Codeforces Round #219

11 years ago, # ^ |

+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 years ago, # |

Good luck to everybody!

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firmsnail

11 years ago, # |

It's a good opportunity to back Div.1

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EnumerativeCombinatorics

11 years ago, # |

+10

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

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_777_

11 years ago, # |

Good oppurtunity to enter Div1

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TsunamiNoLetGo

11 years ago, # |

-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 years ago, # ^ |

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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saba_tavdgiridze

11 years ago, # |

slow judge:((((((((

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ffao

11 years ago, # ^ |

+29

Counting Submissions on Queue is Not Fun

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najim4689

11 years ago, # |

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 years ago, # ^ |

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 years ago, # ^ |

Last one seems good :p

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nurzhands

11 years ago, # |

No hacks round :( (almost)

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11 years ago, # |

← 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 years ago, # |

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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eduardische

11 years ago, # ^ |

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

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11 years ago, # ^ |

← Rev. 2 →

+34

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

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tunyash

11 years ago, # ^ |

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

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11 years ago, # ^ |

+24

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

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11 years ago, # ^ |

+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 years ago, # ^ |

← Rev. 2 →

+45

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

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Rubanenko

11 years ago, # ^ |

Could you describe it briefly?

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kcm1700

11 years ago, # ^ |

+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 years ago, # ^ |

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

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saketh

5 years ago, # ^ |

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 years ago, # ^ |

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

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forthright48

11 years ago, # |

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

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ErvinK

11 years ago, # |

Problem D div2/ B div1 core idea ?

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LashaBukhnikashvili

11 years ago, # ^ |

← Rev. 2 →

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

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ErvinK

11 years ago, # ^ |

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

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LashaBukhnikashvili

11 years ago, # ^ |

← 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 years ago, # ^ |

← 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 years ago, # ^ |

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

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shobhit6993

11 years ago, # ^ |

← 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 years ago, # ^ |

← 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 years ago, # ^ |

← 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 years ago, # ^ |

← 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 years ago, # ^ |

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

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Baklazan

11 years ago, # ^ |

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 years ago, # ^ |

+13

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

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BiMokh

11 years ago, # |

Damn 6 and 12!

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11 years ago, # |

-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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vlyubin

11 years ago, # |

+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 years ago, # |

can you estimate the waiting time before judge ? :D

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firmsnail

11 years ago, # |

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 years ago, # ^ |

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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Xephy

11 years ago, # |

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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eduardische

11 years ago, # |

+16

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

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11 years ago, # ^ |

+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 years ago, # ^ |

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 years ago, # ^ |

Awesome!

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eduardische

11 years ago, # ^ |

Thanks! Cool idea to remember.

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Kianoosh

11 years ago, # |

Counting rectangles was REALLY fun! thanks!

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

11 years ago, # ^ |

← 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 years ago, # |

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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redjohn

11 years ago, # |

Good round! Thanks!

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LiuRunkY

11 years ago, # |

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

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Lazzaro

11 years ago, # ^ |

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 years ago, # ^ |

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 years ago, # |

+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 years ago, # ^ |

← 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 years ago, # ^ |

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

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gen

11 years ago, # ^ |

+19

Huge congrats on the win!

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inspirer

11 years ago, # |

friday13 :) ..that why i did not participate

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kingofnumbers

11 years ago, # |

← 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 years ago, # ^ |

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 years ago, # |

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

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indraep

11 years ago, # |

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

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Rooooookie

11 years ago, # |

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 years ago, # ^ |

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

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Rooooookie

11 years ago, # ^ |

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 years ago, # |

+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 years ago, # |

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

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11 years ago, # ^ |

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 years ago, # |

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

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handlechanged

11 years ago, # ^ |

could someone plz clarify

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Haghani

11 years ago, # ^ |

← Rev. 3 →

DELETED. sorry i was wrong.

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handlechanged

11 years ago, # ^ |

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

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11 years ago, # ^ |

← 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 years ago, # ^ |

← 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 years ago, # ^ |

← Rev. 4 →

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

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DmitriyH

11 years ago, # |

+23

Round statistics, with images.

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Thomas_Ahle

11 years ago, # ^ |

This is super cool!

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11 years ago, # |

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 years ago, # ^ |

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 years ago, # ^ |

← 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 years ago, # ^ |

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 years ago, # ^ |

See my comment above.

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11 years ago, # |

← 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 years ago, # ^ |

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

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11 years ago, # ^ |

← 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 years ago, # ^ |

← 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 years ago, # ^ |

← Rev. 2 →

Ok , I know my solution is wrong !

But my output must be more than Answer ! Ok ?

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DemoVersion

11 years ago, # ^ |

← 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 years ago, # ^ |

-8

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

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aladdin3

11 years ago, # ^ |

← 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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HASP

11 years ago, # ^ |

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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asdfasdfasdfw

11 years ago, # |

-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 years ago, # ^ |

-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 years ago, # ^ |

-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 years ago, # ^ |

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 years ago, # ^ |

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 years ago, # ^ |

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 years ago, # ^ |

+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 years ago, # ^ |

+12

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

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caustique

11 years ago, # ^ |

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 years ago, # ^ |

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 years ago, # ^ |

-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 years ago, # |

+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 years ago, # ^ |

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

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shubham1402

11 years ago, # |

← 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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