of the line segment is O(M) and the operations described above takes O(numberofthelinesegment) each time. The overall time complexity is O(M²)

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nownikhil

11 лет назад, # ^ |

Can you elaborate a bit more, I am not able to understand this.

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saketh

5 лет назад, # ^ |

Actually, it can even be done in $$$O(M \log M)$$$.

Let's maintain $$$F(x) =$$$ maximum happiness we can attain if we end at location $$$x$$$. After $$$f$$$ fireworks have been launched, $$$F$$$ is a polyline with sections having slopes $$$-f, -f+1, -f+2, \dots, f$$$. We only need to know its constant value and the $$$x$$$-coordinates at which its slope changes in order to uniquely represent it.

We can keep a max-heap storing the slope changes left of the constant section and a min-heap storing the slope changes right of the constant section.

When $$$t$$$ units of time pass, we need $$$F'(x) = min_{x \in [x - tD, x + tD]} F(x)$$$. We need to decrease the coordinates of the slope changes to the left of the constant section by $$$tD$$$ and increase the coordinates of the ones to the right by $$$tD$$$. It can be done lazily by keeping a sum of changes to-be-applied associated with each heap.

When a firework is launched at $$$a$$$, $$$F'(x) = F(x) + abs(x - a)$$$. It can be done by introducing two new slope changes. At most one existing slope change will need to shift from the leaf-heap to the right-heap (or vice-versa). The constant term should also be recalculated.

Implementation: 60323635.

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sunzh

4 года назад, # ^ |

What is that you put in the priority_queue left and right?

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forthright48

11 лет назад, # |

I have a feeling that today's system test will take a while...

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ErvinK

11 лет назад, # |

Problem D div2/ B div1 core idea ?

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LashaBukhnikashvili

11 лет назад, # ^ |

← Rev. 2 →

it can be solved by DP(precalculations) in O(n^5) time .

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ErvinK

11 лет назад, # ^ |

I was thinking about some preprocessing idea, but I didn't get it? What is it ?

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LashaBukhnikashvili

11 лет назад, # ^ |

← Rev. 4 →

first of all you need to calculate rectangles sums which first coordinate is 1,1; after this you can calculate sum of any rectangles (in O(n^4)):

rectangle_sum(i,j,x,y)=sum[x][y]-sum[i-1][y]-sum[x][j-1]+sum[i-1][j-1];

if rectangle_sum(i,j,x,y)==0 then we can say that this rectangle consist only zero;

using this you can calculate for any (i,j,x,y) number of rectangles which second coordinate(bottom,right) is x,y and consist only zero (in O(n^4)).

and finally using this you can calculate ans for any (i,j,x,y); (in O(n^5),or O(n^4)).

now you can answer queries in O(1) time.

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11 лет назад, # ^ |

← Rev. 3 →

actually, it can be solved in $\text{[math]}$ time. i think $\text{[math]}$ would TLE.

EDIT: by $\text{[math]}$ i meant $\text{[math]}$ . sorry about any confusion my post caused.

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Thomas_Ahle

11 лет назад, # ^ |

Lyrical got accepted in QNM. Wonder how, my QNM takes 17 seconds :/

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shobhit6993

11 лет назад, # ^ |

← Rev. 2 →

Your solution looks elegant and most importantly small. Could you please explain what does dp[i][j][i+di][j+dj] denote and also provide some intuition behind this "super" recurrence relation

dp[i][j][i+di][j+dj] =
 dp[i][j][i+di-1][j+dj]+dp[i][j][i+di][j+dj-1]-dp[i][j][i+di-1][j+dj-1]+dp[i][j+1][i+di][j+dj]+dp[i+1][j][i+di][j+dj]-dp[i+1][j+1][i+di][j+dj]-dp[i][j+1][i+di-1][j+dj]-dp[i][j+1][i+di][j+dj-1]+dp[i][j+1][i+di-1][j+dj-1]-dp[i+1][j][i+di-1][j+dj]-dp[i+1][j][i+di][j+dj-1]+dp[i+1][j][i+di-1][j+dj-1]+dp[i+1][j+1][i+di-1][j+dj]+dp[i+1][j+1][i+di][j+dj-1]-dp[i+1][j+1][i+di-1][j+dj-1]+check(i,j,i+di,j+dj);

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11 лет назад, # ^ |

← Rev. 5 →

firstly, although the rest of the code looks elegant and small, i want to remind you that i was in a hurry as i submitted this code at 01:59:48, and i apologize that my recurrence relation looks very messy and requires this post for u to understand.

now onto my approach. as you can see i answer queries in $\text{[math]}$ , which means dp[i][j][k][l] stores the number of rectangles which contain only 0s in the subrectangle $\text{[math]}$ . so from here i will refer to (i,j) as the start-point, and (k,l) as the end-point.

di and dj denote the end-point assuming that the start-point is the origin. so if the start-point is (i,j) the end-point would be (i+di,j+dj). the only place to start is a 1x1 square, i.e. dp[i][j][i][j] (where di and dj are both 0), which would be 1 or 0 depending on s[i][j]. clearly we should progressively increase di and dj so that we can use the previous information to compute the currently required info. this is why i fixed di and dj and moved the start-point, rather than fixing the start-point and moving the end-point.

below i have elaborated on all the terms in the recurrence relation. i hope u are able to understand now. if not, send me a message and i will get back to u.

// i want the rectangles that start at or after (i,j) and end at or before (i+di,j+dj)
dp[i][j][i+di][j+dj] =

// rectangles that start at or after (i,j) and end before (i+di,j+dj)
+(dp[i][j][i+di-1][j+dj]+dp[i][j][i+di][j+dj-1]-dp[i][j][i+di-1][j+dj-1])

// rectangles that start after (i,j) and end at or before (i+di,j+dj)
+(dp[i][j+1][i+di][j+dj]+dp[i+1][j][i+di][j+dj]-dp[i+1][j+1][i+di][j+dj])

// i have added rectangles that start after (i,j) and end before (i+di,j+dj) twice, so i should subtract those

// rectangles that start at or after (i,j+1) and end before (i+di,j+dj)
-(dp[i][j+1][i+di-1][j+dj]+dp[i][j+1][i+di][j+dj-1]-dp[i][j+1][i+di-1][j+dj-1])

// rectangles that start at or after (i+1,j) and end before (i+di,j+dj)
-(dp[i+1][j][i+di-1][j+dj]+dp[i+1][j][i+di][j+dj-1]-dp[i+1][j][i+di-1][j+dj-1])

// i have subtracted the rectangles that start at or after (i+1,j+1) twice, so i should add those back

// rectangles that start at or after after (i+1,j+1) and end before (i+di,j+dj)
+(dp[i+1][j+1][i+di-1][j+dj]+dp[i+1][j+1][i+di][j+dj-1]-dp[i+1][j+1][i+di-1][j+dj-1])

// the big rectangle that starts at (i,j) and ends at (i+di,j+dj) maybe a "good" rectangle
+check(i,j,i+di,j+dj);

EDIT: really sorry for this long post, but i hope it served its purpose i.e. everything is clear now! :)

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shobhit6993

11 лет назад, # ^ |

← Rev. 3 →

This is one super case of inclusion and exclusion. Had my eyes tight shut for nearly 10 mins to get a feel of all the inclusion and exclusion taking place. Hats off sir! The fact that you imagined this and coded this up during the closing minutes of the contest really demands respect.

Since the editorialist also seems to have used the same logic and unfortunately, has not cared to explain the intuition behind the recurrence relation, a link to your above explanation should be included in the editorial for others to understand :)

EDIT: No corrections. I overlooked the parenthesis. My bad.

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11 лет назад, # ^ |

← Rev. 2 →

haha thanks. i actually had this logic much earlier than the end of contest, but i was always missing few of the terms, so it was giving me WA on first test on my system. when i finally had the above solution, i submitted it as soon as i saw that it was passing first test (due to time shortage). i tested on second test during the In queue time. i was literally out of breath just after submitting it!

about the correction u posted, what i have written above is correct. there is a - sign outside the bracket which will negate all terms inside.

yes, ur right. i will post a link to the above explanation as a comment on the editorial.

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shobhit6993

11 лет назад, # ^ |

Sorry, I overlooked the parenthesis. My bad. Edited my comment appropriately.

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Baklazan

11 лет назад, # ^ |

Firstly, you compute prefix sums, so you can check whether a rectangle is good in constant time.

You can do memoization for all possible queries.

When you get a query, you chcek the biggest rectangle (rectangle described by query itself) and recursively compute number of smaller good rectangles from answers for smaller queries (using inclusion and exclusion).

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11 лет назад, # ^ |

+13

Or just calculate the answers for all possible queries, and answer them in O(1) time :D

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BiMokh

11 лет назад, # |

Damn 6 and 12!

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KonanMentor

11 лет назад, # |

← Rev. 2 →

Писал "B", написал свою функцию возведения в степень, назвал pow. Долго не мог понять, почему 9 * 10 = 89, отключил cmath — заработало (:

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11 лет назад, # |

-13

Thanks a lot. Very liked problems D and E. Wrong answer 5 in E is thinking same circles good, or I have more errors?

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SC_pRo_ION

11 лет назад, # |

Забыли перевести на русский язык распределение баллов :)

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vlyubin

11 лет назад, # |

+44

Alert at the very end of the contest was not a great idea — 10 seconds left on the clock, I want to send a solution ... when some stupid alert pops up! I panicked a little bit and failed to send it :( Luckily, my solution was buggy anyway, but I think you shouldn't do alerts in the last 5 minutes of the contest (unless it's about extension). It's better to apologize after the contest (if you want to).

Nice round though!

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Dutzul

11 лет назад, # |

can you estimate the waiting time before judge ? :D

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firmsnail

11 лет назад, # |

I wonder why the size of array is not big enough will got a "Wrong Answer", instead of "Runtime Error!" ?

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forthright48

11 лет назад, # ^ |

Because going out of bounds is an undefined behavior in c++. This means anything can happen. http://stackoverflow.com/questions/1239938/c-accesses-an-array-out-of-bounds-gives-no-error-why

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Timur_Sitdikov

11 лет назад, # |

← Rev. 2 →

У кого-нибудь еще страницы загружались 5 минут после клика? В этот раз это сыграло злую шутку. Пишу под Visual Studio — 2008, она ругается на функцию abs от long long. Написал свою abs, послал задачу за 3 минуты до конца. Жду 5 минут, и получаю ошибку компиляции после конца контеста. Спрашиваю некоторых знакомых — говорят ничего, нормально загружается. В чем может быть проблема, или все-таки не я один такой? Одинаково плохо грузит и с Хрома, и с Оперы.

UPD. Самое печальное, что после окончания контеста поправленная версия без написанной ручками abs зашла. Может, все-таки кто-нибудь знает, почему страницы могут грузиться по несколько минут? Они и сейчас медленно грузятся (правда чуть быстрее, где-то 3 минуты вместо 5, опера чуть быстрее, чем хром). Нельзя же самом деле так писать все контесты!

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Xephy

11 лет назад, # |

Oh man, Div2 A remind me sooooo badly of jubeat. I live in Kazakhstan though so I can play it only on BEEP. BTW Japanese version have way more songs. sigh If I could just simply play Evans through with tests in the Sample XD

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11 лет назад, # |

+16

Any ideas for non-quadratic Div 1 D, please?

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11 лет назад, # ^ |

+64

O(NlogN). Lets move two pointers and calculate minimal number of vertex in tree, which contains all of this.

For doing last part, let's have a set of vertexes sorted in order of in-time in dfs. Then sum of distanses between consecutive vertexes (including last and first) is twiced number of edges in minimal tree.

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Endagorion

11 лет назад, # ^ |

O(N log^2 N) gets accepted too (binary search on answer, then two pointers with set). TL is really kind.

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11 лет назад, # ^ |

Awesome!

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11 лет назад, # ^ |

Thanks! Cool idea to remember.

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Kianoosh

11 лет назад, # |

Counting rectangles was REALLY fun! thanks!

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11 лет назад, # ^ |

← Rev. 2 →

yes it was fun, and the best possible end to it as far as i am concerned, because i made a submission on it at 01:59:48 and got it Accepted! :)

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LiuRunkY

11 лет назад, # |

Sorry, but if my solution got WA before the clarification, and i submit the right one at last, can i skip the solutions before and return the lost points of WA? Looks like a silly question :p

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11 лет назад, # |

Good round! Thanks!

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11 лет назад, # |

+11

Какая радость, что в div 1 B (N^4 + Q * N^2) проходит, а (N^4 + 4 * Q * N^2) TLE'ится...

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11 лет назад, # ^ |

Я лично сдал N⁵ + Q

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riadwaw

11 лет назад, # ^ |

← Rev. 2 →

... + QN^2 не у всех проходит тоже, лично ломал.

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11 лет назад, # ^ |

Да, как выше уже написал PavelKunyavskiy официальное было быстрее, но как-то не пришло в голову искать что-то быстрее О(полмиллиарда) на 4 секунды. Может мне надо поправить конвертатор асимптотик во время...

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11 лет назад, # ^ |

заходит (N^5 + Q * N^2). Жаль только, что после контеста)

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ABalobanov

11 лет назад, # ^ |

у меня и N^6 зашло)

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LiuRunkY

11 лет назад, # |

And also i think 10e16 10e15+1 10e9 is a really interesting test for Problem B in Div2

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Lazzaro

11 лет назад, # ^ |

Yeah, what a pity for my solution. Just one neglect of "long long" type's overflow! It's ok after using "unsigned long long".......

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LiuRunkY

11 лет назад, # ^ |

Very Sorry for my hack:( Thought about that... w=w%(k*len)+(w/(k*len)-add)*k*len perhaps this formula is correct:)

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cgy4ever

11 лет назад, # |

+26

Task B,C,D in Div1 are very OI-like. :)

E looks nice: After Inversive transform, it becomes: find sets of pair that share a common middle point (and not parallel). Am I right? Well, I spend 40 minutes to figure out a typo in my code. :(

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jqdai0815

11 лет назад, # ^ |

← Rev. 2 →

+18

After Inversive transform, there could be three points exactly on a line, we should minus it from the answer. TAT

Edit : Oops! I haven't read your post clearly. Sorry. But I think it's a quite tricky case.

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cgy4ever

11 лет назад, # ^ |

Yes, I get AC now. Without that trick, I think it is only worth 1500p.

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gen

11 лет назад, # ^ |

+19

Huge congrats on the win!

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inspirer

11 лет назад, # |

friday13 :) ..that why i did not participate

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Na2a

11 лет назад, # |

-6

Упал с 30ого до 100ого из-за того что написал long long вместо unsigned long long.

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Maxe

11 лет назад, # |

-16

Контест отменят?

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11 лет назад, # ^ |

С чего бы это?

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Maxe

11 лет назад, # ^ |

Из-за проблем с сервисом и проблемами в задаче С во время контеста?

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http://codeforces.me/contest/373/submission/5429866

11 лет назад, # ^ |

Не, вряд ли.

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kingofnumbers

11 лет назад, # |

← Rev. 2 →

div1 C is (fairly standard dp + find max quickly) so I had the idea of it very quickly.

however,A and B required a little bit cleverness so I think a lot but couldn't find the solution.

I think my skills are declining, thanks to baccalaureate for that :(

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ErvinK

11 лет назад, # ^ |

For A it is easy to spot that the best answer is n/2 when after sort, all element from first half is placed in the second half. So the solution is just binary search for each element in the first half, for the smallest elements to be fited in the second half.

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martinezgjuan

11 лет назад, # |

Hadn't participated in months. It was a good contest to get back in :)

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indraep

11 лет назад, # |

ah my correct solution for problem B got skipped by the judge :(

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Rooooookie

11 лет назад, # |

div2 problem B is tricky, calculation would be exceed long long,I noticed this point in contest,but I don't know how to deal with it.And cause me fail system test,also many participant.Then I use double get AC, sad :(

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Elun

11 лет назад, # ^ |

I used __int64, equals to signed long long. Accepted: 5427927. Sorry for my bad english.

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Rooooookie

11 лет назад, # ^ |

Yeah,I see.You are use division and it wouldn't exceed long long.I also think this means but I havn't use it. Then I use multiplication and cause WA.Your means is a good ways too, I think I should consider more, thank you very much! :)

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Sfairat

11 лет назад, # |

+18

You really SHOULD NEVER USE stack with visual studio 2010. If you want a proof of it, just have a look at this and this . VS 2010 — TLE9, GNU c++ 0x — AC 1123ms

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JGalt

11 лет назад, # |

could somebody explain me why this code at local returns "YES" but online is "NO"? (mac, x-code)

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11 лет назад, # ^ |

atoi(&str[j]) this one considers memory from addres of str[j] to closest zero as C-string and converts it to number. Probably next zero will be the end of string str. So you index dig with something big. Next nobody knows what happen.

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handlechanged

11 лет назад, # |

the test case for which my div2 A http://codeforces.me/contest/373/submission/5421738 gives wa gives the correct output on my machine plz look into in

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handlechanged

11 лет назад, # ^ |

could someone plz clarify

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Haghani

11 лет назад, # ^ |

← Rev. 3 →

DELETED. sorry i was wrong.

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handlechanged

11 лет назад, # ^ |

thanx i dint know this!! does int t[12] = {} ensure that all elements are initialised to 0

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11 лет назад, # ^ |

← Rev. 2 →

i don't think so, but

int t[12];
memset(t,0,sizeof t);

makes all 12 elements 0. as u may have guessed, it does not depend on the array size.

however, before you use memset, make a note that it only works when u want all values to be set to 0 or -1, not other integers like 1 or 2.

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Haghani

11 лет назад, # ^ |

← Rev. 2 →

i am not sure what exactly happen if you write "int t[12] = {}" but i submit this code and got accepted.

5433663

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shopnobaj_raju

11 лет назад, # ^ |

← Rev. 4 →

never use fflush(stdin). I commented out your fflush(stdin) lines and it gets accepted.

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DmitriyH

11 лет назад, # |

+23

Round statistics, with images.

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Thomas_Ahle

11 лет назад, # ^ |

This is super cool!

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11 лет назад, # |

I found out what caused me to TLE on C(div1).

Check out 5432941 and 5432965. They only differ in the position of one line: the deque<> is declared once the AC submission, and inside an O(M) cycle in the TLE submission. How come? Shouldn't that just be O(M) additional operations (albeit slow ones)?

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11 лет назад, # ^ |

When the size of a vector (or deque) exceeds its capacity (the size of the buffer which is currently allocated for it), its size is increased and the data moved. This incurs a time overhead. When the size decreases, its capacity is not decreased (even if you clear() it). Thus, when the size increases again but no more than what it had already been before, there is no overhead. However, if you make a new deque (and, implicitly, delete[] the old memory block), then you will pay again. This means that if you reuse the same deque, you will save on memory management. Sometimes it is faster to have a globally declared vector and clear it every time, rather than a locally defined one.

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11 лет назад, # ^ |

← Rev. 2 →

I've considered that, but it's test 29. The times (T[i]) are really close and D = 3, so the deque will only contain at most around 10 elements at once, and so it won't be resized often. It looks like the deque is being allocated much more space than it needs...

Besides, it passes test cases 8 and 9, where the allocations are much more massive.

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xenoslash

11 лет назад, # ^ |

I don't know how deque is implemented in cpp, but if there are array resizings, I can see how if you declare the dequeue exactly once, it will never shrink back, and have to reexpand. But if you declared the dequeue multiple times, you would restart with the default initial size and have to reexpand.

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11 лет назад, # ^ |

See my comment above.

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11 лет назад, # |

← Rev. 2 →

At problem A , this code got Wrong Answer !

sort(a+1,a+n+1);
int l = 1 , r = n/2+1;
while(l <= n/2 && r <= n){
    if(a[r] >= a[l]*2)
        l++;
    else
        r++;
}
cout<<n-l+1;

but when I changed it to

sort(a+1,a+n+1);
int l = 1 , r = n/2+1;
while(l <= n/2 && r <= n){
    if(a[r] >= a[l]*2)
        l++ , r++;
    else
        r++;
}
cout<<n-l+1;

it got Accepted ! But I can't understand why this is possible ! If anybody knows that , please help me ! :)

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adelnobel

11 лет назад, # ^ |

Because if you don't increment r then you may take an element more than once!

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11 лет назад, # ^ |

← Rev. 2 →

-8

Output : 250335

Answer : 250336

if I take an element more than once my Output must be more than Answer !!

This is not my wrong , I am sure :)

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zylber

11 лет назад, # ^ |

← Rev. 2 →

The output decreases as the number of pair increases, you are taking an element more than once.

Take this example: 3 4 5 10. Your code matches both the 3 and the 4 with the 10.

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11 лет назад, # ^ |

← Rev. 2 →

Ok , I know my solution is wrong !

But my output must be more than Answer ! Ok ?

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DemoVersion

11 лет назад, # ^ |

← Rev. 2 →

No.

Youre Output is less than Answer because after you put an kangaroo in another. YOU DO NOT SKIP BOTH OF THEM and it can cause another PAIR , so the Increment of Variable l Because The final result is n-l-1.

The Error Pair cause Decrement in output.

I can hack your first solution with

4
1 1 1 3

Answer = 3

Output = 2

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11 лет назад, # ^ |

-8

I got it ! you are right ! Thank guys :X

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aladdin3

11 лет назад, # ^ |

← Rev. 4 →

As far as I see, if you take an element more than once, as the example given by zylber "3 4 5 10", you say that you can input both 3 & 4 within the 10 .. which means the number of hidden is 2 (l = 3) & the number of visible is 2 (n - l + 1) .. while in fact the number of visible should be 3. So your output will be less than Answer, not the other way around.

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11 лет назад, # ^ |

It's problem 'A'. If you increment 'l', well, you use a[l] and a[r], else you don't use a[l] and a[r].

Sorry for my bad english.

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11 лет назад, # |

← Rev. 3 →

Почему в претесте #6 задачи А ответ не 45? Вот цепочки, которые я формирую

40 - 82 - 201 - 405 - 814
45 - 94 - 204 - 409 - 819
48 - 97 - 205 - 433 - 886
54 - 109 - 222 - 449 - 911
60 - 125 - 271 - 549
83 - 237 - 474 - 948
110 - 240 - 484 - 972
130 - 271 - 570
132 - 276 - 576
134 - 282 - 579
139 - 288 - 589
139 - 288 - 590
141 - 303 - 618
142 - 314 - 636
145 - 330 - 666
146 - 336 - 682
150 - 341 - 702
154 - 347 - 735
242 - 520
245 - 535
249 - 601
364 - 754
384 - 775
438 - 896
441 - 911
454 - 916
601
604
638
647
655
778
783
804
807
836
841
842
845
923
930
938
952
957
974

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11 лет назад, # ^ |

Нельзя цепочки длины больше двух.

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11 лет назад, # ^ |

← Rev. 2 →

Ну чтож, это просто ЭПИК)))

Спасибо, а то я бы еще долго упарывался

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asdfasdfasdfw

11 лет назад, # |

-8

Problem A (div2).

I got WA on this submission because of test #13. The error is because of line 9: char row[4];, which should have beenchar row[5];. I know why I need 5 and not 4 (that's because of '\0').

On my computer, the test #13 gives the correct answer and I do not receive any errors. Can anyone explain me this weird behaviour and how can I avoid it in the future? Does it have anything to do with the compile and run command?

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ifsmirnov

11 лет назад, # ^ |

-11

In C++ accessing out-of-bounds elements of array is undefined behavior because arrays are only positions in memory. Sometimes memory after the array is not used and your code passes, sometimes it is occupied, so you do not get a correct value. You cannot check it during runtime someway.

The general advise is always to create a bit bigger arrays than you need (e.g. 100010, when n <= 100000).

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permin

11 лет назад, # ^ |

-16

The general advise is always to create a bit bigger arrays than you need (e.g. 100010, when n <= 100000).

My general advise is to use vector instead of static array :) In that case you will find out of range even in small cases.

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yeputons

11 лет назад, # ^ |

Nope, you won't. It also depends on compiler settings. For example, you'll get exception if you use at() or specify a special debug mode define, but operator [] does not check anything by default.

So, if you access items around size() + 2, you probably won't get an error without debug.

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Dixtosa

11 лет назад, # ^ |

I don't like that "general advise". If you create an array the size of which you choose arbitrarily, it means you don't rally understand your solution which is the worst thing you never want to face.

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gen

11 лет назад, # ^ |

I think that kind of thinking is ineffective. If you set the size of an array +5 of what you are expecting, you don't get a worse situation. In contrast, creating an array of a smaller size that is needed based on an incorrect judgement, you may get punished by WA, while the solution basically is still correct.

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darnley

11 лет назад, # ^ |

+13

It was a very wise comment, it's a shame it got negative feedback.

Indeed, arrays of size 10010 is an ugly code. "Competitive programming" could do better than encouraging bad style.

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darnley

11 лет назад, # ^ |

+12

Vanya, I hope you don't teach this to school kids in LKSh, do you?!

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caustique

11 лет назад, # ^ |

Why not?

This is the only possible approach to guarantee that there's no +/-1 errors in C++ while keeping speed of the program (not using vectors + .at() or dynamic arrays).

As far as I know, you write in Java and there you can create arrays like that:

int[] a = new int[n];

but in C++ it's not the case.

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darnley

11 лет назад, # ^ |

Nope. The only possible approach to guarantee that there's no +/-1 errors in C++ is to write code that doesn't have +/-1 errors :)

Despite the smile it is not a joke at all.

If you believe that your code needs a +10 in arrays size "just in case", would you trust this code to solve some dangerous real life problems?

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gen

11 лет назад, # ^ |

-8

The goals in competitive programming and production somewhat differ. In ACM-style contests, the goal is to get Accepted for a problem minimizing the wrong submittion count and the time spent on it. In production, the time and "trial-and-error" constraints are not so strict, the main goal is the correctness and soudness of the code. That's why some hacky tricks are very useful in CP, while not welcome in production. So I think this case is perfectly fine and should be encouraged in CP and discouraged in production.

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Darkstalker

11 лет назад, # |

+16

When will rating updated ? No one ask about this. Am I lack of information such a there's some trouble or cheat issue or unrated ? Thanks

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11 лет назад, # ^ |

I guess they are busy answering requests for skipping some submissions in Div2-B.

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abistrigova

11 лет назад, # |

Подозрительно похожи посылки 5431127 и 5431290

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savinov

11 лет назад, # ^ |

я бы сказал они идентичны с точностью до замены cin -> scanf, cout -> printf

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11 лет назад, # |

А как решалась C div2?

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11 лет назад, # ^ |

Наверное, самый простой способ — бинарный поиск по ответу. Пусть на текущем шаге мы бинарным поиском нашли некоторое число M, тогда мы должно проверить, что в отсортированном массиве размеров мы можем запихнуть первые M кенгуру в последние М кенгуру, если можем, то двигаем левую границу, иначе — правую.

Вторая идея — 2 указателя. Первым указателем итерируемся от 0 до N / 2, а вторым указателем подбираем ответ. Можешь в моих посылках код посмотреть, для наглядности.

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Tima

11 лет назад, # ^ |

Понятно, что нам нужно максимизировать количество кенгуру-пассажиров. Очевидно.что количество кенгуру-пассажиров не больше чем n/2. Сделаем сортировку s[i]. Думаю , дальше так будет понятно :

d = n/2;x =0; m = n;
while (d > 0 && n > d){
		if (2*s[d] <= s[n]){
			x++;
			d--;
			n--;
		}
		else {
			d--;
		}	
	}

Тогда ответ будет m-x;

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11 лет назад, # ^ |

← Rev. 2 →

Аналогичное решение дает TL24 http://pastebin.com/Jh7fQEN2

UPD: нашел баг

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shubham1402

11 лет назад, # |

← Rev. 2 →

problem B why did it gave TLE for test case to me ? test case 10000000000000000 1 1 code:

long long nod(long long a) { long long b = 0; if(a == 0) return 1; while(a!= 0) { a = a/10; b++; } return b; } int main() { int t; //cin>>t; t = 1; while(t--) { long long w,m,k; cin>>w>>m>>k; long long op = 1; long long length = 0; while(1) { long long temp = (long long)nod(m); long long temp1 = temp * k; long long temp2 = pow(10,temp); long long temp3 = temp2 — m; //cout<<"temp: "<<temp<<" temp1: "<<temp1<<" temp2: "<<temp2<<" temp3: "<<temp3<<endl; //cout<<"w: "<<w<<endl; //cout<<"length: "<<length<<endl; if(temp3 <= (w/temp1)) { length += temp3; w = (w % temp1) + (temp1 * ((w/temp1) — temp3)); m = temp2 ; } else { length += (w/temp1); break; } } cout<<length<<endl; } return 0; }

it is working very fine on my machine ?? any one to help ??

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