Hello Codeforces!

On Jan/29/2020 17:35 (Moscow time) Educational Codeforces Round 81 (Rated for Div. 2) will start.

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be **rated for the participants with rating lower than 2100**. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given **6 problems** and **2 hours** to solve them.

The problems were invented and prepared by Roman Roms Glazov, Adilbek adedalic Dalabaev, Vladimir vovuh Petrov, Ivan BledDest Androsov, Maksim Neon Mescheryakov and me. Also huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Good luck to all the participants!

Our friends at Harbour.Space also have a message for you:

*Hi Codeforces!*

*As a special prize for the Educational Round 81, we would like to invite the top participants to take part in our Hello Muscat ICPC Programming Bootcamp, which will take place in Oman, from March 19 to March 25, 2020. The prize will cover the participation fee, accommodation, and half-board meals for the entire duration of the bootcamp (except flights)!*

**There are three requirements to satisfy:**

*You took part in at least 10 rated contests on Codeforces**Your max rating should be less than 2400**You should be eligible for ICPC and/or IOI 2020+*

*Good luck to everyone!*

*We are also excited to announce that we are working with our partners to provide free participation (flights not included) for the teams going to the ICPC World Finals 2020, which will take place in Moscow. If you and your team qualified for the Finals, fill out the form below to see if you’re eligible.*

Congratulations to the winners:

Rank | Competitor | Problems Solved | Penalty |
---|---|---|---|

1 | nickluo | 6 | 177 |

2 | KrK | 6 | 184 |

3 | NoLongerRed | 6 | 185 |

4 | Anadi | 6 | 200 |

5 | neal | 6 | 204 |

Congratulations to the best hackers:

Rank | Competitor | Hack Count |
---|---|---|

1 | GiantTornado | 58 |

2 | rachit_raj | 51:-1 |

3 | harshraj22 | 41:-5 |

4 | Ahmad__ | 35:-1 |

5 | sadik | 33:-2 |

And finally people who were the first to solve each problem:

Problem | Competitor | Penalty |
---|---|---|

A | antontrygubO_o | 0:01 |

B | neal | 0:05 |

C | IgorI | 0:07 |

D | NoLongerRed | 0:06 |

E | A_Fan_of_the_AK_King--lk | 0:14 |

F | Combi | 0:23 |

**UPD:** Editorial is out

Educational Rounds always have good problems. Hope for a good contest.

Finally a contest ! Hope the gap between rounds will decrease

It feels more longer to users whom Div.3 contest is unrated for.

Wish not.

I see this happening to me.

Petition to MikeMirzayanov to decrease the gap between forthcoming contests.

Looking forward to interesting problems

vovuh are you the Vladimir Petrov who won IMO Gold in 2018?

Unfortunately, no :(

Lets hope that the frequency of contests increases.

Problems are difficult!

I don't think so.

B and C are difficult.

B and C are difficult for

.NewbiesEven if the problem is not original, I think providing links to such problems and solutions

during contestviolates the CF round rules. Let us keep silence until the end of the round.Finally solved the first problem. Heading for second. So much to learn...!!

I guess, someone accidentally swapped B and D :)

D is easy if you know some number theory, but B is just casework, I don't think it's harder.

bilal saeed hoooooooooooooooo!!!!!!!!!!!!!!!!!!!

aaaaaaaaaaaaaaa!!!!!!!!!!!!!!

A Problem of APIO 2016:Boat

A harder version of problem F.

C was similar to https://atcoder.jp/contests/abc138/tasks/abc138_e

I consider C as a problem which I've solved it before.Unfortunately,I couldn't remember it exactly while the round.Now I can get it.

how to solve C ?

How to solve D?

Find g=gcd(a,m). Divide m by g, i.e. let f=m/g. Find Euler Totient Function value of f i.e. phi(f).

Why?

Okay, we know gcd(a, b) = gcd(b, a%b)

So gcd(a+x,m) = gcd(m, (a+x)%m) and value of x lies between [0,m-1] that means possible values of (a+x)%m are [0, m-1] because (a+x)%m keeps on incrementing till it completes the full modulo cycle.

do u have any proof ?

Let's say gcd(a, m) = g

We know gcd(a, m) = gcd(g*p, g*q) where gcd(p, q)=1 and (0 <= p < q). So we have to choose such x those satisfy gcd(a+x, m) = gcd(g*p, g*q). So we all have to do is find the number of co-prime of q those are less than q. So we have to call phi(q).

sorry for my bad english...

Thanks for your explaination. I still don't understand the final step. I am only able to get phi(q+a/g)-phi(a/g) by the above steps instead of phi(q).

gcd(a+x,m)=gcd(m,(a+x)%m)=gcd(g*q,g*p) note that (a+x)%m will always be less than m and can take up any value from 0 to m-1. So finding phi(q) will suffice.

Still didn't get it.Can you please elaborate more?I have the same doubt as tycw.

What we think about should be (a+x)%m instead of a+x. So the range of answer becomes [0,m).

Okay, but how its ensured that all the numbers we get will be >=a,in phi(m/gcd)?

why

How to solve E?

keep segtree on array where arr[i]=cost if you split at i, start by putting everything on the right and leaving the left empty, so arr[0]=a[0], arr[1]=a[0]+a[1]... then start by putting 1 on the left, then 1 and 2, then 1,2 and 3 and so on. when you put a new number on the left the cost for all splits with that number already on the left decrease by a[position[i]], the cost of all the splits without that number (the ones with i<position[i]) increase by a[position[i]]. each time get the minimum in range (0,n-2) (because you can't split at last element).

Thx a lot!

Another segment tree solution that doesn't require range update/lazy propagation:

Represent each segment by the triple $$$(suml, sumr, cost)$$$. Define the combination of two segments as follows: $$$A + B = (suml_A + suml_B, sumr_A + sumr_B, \min(cost_A + suml_B, sumr_A + cost_B))$$$ (this operation is not commutative, but it is associative)

Start with $$$tree[p_i] := (0, a_i, 0)$$$. Iterate $$$i$$$ over $$$[1, n-1]$$$ by updating $$$tree[p_i] := (a_i, 0, 0)$$$, then grabbing the $$$cost$$$ value from the "sum" of the entire segment tree. The answer is the minimum $$$cost$$$ value retrieved.

Explanation: Fix a split of the permutation to sets $$$L$$$ and $$$R$$$. $$$suml$$$ represents the sum of the costs of $$$L$$$ in that segment, likewise $$$sumr$$$ for $$$R$$$. Now consider how to combine the statistics for segments $$$A$$$ and $$$B$$$. Combining the sums is obvious, but what is $$$cost$$$? Consider there is a point $$$x$$$ within the segments where we want to move all points $$$i > x$$$ from $$$L$$$ and $$$i < x$$$ from $$$R$$$. If $$$x$$$ is in the left segment ($$$A$$$), the new cost is $$$cost_A + suml_B$$$, as we must move all members of $$$L$$$ in the right segment ($$$B$$$). If $$$x$$$ is in $$$B$$$, the new cost is $$$sumr_A + cost_B$$$ as we must move all members of $$$R$$$ in $$$A$$$. We thus pick the minimum of these two scenarios. The updates progressively move elements from $$$R$$$ to $$$L$$$ as the initial split. Be sure not to move the last element, or your answer would always be spuriously $$$0$$$.

Sample: 69829539

Am I stupid? How do you solve D? I got to count numbers $$$b$$$ with $$$0 <= b < m$$$ with $$$gcd(m, b) = gcd(m, a)$$$ but made zero progress afterwards

Edit: As I look at my previous claim I realize that my proof in-contest is flawed. Does anyone know if the statement is actually correct? Looking at the solution given, my statement should be true but I am unable to prove it.

Edit: Just kidding, not flawed. $$$gcd(a, b) = gcd(b $$$%$$$ a, a)$$$ which easily applies to $$$gcd(m, a+x) = gcd((a+x)$$$%$$$m, m)$$$ since $$$a+x$$$ takes on m consecutive unique values, we know it takes on every integer $$$[0, m)$$$ exactly once.

Let $$$d = gcd(m, a)$$$ and $$$m_1 = m / d$$$, $$$a_1 = a / d$$$, $$$b_1 = b / d$$$. Then $$$gcd(m_1, a_1) = gcd(m_1, b_1) = 1$$$.

In other words, you have to count numbers $$$b_1$$$ with $$$0 \le b_1 < m_1$$$ which are coprime to $$$m_1$$$. This is actually an Euler function, $$$\varphi(m_1)$$$.

Pretty neat approach.

Yes, but there is one more condition on b, that b >= a. It can be easily done by just altering the algorithm for calculating the Euler Totient Function.

There must be some reduction to get this statement with b. At least unmodified totient function works: 69784007

We can reduce the problem to couting the number of numbers relatively prime to m on interval [0..m/gcd(a, m)]. Let's say that m = m/gcd(a, m). Firstly you need to factorise m and store vector of unique prime factors. Then you can use the principle of inclusion and exclusion to calculate the number of such numbers that have common prime divisor with m.

I solved it in a similar way.

I've done it kinda differently compared to the Euler Totient guys here. Using the statement notations, we have a, m, and x. a = gcd*u and m = gcd*v, where (a, m) = gcd so (u, v) = 1 gcd must divide (a+x) so gcd must divide x therefore we have x = gcd*y

Which brings us to an equivalent problem: find the count of y for which: (u+y) and v are coprime, where y is in range [0, v), aka their gcd is 1.

For that, I used the principle of inclusion/exclusion because v had something like 12 factors maximum. And because I was in full blown panic mode and couldn't think of Euler.

Can you explain your code how are you computing number of y's so that gcd(u+y,v)=1.??

Do you know about the Inclusion and Exclusion Principle? I'll count the values of y for which the gcd will be different than 1: divisible with p1, p2, ... or pk, where p[1...k] are all prime factors of v. That's done with above mentioned principle: you basically consider all subsets of {p1, p2, ... pl} and you either add or remove their count from the result depending on the parity of its size. https://codeforces.com/contest/1295/submission/69761793 I'm hoping my code is actually correct, as in it doesn't fail after the hacking phase. It's really short once you remove the comments.

Hey bro can you tell how is

`u + y == v (mod x)`

, and what computation you are doing in your solve function after taking all the numbers from the set.Great minds think alike!

Nice. But if the range of x is: 0 <= x < 2m Then Euler Phi does not work I guess. Like if a=8, m=12 and for 0<=x <m , The answer is 2 like Phi(3) but for 0 <=x <2m , answer is 4 i guess. Which is not Phi(3) but it is 2*Phi(3). Also if x was: 0 <=x <= 2m , Then the answer is 5. I don't know how to explain that. But if you see, for ==> gcd(a+x,m) = gcd((a+x)%m),m) (according to many comments) which says that no matter what is x's range, we only need to find the Phi(m/gcd(a,m)) which is incorrect for x's various range.

How did you come up with gcd(m,a) = gcd(m, b). Couldn't understand this part.

gcd(a+x,m) = gcd((a+x) mod m, m), so in fact you just need to check which remainders of m have the same gcd as a.

we have to check no of coprime with m in range(1,m-1)?

Not exactly, because you are looking for the numbers in range(1, m-1) which have the same gcd with m as

ahas (which is not necessarily 1).How do you make GCD (a, m) = GCD (b, m)

Let X=GCD(a,m).

So if GCD((a+x)/X,m/X) equals to 1, GCD(a,m) will equal to GCD(a+x,m).

I'll bite, how to solve F? :P

GG, thanks for all the awesome problems! It was my first time participating in such a coding contest and it did take me some time to get used to many things on Codeforces. As a first-timer, Codeforces is like a gold mine to novice programmers like me xD.

Can someone give me a few pointers on how to improve as I found some questions quite challenging and there are always some nuts who can solve them in like 10min :D

Cheers~

One who uses hakase as profile is always on a roll :)

xD ikr

How to solve C? My idea was to use hashmap (Key: Value being Char: List of Index) for characters of string S. But got TLE

my_submission

and iterate through string t and solve problem greedily

Store indexes for every letters 'a' to 'z' from String s. After storing, sort the indexes of individual letters. At first make a variable idx set to 0 before iterating String t. set ans to 0. While iterating on String t, you can check if current character happens from idx to afterwards in String s by just doing binary search on that character's stored indexes. if idx is not found in stored indexes, set idx to 0 again and now it must be found. just increment the ans when idx is not found.

just in case when at least one letter in String t is not in String s, then the ans is -1.

https://codeforces.com/contest/1295/submission/69785885

Solved it, but I had to use binary search. But the previous try was in Python, so maybe because it is not required.

See https://codeforces.com/contest/1295/submission/69759300

Although this algorithm is slow,it can avoid using binary search.

This question was copied from here not fair:( https://stackoverflow.com/questions/53065084/minimum-no-of-operations-required-to-create-string-a-by-appending-subsequence-of

These are some characteristics which education rounds follow.

Ok so indirectly u want to say that in educational rounds before solving any question we should Google it??nice

Yes you can google and stay green.

Ok Mr Red xD

Yes you can do it.

But are you sure that you can solve the problems even if you google it?

So that's the reason why you are staying green.

If you still want to stay green,you can google the problems,even if it violates the rules.

xD

Dwas similar to https://www.codechef.com/problems/CDZ14DC was copied from here https://stackoverflow.com/questions/53065084/minimum-no-of-operations-required-to-create-string-a-by-appending-subsequence-of

It might be a very weird thing to ask as an author, but how to solve F?

It seems that my model solution is a lot more complicated that the ones from the participants.

Short solution summary (div1 500)

Thank you! It's amazing how the combinatorial approach to the problem can make the solution a lot easier. Unfortunately, I didn't even think about that, and the probabilistic way lead me to maintaining and interpolating the probabilities as polynomials (which is way harder, obviously).

Hi, can anybody make a hack? I receive an error after Hack button..

Me too. The error message "Illegal Contest ID" is shown

Open the solution by clicking on its id and the click

hackitThank you!

Me too.

Same

Did you really think this B was appropriate?

ya, yep, yes, yo

I just found a cheat in this contest;

At "Hacks" page of this contest. I found ma_da_fa_ka and InsaneNerd hacked each other on problem B. Since it's very funny, I looked into their solution and only to discover they copied each other and the both get accepetd in their latest solotion!

By the way, ma_da_fa_ka's code even didn't have indentation.

By the way, ma_da_fa_ka used bad word as his handle. (lol)

And one of the funniest thing is that ma_da_fa_ka even tried to

hack himself,although that hack wasunsuccessful. xDSo why

Codeforcescan't stop them from cheating?I think that will be a big problem.

Wish MikeMirzayanov can manage to stop them.

How do I know if D Euler(m / gcd(a,m) )

Instincts....

Let $$$d = gcd(a, m)$$$.

Since $$$0≤ x < m$$$ and $$$1 <= a < m$$$, we easily conclude that $$$(a + x)$$$ mod $$$m$$$ can take values in the range $$$[0, m - 1]$$$ and we know that $$$gcd(a + x, m) = gcd(m, (a+x)\ mod \ m) = d$$$

Let's look at the case when $$$d = 1$$$, let $$$b = (a + x)\:mod\:m$$$, our goal is to find all the values of b where $$$gcd(m, b) = 1$$$ which is by definition $$$\phi(m)$$$ given that we concluded earlier that b can take any value in the range $$$[0, m - 1]$$$.

Now, let's consider the case where $$$d \neq 1$$$. Clearly, the values of b that we are looking for in the range $$$[0, m - 1]$$$ must be multiples of $$$d$$$, so let's enumerate all the $$$m / d$$$ multiples of $$$d$$$ in this range as follows: $$$ 1, 2, ..., (m / d)$$$. Let $$$i$$$ be the enumeration of the $$$i^{th}$$$ multiple, we want to count $$$i$$$ such that $$$gcd(i * d, m) = d$$$, but since $$$d \ | \ m \ and \ d \ | \ (i * d)$$$, we really want to count $$$i$$$ such that $$$gcd(i, m / d) = 1$$$, which turns out to be $$$\phi(m / d)$$$.

Thank you very much

Problem is saying 998244353 decimal digits. What i was thinking number should be less than 998244353. Can anyone suggest what should i do to avoid such type of mistakes ???

Learn to read.

Thank you so much.

read the entire statement, it also mentioned Note that the answer may not fit in the standard 32-bit

or 64-bit integral data type.Such a bug B!This round shoule be unr!

Agree, I would have spent years trying to debug my solution if I did not decide that solving C first is a better idea...

how to solve B?

You can create an array $$$cnt_0$$$ and $$$cnt_1$$$ in which $$$cnt_x[i]$$$ is the number of characters $$$x$$$ are from $$$0$$$ to $$$i$$$. Also, you can calculate $$$delta[i]$$$ which is $$$cnt_0[i] - cnt_1[i]$$$. Then, when the string is repeated again, the same pattern of $$$delta$$$ is going to occur but increased by a constant, because the number of $$$0$$$ and $$$1$$$ that you have from the previous string are accumulated. I'm going to concatenate $$$s$$$ two times to make the pattern visible.

Let's analyze the following input

$$$n = 6, x = 10, s = 010010$$$

$$$cnt_0: 1, 1, 2, 3, 3, 4 | 5, 5, 6, 7, 7, 8$$$

$$$cnt_1: 0, 1, 1, 1, 2, 2 | 2, 3, 3, 3, 4, 4$$$

$$$delta: 1, 0, 1, 2, 1, 2 | 3, 2, 3, 4, 3, 4$$$

Every time a string has finished, delta is going to change in a constant. Let's called that constant $$$d = delta[n] - delta[0]$$$. For each position $$$i$$$ between $$$0$$$ and $$$n - 1$$$, you know that you can reach all the numbers of the form $$$delta[i] + kd$$$, in which $$$k \geq 0$$$ because it is the number of times that the string is going to be repeated. So, for each $$$delta[i]$$$, $$$i$$$ between $$$0$$$ and $$$n - 1$$$, you have to check if it possible to obtain a non negative $$$k$$$ such that $$$x = delta[i] + kd$$$. The only corner cases are in which $$$d = 0$$$, then, if you have at least one solution, you are going to have infinite solutions of this form. Finally, check the case of the empty string if the solution is no infinity. The time complexity is $$$O(n)$$$.

Thank you for the clear explanation.

I have somewhat awkward question:

Have you seen similar problem before? In other words, do you feel that your experience has helped you to solve it?

I was trying really hard to solve it for 90 minutes, but get lost in own thought process. So I wonder if I need more practice or maybe this CP thing is simply not for me ;-).

I don't like problems like today's B where you have to be so careful with corner cases. There isn't any fancy idea behind, just annoying implementation ._.

take '0' as $$$1$$$ and '1' as $$$-1$$$ and calculate the prefix sum $$$pre$$$.

then for each $$$pre_i + x\cdot pre_n = balance$$$_$$$value$$$, if there is a non-negative integer $$$x$$$ satisfying the formula, add 1 to the answer. if balance_value is zero then add another 1 to the answer.

if $$$pre_n=0$$$ then if there exists an $$$pre_i=balance$$$_$$$value$$$, then answer is -1; else answer is 0.

why add 1 when balance_value is zero?

Because empty prefix can be a prefix too. So the balance of empty prefix is 0.

VERY VERY WEAK test cases for B :(

Author is fond of strings i guess.

D can also be solved with mobius function(Inc-Exc) 69790210. Let g = gcd(a,m). gcd(a+x,m) is g if and only if g|x. gcd(a+x,m) = gcd(g*c1 + g*y, g*c2) = g. gcd(c1 + y, c2) = 1. Use inc-exc now — neglect non-square free numbers for others if a divisor is prime product for even number of primes than subtract such numbers, for odd add them using the observation that if mu[n] = mu[d]*mu[n/d].

Anybody with a good hack for B?

2 6 2 001001 3 0 011

Answer should be 3 3

Thanks, some of the hacked solutions fail on this test indeed. Interesting

How to solve C ?

See 69759300

Let's store index of every letter (in vector, for example). Then, just simulate building of string with fast find of needed index — lower_bound is good for this purposes. 69790270

Problem "A" was similar to this problem. Link: www.hackerrank.com/contests/programming-contest-4-cse-ian-of-bd/challenges/number-with-sticks

I am not sure about my solution,can anybody hack it.

It is correct for sure.

that was one of the worst contest in all of the codeforces rounds. the problems were really bad. i hope to see better contest in future.

why is it really bad? is it because you didn't do well in this contest.

what a good contest for you would look like?

for starters, a good CF contest wouldn't have more people solving C than B

so what? the problems in edu rounds are equal (there are no points for problems, ICPC style). so solving C more than B is just a matter of swap. so people should see the number of participants who solve the problem and move to that problem. and being stuck in a problem is no good skill. people need to learn to move on.

and even if the problems were given points and people solve c more than b, it does not yield the entire contest bad.

you guys just talk trash about the contest while you do not know the efforts made to prepare the rounds. It is a lot harder to estimate how the contest would go for the participants.

That might've worked if they had managed to prepare B well enough to avoid errors in the testcases

Bruh salty. Edus are getting better ever since they reduced it to 6 problems.

BRs82 was one of the worst person in all of the codeforces users.The comments were really bad.i hope to see better user in future.

Can someone explain the idea "phi value of m/(gcd(a,m))" for problem D? I mean how we got it?

This has already been explained in previous comments.

Let $$$gcd(a, m) = g$$$

We have $$$gcd(a, m) = gcd(a+x, m) = gcd((a+x) \ \% \ m, \ m)$$$

Notice that as $$$x$$$ increases, $$$(a + x) \ \% \ m$$$ will end up taking all values in $$$[0, m)$$$

So we only need to find the number of $$$k \in [0, m)$$$, such that $$$gcd(k, m) = g$$$

Since $$$gcd(ca, cb) = c \iff gcd(a, b) = 1$$$, we can take any integer coprime to $$$m/g$$$ and multiply it by $$$g$$$ to find a suitable $$$k$$$.

Got the idea.

Thx a lot

will there be points for successful hacking during the hacking phase?

[Nothing bruh]

sorry?

Educational CF Round and Div.3 Round have no points for successful hacking attempt.

69796073

Is it possible to see the hack case for my solution after preliminary result? I really have no idea of the counter example :(

I solved E without any range updation datastructure , just use map of values which have arrived and changes the current cost with coming elements and decrementing the cost if already added to the initial cost, it gives me wrong answer on testcase 10, can anybody look into it my submission https://codeforces.com/contest/1295/submission/69796456

Can anyone please tell me what is wrong with my code for problem B. https://codeforces.com/contest/1295/submission/69788305

The EMPTY STRING is also a valid prefix.

It was ensured in the announcement.

A was very similar to 101612A - Auxiliary Project

What is "Unexpected verdict"? Hack #613188

It means that one of the problem makers' solution is incorrect. (The official solutions had a different answer to each other.)

Didn't know much about such a nice function (Euler Totient Function)

Thanx to problem D for increasing my knowledge.

I think so many TLE codes are "accepted" in Problem C.

A very simple testcase in Problem C was hacking testcase for many solutions.

1

aaaa....a(length is 100000)

aaaa....a(length is 100000)

If I woke up early, I could do more hacks.

Why can such codes pass all the test data? Is all the test data random?

Yes, Testdata is too week. But it is no matter, because this is an "Educational Contest" which is for not only algorithm ranking but also hacks ranking.

I think if you hacked at least one problem with this test case, it might be added to final test.

True. So the codes get TLE finally anyway. but my hack score ...

Editorial please?

For problem A,the number has two forms:

111...111

or

7111...111

If n is even,the first form is better,otherwise,the second form is better.

The editorial of B,C and D are showed in former discussions.

I haven't solved problem E,F yet,so you can ask tourist . :D

deleted

Am I the only 8ne who solved C with a precomputed DP

Nope,but INMAK also uses DP.

When will the system testing start?

It's pretty.

where's the rating Lebowski

For question B: Can anyone explain why the output of

is 2 ? Please help, my solution is hacked and I don't know why

It's 3.

I get it, thanks

There are three prefixes that satisfied the condition:""(EMPTY STRING),"01","0110".

No matter how hard I try, I could not understand the solutions given for problem D. It would be great if someone could explain me how to do in simple and detailed steps. I would be really grateful to them.

Thank you!

Let X be gcd(a,m),so gcd(a/X,m/X)=1.

If gcd(a+x,m)=X,then (a+x)%X equals to 0,x%X also equals to 0.

So gcd((a+x)/X,m/X) equals to 1.

Find all the x in eular algorithm.

Finally! got it thank you so much :)

when will editorial be published?

Waiting for the Editorials!!

Has the editorial for this contest been published yet? Or there won't be one?

It would surely be published. Not sure about the release time.

expert in a one contest

Auto comment: topic has been updated by awoo (previous revision, new revision, compare).problem A What's wrong in this?

## include<bits/stdc++.h>

using namespace std; long long n; int main(){ int t; cin>>t; while(t--){ cin>>n; if(n%2==0){ for(int i=0; i<(n/2); i++) cout<<"1"; } else{ cout<<"7"; if(n!=3){ for(int i=0; i<((n-1)/2); i++) cout<<"1"; }

cout<<endl; }

}

In your loop for(int i=0; i<((n-1)/2); i++), it should be i < (n — 3)/2 instead, as 7 takes 3 resources.

thanks buddy!!

Why is this solution so slow? (Problem C):

http://codeforces.com/contest/1295/submission/70937323

As for me, the complexity of this one is O(T*|t|*log(|s|))