shishyando's blog

Codeforces Round #671 Editorial
By shishyando, history, 4 years ago, In English

All problems were prepared on Polygon by me, so if you faced any problems (we know about A) I am really sorry about it.

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Idea: shishyando

Jury solution: pastebin

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Idea: Artyom123

Jury solution: pastebin

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Idea: shishyando

Jury solution: pastebin

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Idea: Artyom123

Jury solution: pastebin

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Idea: shishyando

Jury solution: pastebin

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Idea: isaf27 + Kotehok3

Jury solution: pastebin

  • Vote: I like it
  • +223
  • Vote: I do not like it

Comments Comments (171)
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4 years ago, # |
  Vote: I like it +22 Vote: I do not like it

Good job!

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4 years ago, # |
  Vote: I like it -17 Vote: I do not like it

Fast blazingly fast

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4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Thanks for quick editorial.

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I see people down-voting, I just liked the editorial and i was thankful for the contest and the quick editorial, now ,whom did I hurt or what was wrong in this comment for these downvotes?

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4 years ago, # |
  Vote: I like it -11 Vote: I do not like it

Thanks for the speedy Tutorial man!

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4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Thanks!

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4 years ago, # |
  Vote: I like it -10 Vote: I do not like it

For D2 My solution got hacked for testcase 67 Can anybody help me? This is my submission link: https://codeforces.net/contest/1419/submission/93235884

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    That was a system test, not a hack test

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Um, for your information, there is a button called "Click here for more information" at the bottom of the page of every submission. Click on it and you can see all tests and know what kind of test data you fail on.

    BTW it's really important to make good index calculation :) And not don't blame hacks, though this solution is hackable.

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4 years ago, # |
  Vote: I like it +4 Vote: I do not like it

I'm still struggling with B...

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    4 years ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    try making out patterns on paper and see how the pattern goes, they gave un hint in sample that answer was going to be in 1 to 30 only. so you could simply precompute these 30 values.

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    You can solve it through recurrence. The first stair case (a single green square in the picture in the problem statement) with will have $$$1$$$ cell. Lets denote the number of cells in the first stair case with $$$a_1$$$ so $$$a_1 = 1$$$ cell. The next possible stair case (the second possible stair case) will have $$$a_2=6$$$ cells. It's made out of two green cells (two $$$a_1$$$s) and one yellow square which has $$$2^2$$$ cells (as in the picture in the problem statement) so $$$a_2=2*a_1+2^2$$$. The third possible stair case will have $$$a_3 = 2 * a_2 + 2^4$$$ cells. I hope that by this point you can see that $$$2^4$$$ corresponds to the red square from the picture in the problem statement? So now the general formula for the number of cells in a nice stair case is $$$a_i = 2 * a_{i-1} + 2^{2*(i-1)}$$$. Now all you need to do is count $$$a_i$$$ up to some value so that you will meet the problem constraints. Then count a prefix sum array of the $$$a_i$$$s and then you can greedily iterate over it up to a given $$$x$$$. Here's my code 93245756.

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      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thanks, I get your explanation, not sure if its really a proof but close enough. I wasnt confident enough in the contest to procede with this idea

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        4 years ago, # ^ |
          Vote: I like it +9 Vote: I do not like it

        Don't forget to use 1ll if you want to make some power of two ( use 1ll<<x instead of 1<<x), that cost me about 40 min to debug during the contest...

        GL

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          4 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Bro what does this 1LL does and what do you mean by 1ll<<x and 1<<x

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            4 years ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            1LL means 1 but in type long long. Just the same as (long long)1

            1ll<<x means shift the bits right x times in binary. This is how I do for getting powers of two in C++. Bit operations are useful to me :v

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              4 years ago, # ^ |
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              Author's solution is confusing me can you help me bro

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              4 years ago, # ^ |
                Vote: I like it 0 Vote: I do not like it

              hi pls can you tell me why do we need to use 1LL and not just 1?

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                4 years ago, # ^ |
                  Vote: I like it 0 Vote: I do not like it

                So that there is no overflow of number in int

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                  4 years ago, # ^ |
                  ← Rev. 2   Vote: I like it 0 Vote: I do not like it

                  just out of curiosity i have written the code without using 1LL but it still giving the desired outputsolution without using 1LL pls can you tell why this is showing accepted then

                  thanks

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                  4 years ago, # ^ |
                  ← Rev. 2   Vote: I like it +1 Vote: I do not like it

                  Because you have i already ll long long int type One side of expression is long long int type it is enough

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      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Why are we doing prefix sums?

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        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        You don't necessarily need it. We are searching for the smallest n such that a1 + a2 + ... + an < x(the input val), so if you do prefix sum, you can search more conveniently. Again, this is not necessary.

        Pls upvote if this helps :)

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4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Jury's solution for D2 can't opened..

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4 years ago, # |
  Vote: I like it +23 Vote: I do not like it

Please make strong pretests from next time. Nothing is more painful than getting your problem A wrong after the system testing :/

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    4 years ago, # ^ |
      Vote: I like it +53 Vote: I do not like it

    First time creating a contest is hard, some mistakes may be made.... We will be even more careful next time.

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      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      ㅤ

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      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      how u come up to 2^k-1 diectly..in div2b can u explain a bit. Thnx

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        4 years ago, # ^ |
          Vote: I like it +27 Vote: I do not like it

        There's a really useful tool called messing around with the numbers

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        4 years ago, # ^ |
        ← Rev. 2   Vote: I like it +12 Vote: I do not like it

        First, one can easily see the basic solution is one stair (after understanding the problem, which took me something like 10 min).

        Then, the sample input explanation tells you that another nice stair is which have 3 stairs.

        Lastly, (maybe) you would notice that the image in the description has 7 stairs, and there are only squares with a side length of powers of two.

        So I assumed that nice stairs must consist only squares that have side length of powers of 2.

        Next is to prove it (you can choose to prove it by AC, but that's not what I'm gonna do here) using mathematical induction (or something similar): 1. assume that a solution with n stairs ( a_n ) exists. then you can see that by duplicating this solution and attach them to a bigger square in the middle (like the yellow square in the description), you can make solution ( a_{n+1} ), as you just created a new nice stair contains (2n+1) squares and (2n+1) stairs.

        Since the base case is 1 stair with 1 cell, the only solutions must be something in this series.

        I'm lazy so I'm not gonna write down the formulas. Hope this make you understand better.

        And, make sure to use 1ll instead of 1 when using 1ll<<x for powers of two. This cost me 40 min approximately during contest, which results in not having enough time for implementing pE and broke my dream to 1900+.

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          4 years ago, # ^ |
            Vote: I like it +3 Vote: I do not like it

          This proof is not satisfactory enough. At best, it says that it works for 2*a_i +1 if it works for a_i. How do you still know there aren't any other numbers in the middle for which it works?

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            4 years ago, # ^ |
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            Oops my bad. I forgot to mention that this is what i went through during the contest. As you say, this do not prove that there is no other solution, but I thought this was correct and gave it a try. If anyone knows the full proof pls let me know ><

            Thanks for pointing that out :)

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4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Please make tight pretests I realized my mistake after the contest in Problem A. Please try to set some pretests because passing two pretests I can then console the answer as well

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4 years ago, # |
  Vote: I like it +10 Vote: I do not like it

My contest was ruined because of B. I will look for pattern and think less next time .

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    4 years ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    I found the pattern in the "sum of first n numbers formula" during the contest. I just figured that we have to increase n by a power of 2 in this formula (n*(n+1))/2 and subtract it from x until you run out of cells. I just made this conclusion from the first two good staircases and tried my luck. here's my submission: https://codeforces.net/contest/1419/submission/93215283

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4 years ago, # |
  Vote: I like it -22 Vote: I do not like it

"If the single last digit is odd, then Raze wins, else Breach wins." It is not mentioned that If the single last digit is even then Breach wins. How can you assume that ?

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    4 years ago, # ^ |
      Vote: I like it +17 Vote: I do not like it

    If the single last digit is odd, then Raze wins, else Breach wins

    If the last remaining digit is not odd, it is even. That should be obvious.

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +1 Vote: I do not like it

    I didn't understand your question. Any digit is either odd or even. Hence, if the single last digit is not odd, then it is even.

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    4 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    My bad, I misinterpreted it -_-.

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Great Contest !! Good work Guys

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

long statements were obvious and low pretests are obvious too because of this much of WAs after the contest

but out of those, problems were good

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4 years ago, # |
  Vote: I like it +2 Vote: I do not like it

On question e: So, I don't know why, but I misunderstood the meaning of coprime (yeah I know) to be divisible, so two numbers that are coprime means (what I thought) a number is divisible by the other.

So I was solving a more restrictive case of the question. Well, not that it matters. I wasn't able to get it in time anyway.

Thanks for the fast tutorial!

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I am unable to get the explanation for B. I know the pattern , but can someone prove it please ?

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    4 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    See, if you just try it out by yourself, you'll notice that stairs are only possible for n values: 1,3,7,15, 31....., i.e 2i+1. And also the cells needed for a particular value of n, is actually n(n+1)/2 . So one could just loop while sum is less than x and updating n with 2n+1.

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4 years ago, # |
← Rev. 3   Vote: I like it +13 Vote: I do not like it

https://artofproblemsolving.com/community/c6h34314p213007

A was too hard though :)

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4 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

What's the implementation idea for E? Even with the editorial, I still don't see how to efficiently sort all divisors into these "buckets" based on what prime divisor they have. Or is O(number of divisors * number of prime divisors) good enough?

Edit: thanks to everyone who responded. I forgot just how sparse primes are. Turns out, you don't even need to precompute primes, just checking for primality with a simple $$$O(\sqrt{n})$$$ function is enough: 93278287

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    4 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    The maximum "number of divisors" for $$$n \leq 10^9$$$ should be around 1000 so it is enough to pass

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it

    Before solving any queries, precompute all "small" primes up to $$$\sqrt{10^9}$$$. Then any number up to $$$10^9$$$ can have at most one "large" prime factor.

    Now to solve each query:

    For each $$$n$$$, iterate through the "small" primes to find out which "small" primes are prime factors of $$$n$$$. You will not TLE because there are only a few thousand "small" primes.

    Using this you can find the "large" prime factor if it has one. There will be at most $$$9$$$ prime factors in total because the product of the first $$$10$$$ primes is more than $$$10^9$$$.

    Then, iterate up to $$$\sqrt{n}$$$ to find all factors of $$$n$$$. For each factor, you can check all the prime factors. Again, you will not TLE because of the limit on the total number of factors.

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    4 years ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    The problem guarantees that

    the total number of divisors of n for all test cases does not exceed $$$2⋅10^5$$$.

    And number of prime divisors is a small number, as it is at most 10 (2*3*5*7*11*13*17*19*23*29 = 6e9)

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    4 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Number of prime divisors is less than 10. And number of divisors is bound roughly by square root of n (it is only a loose bound, but already enough).

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    4 years ago, # ^ |
    ← Rev. 5   Vote: I like it +5 Vote: I do not like it

    More Simple (Easy to Understand) Solution for E:

    Insert all the factors of n in a set.

    Set the first value as s.begin()

    After that for each index i, find the first number from current set (using just simple iteration) whose gcd with ans[i-1] is not 1. And put that particular value at the ith position.

    That's it. You will be able to fill the whole array this way. No matter what is the value of n.

    It can be proved intuitively, mathematically, and with examples too. (Left to readers)

    For the number of insertions,

    The whole array won't need any insertions. Because we filled it in a way, that needs 0 insertions, for such arrangement.

    So, just check gcd(ans[0],ans[no. of factors-1])

    If it's 1, then insertions req = 1

    Else, insertions req = 0

    I solved it, this way.

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      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it +3 Vote: I do not like it

      It can be proved intuitively, mathematically, and with examples too. (Left to readers)

      I'm really curios about this, can you at least provide some intuition?

      This is definitely much easier to implement, but I don't see why this process will never get itself in a corner. You essentially have a greedy algorithm, and it's a bit surprising to me that it works.

      I think there are two big questions: (1) why is there always a non relatively prime number available, and (2) why will the last one never be relatively prime with the first number (the first one must be prime, so basically why is the last number a multiple of the first one?)

      Edit: Ok, I see why (2) is correct: the last number you pick will always be n, so it's of course a multiple of the first divisor.

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        4 years ago, # ^ |
        ← Rev. 2   Vote: I like it 0 Vote: I do not like it

        Because n always come at the last position as you said and it is not coprime to any of the earlier numbers.

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        4 years ago, # ^ |
        ← Rev. 2   Vote: I like it 0 Vote: I do not like it

        I got a simple intuition while solving,

        It was like:

        If I take 1st element as smallest number (It will always be prime), and 2nd element as the smallest available non co-prime number with the 1st element, then, there will be a number in the factor list which will be non-prime with the 2nd element, always.

        Let's take an example:

        n = 6 Factors = 2,3,6

        ans[0] = 2 (We will set this)

        ans[1] = 6 (gcd(2,6) != 1)

        Now, if we have 2 & 6 in the factor list, one can easily say that 3 and maybe some other multiples of 3 too, will be there in the factor list, which will be non co-prime with 6.

        So, this will be valid for filling out the whole list.

        This was my intuition and an informal proof, for this solution.

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          4 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          can you please explain, how will there always be a non co-prime number in the divisor set available? can this be proved? because this method is really great!Thanks!

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    4 years ago, # ^ |
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    I don't know why this simple solution works 93280623. Is test cases are Weak?

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I have used the same approach as mentioned in the editorial for Problem A & C but I am getting wrong answer.
Can anyone please help me out?
Problem A my submission.
Problem C my submission

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Problem A: "num" can be many digits long so it cannot fit in the int data type. You should store "num" as a string instead.

    Problem C: You missed out one case. From the editorial: "If at least one account is already infected we can infect all other accounts in a single contest."

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4 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

In question A it is not given that positive integer is string(it could be taken as int type too).

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Which type you use is your personal affair However, it is given that it has up to a thousand digits, which does not leave much choice

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It's just stdin. You can read it as int or string but since n <=10^6, that integer value as a whole will be huge!

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4 years ago, # |
  Vote: I like it +19 Vote: I do not like it

nice div2E

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4 years ago, # |
  Vote: I like it +20 Vote: I do not like it

I think you overcomplicated D2... It's literally D1, just a tiny bit different in calculating the number of cheap ones.

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4 years ago, # |
  Vote: I like it +12 Vote: I do not like it

why i can't see jury solution..?

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yeah, it's a real problem. Fix jury's solutions, please!

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4 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

In Problem C, I thought rating can't go beyond [-4000,4000] so messed up :(

I am curious though, how would we solve this problem if rating can't go beyond this range.

Any ideas?

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4 years ago, # |
  Vote: I like it -9 Vote: I do not like it

After seeing the number of successful submissions in D2 and E, I really wish I hadn't tested this round and would've given it live. Would've been a really great opportunity to go back to purple. ;__;

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

https://codeforces.net/contest/1419/submission/93253990 Can someone please tell why this code is failing at test case 2. It is giving the right answer when i put in those individual test cases, but for some reason when i am putting them collectively, its failing to give the right ans.

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4 years ago, # |
  Vote: I like it -8 Vote: I do not like it

one of the best codeforces round i have taken part in .. great problems, enjoyed a lot :) :)

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4 years ago, # |
  Vote: I like it +9 Vote: I do not like it

There was not a necessity of including the problem D1 in this contest.

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Problem D is solved more than Problem A.. weird..

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It's not weird XD this problem indeed has the same difficulty as A.

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4 years ago, # |
← Rev. 3   Vote: I like it +20 Vote: I do not like it

D2 can be solved in a very simple greedy fashion. I simply sorted a[] and first filled all even positions for the final order and then filled all odd positions in the increasing order of a[i]. See my submission for better understanding — https://codeforces.net/contest/1419/submission/93275649

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    4 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    I think sorting takes O(nlogn). So solution can also be O(nlogn). Right?

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      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Yup, but i found this to be very straightforward and elegent.

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I am practicing after contest and I also came up with this solution intuitively but I wonder how to prove it, do you also have a proof?

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yes, seeing many of submissions use greedy. Why this greedy approach can work? Any proof or idea can help me better understanding on this? Thanks!

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4 years ago, # |
← Rev. 2   Vote: I like it +7 Vote: I do not like it

On clicking the link for jury solution of E, it shows "You are not allowed to view the requested page." P.S. It is fixed now.

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4 years ago, # |
  Vote: I like it +5 Vote: I do not like it

When I click on Jury Solution I get an error "You are not allowed to view the requested page".

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4 years ago, # |
  Vote: I like it +20 Vote: I do not like it

I have a question about D2.

My solution is to just order n/2 biggest elements at odd position in increasing order, and n/2 smallest elements at even position in increasing order.

After I solved E, just about 10 minutes were left so I submitted this solution without any hope. But it passed system test .. Can someone prove my solution?

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    4 years ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    Let's call a number "big" if it's bigger than the median, "medium" if it's equal to the median, and "small" if it's smaller.

    You can easily see that an optimal solution can be constructed that consists only of small and medium numbers. In your solution, all the small numbers are automatically cheap. And the number of cheap medium numbers is bounded by the number of large numbers (minus 1), which your solution also gets.

    So it's impossible to do better.

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Assuming that your proposed solution is 1-indexed, consider the following. On placing the smallest (n / 2) elements in ascending order in the array at even positions, it's ensured that the two smallest elements in the larger (n / 2) elements will be matched against the smallest element in the entire array, In case of any other ordering of the two smaller elements in the array, against which an element from the smaller n / 2 elements may be compared, there is always an alternate ordering where the comparison is never worse.

    For instance, if the two smallest elements in the larger (n / 2) array are x and x + y, and they are currently compared with some s, where s is not the minimum of the array, replacing s with s — t (t > 0) never yields a worse comparison result.

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    4 years ago, # ^ |
    ← Rev. 9   Vote: I like it 0 Vote: I do not like it

    I also solved this way

    My thoughts

    The array will look like this

    \/\/\/\/\/\/

    So obviously we will not get more than n // 2 items and we should consider n // 2 smallest

    Also, for n < 3 the answer is 0

    Now, let's look at elements [n // 2], [n — 1] and [n]

    If [n - 1] > [n // 2],

    then everything is fine, we captured the most expensive item we could hope for, there is no better use for [n — 1] and [n]. We can add one to the answer and forget [n] and [n // 2], effectively reducing the problem to size (n — 2). We don't forget about [n — 1], but on the next step we will have to place it at exactly the place where it already is.

    If [n - 1] == [n // 2]

    Let's consider [k] such that [k] is the rightmost element less than [n // 2]. Then all elements in range k + 1... n // 2 are absolutely unreachable no matter what we do as they're equal to [n — 1]. And all elements in range 0... k are reachable by elements n // 2... n because these elements are certainly bigger

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4 years ago, # |
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Though I liked the questions , and also the editorials were fast , I would like to request you to make better pretests , as it hurts man when one of your solution which you assume has been passed gets failed, it really hurts man. Today I have seen many of them for question 1st and 5th , hope this doesn't happen again , as it really really hurts

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4 years ago, # |
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There is no access to jury's solutions

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4 years ago, # |
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Can someone help me with the search for errors in my solution of E? I can't understand tester's comment on WA2.

wrong answer Integer parameter [name=divisors[1252]] equals to 0, violates the range [2, 735134400] (test case 1)

Where is the problem in my answer for this test? I can't see full correct answer, I can't see jury solution right now, I don't know where is error in my program :(

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    4 years ago, # ^ |
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    It means you have not output all divisors of 735134400. There are 1344 in total. The judge has read the zero in the second line as a divisor not as the number of moves. That is why you get this error message.

    Edit: Since 1 is coprime to all numbers, you only have to print 1343 divisors.

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4 years ago, # |
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Its nice to have these kind of "adhoc only" round sometimes. Even the idea for F problem is very easy to come up with (implementation is definitely tedious). Nice round!

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4 years ago, # |
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Thanks for soooooo fast tutorial! And also thanks for the great contest, it improved my coding skills greatly lol.

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4 years ago, # |
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The author's solution E is not correct, in particular, for example 12. It will make 2, 12, 6, 3. And the answer will be 1, although it is clear that it is 0 (that is, for 2 prime and at least 4 divisors)

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    4 years ago, # ^ |
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    They haven't explained how the following case will be handled exactly:

    $$$ {p_1}^{q1}.{p_2}^{q2} $$$ where ($$$ q1 > 1 $$$ or $$$ q2 > 1 $$$). 12 falls in this category, although they have handled this case in their jury solution.

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    4 years ago, # ^ |
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    This can be easily handled if after making the structure mentioned in the editorial, one starts filling up divisors from the last prime $$$ (p_{n}) $$$, this way we can ensure that there will be always something in between $$$ p_{n} $$$ and $$$ p_{1} $$$.

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4 years ago, # |
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I fst *2...

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4 years ago, # |
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not a good contest

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4 years ago, # |
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It's also possible to solve problem F with the observation that, given any minimum spanning tree of a graph, that graph is connected by edges of length at most $$$t$$$ if and only if that minimum spanning tree has no edges longer than $$$t$$$. (This is equivalent to saying that Kruskal's algorithm works.) Adding one vertex and at most four edges to a graph doesn't change its minimum spanning tree very much, only removing at most three of the edges in the original minimum spanning tree, so it's possible to calculate the minimum value of $$$t$$$ for any plausible relevant detachment location in $$$O(5^2)$$$ with a modified Prim's algorithm on an 5-vertex graph associated with that location. (See 93293641.)

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4 years ago, # |
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Framing of B question was very confusing. It took me more than 1 hr to just know what the question is saying.

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4 years ago, # |
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In B, the problem is basically choose the maximum count of numbers from (1, 3, 7, 15 ....) and form any number less than equal to x. Why can we solve it greedily? Shouldn't we be applying dp?

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    4 years ago, # ^ |
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    Because we need to find the maximum different nice staircases using 'at most' n cells, and if we use from small nice staircases to big nice staircases It always have the correct answer. Using dp, if he ask 'how many ways' using exaclty n cells.

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      4 years ago, # ^ |
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      I still don't understand. Isn't the concept similar to coin change dp problem?

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        4 years ago, # ^ |
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        In this specific problem, you need dp if you have to use exactly $$$n$$$ cells.

        Greedy is enough if you have to use at most $$$n$$$ cells.

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4 years ago, # |
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Am I the only one who thinks that C could have been phrased a little better ?

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4 years ago, # |
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Loved the contest! I think D questions were a bit too easy, but hey I'm a newbie so I shouldn't complain!Also a quesion on Omen could've been more interesting ;)

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4 years ago, # |
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In problem stairs:

If we take staircase of height 6, then it can be an answer, with 1 square of size 3, 2 squares of size 2 , and rest will be squares of size 1, they will be disjoint , only corners of same size square will touch.....check if this case is right or wrong.

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4 years ago, # |
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who can tell me the train of thought ( is it right? ) of F? ( or tell me where it is thanks )

i can't understand the solution

thank you so much

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4 years ago, # |
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shishyando: Although the contest was good, the problem statement of C was not really clear. After reading the tutorial only, I figured out that if I have one infected account, I can infect all others by making all others equal to infected one, then change the infected one. Reading the problem statement, I didn't find any clue of that, what I interpreted while reading was, that newly infected accounts and the previously infected account (from which new infection occurs) should have same ratings even after the contest. I think mentioned word "instantly" didn't really justified the phenomenon you were trying to convey. It may be possible that only I found this thing to be difficult to understand. I hope you take this as a constructive criticism.

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4 years ago, # |
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Ok but E was just USAMO 2005 right?

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4 years ago, # |
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https://codeforces.net/contest/1419/submission/93303906 Please tell what is wrong here.

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    4 years ago, # ^ |
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    Why are you not getting Segmentation fault in these lines?

    for (ll i=0;i<n;++i){
          if(a1[i]>a1[i-1]&&a1[i]>a1[i+1]){
              ++cnt;
             
          }
      }
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      4 years ago, # ^ |
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      can you please explain further

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        4 years ago, # ^ |
        ← Rev. 2   Vote: I like it 0 Vote: I do not like it 
         for (ll i=0;i<n;++i){
          if(a1[i]>a1[i-1]&&a1[i]>a1[i+1]){

        1. Your if condition is wrong. Count will increase if a1[i] is smaller than a1[i-1] && a1[i+1].

        2. for i = 0, a1[i-1] is not defined, and for i = n-1 , a1[i+1] is not defined.

        Correction : 

        for (ll i=1;i<n-1;++i){
          if(a1[i]<a1[i-1]&&a1[i]<a1[i+1]){
              ++cnt;
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          4 years ago, # ^ |
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          i did as you said but now it is failing at the very first test.

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4 years ago, # |
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I have an alternate solution for D2. Sort the array, place the first half on odd indices and second half on even indices. For example if the sorted array is 2,4,6,8,10; then make the array as 6,2,8,4,10. After this, count the number of elements which are less than their neighbours. If there are repeated elements, it would still be a valid arrangement to find maximal no. of cheap spheres.

This is in a way similar to your solution, where I am using the same concept that if we can find x cheap, we can definitely find x-1 cheap spheres too. By this method, you can directly get the maximum amount of cheap spheres possible.

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    4 years ago, # ^ |
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    Yes, most people who did greedy used this. It's easier to see.

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4 years ago, # |
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Still can't understand problem B. Can anyone help me?

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    4 years ago, # ^ |
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    If you managed to find the pattern, then the problem becomes pretty easy. Here's my submission- 93223145

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4 years ago, # |
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what does this line mean ? A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells.

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    4 years ago, # ^ |
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    It means if a staircase can be completely covered with n non-overlapping squares that don't go out of the staircase, it's called nice. (look at example)

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4 years ago, # |
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In E, what if k=2 and q1+q2>2, n=p1^q1+p2^q2. Then p1*p2=pk*p1, so it doesn't work. For example: n=24, 2 4 8 24 6 3 6

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4 years ago, # |
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Why simple sorting and greedy placing did worked for me in problem D2 submission

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4 years ago, # |
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Could someone please provide a formal proof of why the greedy solution in D2 works. I understand till the idea that picking the smallest (n-1)/2 elements is the most beneficial way as it's always better to keep the possible candidates for the center as low as possible. But I'm not able to proceed beyond that in the proof. Thanks in advance.

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4 years ago, # |
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https://codeforces.net/contest/1419/submission/93311817 can anyone please tell me why I am getting wrong Answer

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4 years ago, # |
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Can someone tell me what is the complexity of this solution for problem E? 93314845

During the contest I was sure this would TLE, but voila!!!

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4 years ago, # |
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I think the statement of the ahead two problems is a little bit long.....

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4 years ago, # |
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Why to make D2 complicated (by using binary search) if this simple approach is enough.
Solution

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4 years ago, # |
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Please somebody tell me what am I missing in C
https://codeforces.net/contest/1419/submission/93322918
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    4 years ago, # ^ |
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    In the case in which all the elements are equal in your code you have 2 options, if they are equal to x the answer is 0 but in the opposite case the answer must be 2 because there is no way that the sum of the differences is 0, so you have to infect n-1 and leave one uninfected to make up the sum for what you need 2

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4 years ago, # |
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Thanks for nice problem set and quick editorial.

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4 years ago, # |
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https://codeforces.net/contest/1419/submission/93326069

I am failing test set 10 on problem E because of incorrect number of transformations on case 25, yet when I compare it with correct solutions, it gives the same number of transformations. Is there a bug? Or am I missing something?

Thanks in advance

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4 years ago, # |
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https://youtu.be/fHyJObr8EdI My solutions ( A-D )

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4 years ago, # |
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Hi everyone! I am not able to understand the logic of the part ~~~~~ else if (cnt > 0) { cout << 1 << '\n'; } ~~~~~ in the jury solution of the problem Killjoy at line number 21. If cnt represents the number of ratings equal to x initially, how does it being greater than 0 helps in determining the contests to be 1?

E.g. If we have x = 70 and the ratings as [70,72,74], the answer should be 2 contests but the solution should show it as 1.

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    4 years ago, # ^ |
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    The answer is 1. Imagine we sum the difference to 70 for all numbers, we call this number k. k = (70 — 70) + (72 — 70) + (74 — 70) so k = 6. It may seem that you need another contest but because 70 = 70 we can infect this account before the contest start. And now we can sum or substract to this k. The process will be: 1.Infect account with rating 70. 2.After the first contest accounts will be [76, 70, 70].

    If (cnt > 0) it means that we can inffect one account before the contest. So we will need just one contest to infect the rest.

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      4 years ago, # ^ |
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      Hi, thanks for the explanation. Also did you mean that in step 2 of the process, after the first contest, the ratings will be [76,70,70] i.e. we will compensate k=6 by adding the already affected account by 6 and subtracting it from the disinfected accounts?

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4 years ago, # |
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CAN ANY BODY TELL THAT IN PROB C OUR NEW RATINGS CAN GO BEYOND -4000 OR +4000 ???

PLEASE tell

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    4 years ago, # ^ |
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    New ratings can be any integer. so the answer is YES , it can go beyond -4000 and +4000

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4 years ago, # |
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My code for problem D accepted D2(93375350) but failed in D1(93375360)

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4 years ago, # |
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I'm afraid the problem E is a bit easier than common...

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    4 years ago, # ^ |
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    And many of my friends said that they have difficulty understanding the problems...

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4 years ago, # |
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neal actually make D2 a greedy problem, how to prove it, can someone help me ?

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4 years ago, # |
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Why are the limits in F so tight? Even after 4 TLEs and a bunch of constant optimisation, my $$$O(n^2 \log 10^9)$$$ (93436282) barely passed in 1996ms.

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    4 years ago, # ^ |
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    Jury's solution passed in 46ms, so we think the limits are ok.

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4 years ago, # |
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Thx

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4 years ago, # |
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There exist a solution of F with time complexity of $$$O(2^m n\log^2 n\log T)$$$ where $$$m$$$ is the number of components that is $$$\leqslant 4$$$ .

It uses scanline and Segment_tree Divide (sorry for cannot explain it in English excatly) to count if there is a point coverd by all the components.

Upd : I've improved my algorithm to $$$O(2^m n\log n\log T)$$$ which allows n up to $$$10^5$$$. We can just turn range modify to range query so that we don't need segment_tree Divide.

code

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4 years ago, # |
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Hello!

I have solved F in the following way and am getting WA6.

Algorithm: The difference with the solution in the editorial lies in checking whether certain value is valid. The way I check it is that after performing a DFS on the graph with a limit $$$T$$$ on the distance, we get a list of components. I leverage the fact that the point to be added must share either $$$X$$$ or $$$Y$$$ coordinate with each of the components. So I pick the smallest (in number of vertices) component and iterate through the $$$X$$$ and $$$Y$$$ coordinates of its points. For each, x0, I compute the possible intervals for $$$Y$$$ coordinates. To do that I compute such intervals for each component one by one and intersect these sets of intervals. Before computing the intervals, I've first sorted the points in the components based on its $$$X$$$ or $$$Y$$$ coordinate depending on for which coordinate I'm computing the feasible intervals. This results in the intervals being sorted in ascending order. In turn, this makes it easier to intersect the intervals in linear time. In the end if the intersection of these intervals is non-empty, then we can find such point, otherwise we can't.

Can anyone see any flaws in the approach or is it rather implementation error?

Any feedback highly appreciated!

Thank you!

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    4 years ago, # ^ |
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    YLWang does this sound similar to your approach?

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    4 years ago, # ^ |
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    shishyando Have you considered or come across a solution similar to the above? Thanks

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    4 years ago, # ^ |
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    Found my mistake, it was extremely subtle typo, copy-paste error. In one place, instance of somePoint.x, I had somePoint.y. Took me hours of debugging...

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    4 years ago, # ^ |
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    I think the main idea of your approach is right. Maybe details got wrong.

    You can use generator and brute-force to check your solution.

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      4 years ago, # ^ |
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      Thanks!

      I sorted it out as mentioned in my last comment. The bug was that when sorting points based on $$$Y$$$ coordinates, when y's are equal, I should compare x coordinates, but I was comparing x coordinate with the other point's y coordinate. Silly typo. Once that's fixed, passed all tests in 249ms in Java 11. Here's the accepted submission

      In terms of finding the error, I was expecting it to be a genuine error so was reading through more complex parts of the solution :) The way I found it was to take fully visible tests from CF tests. I had to run on 30 tests before finding a failing example. For this problem, to generate valid test cases where the answer isn't -1 requires a bit more careful impl than just generating random points. I was reluctant to do that.

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4 years ago, # |
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The harder version of Sage's Birthday got solved by greedy approach.

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3 years ago, # |
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The problem D2 is rated 1500 when I've seen the same problem in Div2A at 800-1000, is there a reason why, Its definitely not a 1500 level problem if I was able to solve it in less than 10 mins (on the time of writing this I am rated 1331).

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