awoo's blog

By awoo, history, 3 years ago, translation, In English

1569A - Balanced Substring

Idea: BledDest

Tutorial
Solution (awoo)

1569B - Chess Tournament

Idea: BledDest

Tutorial
Solution (Neon)

1569C - Jury Meeting

Idea: BledDest

Tutorial
Solution (Neon)

1569D - Inconvenient Pairs

Idea: BledDest

Tutorial
Solution (adedalic)

1569E - Playoff Restoration

Idea: BledDest

Tutorial
Solution (BledDest)

1569F - Palindromic Hamiltonian Path

Idea: BledDest

Tutorial
Solution (awoo)
  • Vote: I like it
  • +108
  • Vote: I do not like it

| Write comment?
»
3 years ago, # |
  Vote: I like it -48 Vote: I do not like it

Fastest editorial I've seen

»
3 years ago, # |
Rev. 2   Vote: I like it +14 Vote: I do not like it

Lesson learned: Don't use segment trees unless required.

(My segment tree implementation for D gave TLE on system testing.)

  • »
    »
    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Interesting. I used segment trees and maps as well, and the highest runtime was ~1 second.
    My submission 128269298

»
3 years ago, # |
  Vote: I like it -7 Vote: I do not like it

code and explanation for problem D

Spoiler
»
3 years ago, # |
  Vote: I like it -7 Vote: I do not like it

I am facing some difficulty in understanding the equation in Problem C: $$$A_n^{n-k-1} = \frac{n!}{(k+1)!}$$$ What is $$$A$$$ here and how did we get $$$\frac{n!}{(k+1)!}$$$?

  • »
    »
    3 years ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    $$$A_x^y$$$ is the number of ordered ways to choose exactly $$$y$$$ different objects from $$$x$$$. So, it's like a binomial coefficient $$${x}\choose{y}$$$, but with also considering the order in which we choose the object. Hence the formula for $$$A_x^y$$$ is $$${{x}\choose{y}} \cdot y! = \frac{x!}{(x-y)!}$$$: there are $$${x}\choose{y}$$$ ways to choose exactly $$$y$$$ objects out of $$$x$$$, and $$$y!$$$ ways to order them.

    • »
      »
      »
      3 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      why to select n-k-1 from n we know which elements are n-k-1( n-k-1 elements are those which are not equal to ax or ax-1,where ax is the maximum element);

      • »
        »
        »
        »
        3 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        It is like considering the positions; instead of the elements. There are in total n positions. We wish to pick n-k-1 positions among them for the n-k-1 elements. There will be k+1 gaps after that. The last gap is fixed for the max element. The rest of the k elements goes k! ways into the rest of k gaps. I was also confused at first. Hope this helps.

  • »
    »
    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    nPr formula is n!/(n-r)! where nPr represents permutating r numbers in n places. It is basically n p (n-k-1).

    • »
      »
      »
      3 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      is it true? nPr= rAn ? bcoz n should be greater than r.

»
3 years ago, # |
  Vote: I like it -7 Vote: I do not like it

Problem C : if (cmx == 1) ans = (ans — sub + MOD) % MOD; I don't know why we need plus MOD. I thought we just need : ans = ( ans — sub ) % MOD Thanks for someone explain this (^^)

  • »
    »
    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It's because ans and sub are both %MOD, and if (ans%MOD — sub%MOD) < 0, so ( ans — sub ) % MOD will be < 0 too, that's because the % of negative numbers is also negative.

    for example, if MOD = 1e9+7, sub = 5 and ans = 1e9 + 8 (but in your code ans = 1, because you do ans%= MOD): ans = (1 — 5)%MOD = -4%MOD = -4

    But if you add MOD this never you be negative, because sub < MOD.

    I hope you understood and sorry for my bad english, I'm not fluent.

    • »
      »
      »
      3 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I get that you are doing this because ans-sub can be negative but how does MOD+(ans-sub) give the right answer when ans-sub is negative. Is this a property of MOD?

      • »
        »
        »
        »
        3 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        The "taking value modulo MOD" operation basically tells you the distance between your number and the previous number that divides MOD. So in case of negative numbers (in range (-MOD; 0)) you need to find the distance between -MOD and your number. If you shift both values by adding MOD it becomes the distance between 0 and (your number + MOD) which is of course (your number + MOD).

  • »
    »
    3 years ago, # ^ |
      Vote: I like it -7 Vote: I do not like it

    By the way, answer can be simplified to n! * k / (k+1). That way you don't need to subtract modulo MOD :)

»
3 years ago, # |
Rev. 3   Vote: I like it -7 Vote: I do not like it

In problem C my approach giving the wrong answer on test 2 but I couldn't understand why it is giving this error.

My Solution
Code

I tried to explain it well. I hope you got it and you can help me.

UPD: I added accepted solution so you can check my solution to understand the problem.

  • »
    »
    3 years ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    It's because of this int x = (fact(n) / fact(mp[maxi] + mp[maxi-1])) % mod;

    fact(n) is divisible by fact(mp[maxi] + mp[maxi-1] if you don't use mod, but with mod it doesn't, for example:

    16 is divisible by 8 but if i use mod 3 (16%3 = 1 and 8%3 = 2): 1 isn't divisible by 2

    To fix this we can use module inverse, you can see it here or here

    I hope you understood and sorry for my bad english.

    upt: I don't know if the rest of your solution is correct

    • »
      »
      »
      3 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thanks for your answer I got it. I didn't consider errors about mods.

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

For problem A you can use two pointers.

Spoiler
  • »
    »
    3 years ago, # ^ |
      Vote: I like it +23 Vote: I do not like it

    Is it worth ?

  • »
    »
    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    There are 4 methods for this question. 1. O(n) — simple traversal as described in editorial. 2. O(n^2) — brute force — generating all subarrays 3. O(n) — using two pointers. 4. O(n) — using traversal + map used to store sum till particular index

    • »
      »
      »
      3 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      What about finding count of 'a' and 'b' for each range using segment/fenwick trees ? Isn't that useful ?

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

I have a different implementation for E , I am traversing on 1 match per function call by finding the people who haven't lost yet . It was giving TLE until I figured out ( yes figured out , because I didn't Google it ) how to traverse on Only set bits of a number in a given range ( range of bits ) . My submission

»
3 years ago, # |
  Vote: I like it +17 Vote: I do not like it

I think there exists a method for question D that is a little faster than the tutorial. The method used in the tutorial is to enumerate points, but I think it would be faster to enumerate edges. We divide the points into two groups, on x and on y, and discard the ones that satisfy both, since they cannot be composed. This way a pointer can be used to model the points that are between the two edges. Here the practice of my idea link

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

Your codes for D/E/F are soo looooong

»
3 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Easy D approach -

Count no of points between each adjacent vertical lines. Let this count be n. Total no of pairs will be n*(n-1)/2. Subtract those pairs which lie on same horizontal line. This can be easily done using map.

Do the same for each adjacent horizontal lines

128428875

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

Hello, Can anybody please explain what does (1<<(1<<k))-1 represent in problem E??

  • »
    »
    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It means $$$2^{2^k}-1$$$. The << (bitwise shift) operator shifts the bits in the left number by the right number of times.

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

[deleted]

  • »
    »
    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Both $$$m$$$ and $$$n$$$ can be upto $$$2e5$$$. So yes, $$$O(mn)$$$ will exceed both memory and time limits.

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

Hi I was upsolving the C problem, can someone tell me why this wont pass the second test case? Thanks in advanced! my solution

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

Can someone help me with my solution for problem C. So I came up with the same idea as the solution that a nice permutation must have at least 1 pair of i, j that i < j, a[i] is the largest number of the array and a[j] = a[i] — 1. I will call the largest number M. So using that, I will choose 2 position for M and M-1 , so there will be n*(n-1)/2 ways to place M and M-1. Suppose there are k number equal to M-1, there will be k*n*(n-1)/2 such ways. The remaining element can be placed anywhere so there will be (n-2)! ways. So in total there will be k*n*(n-1)/2*(n-2)! pair of nice permutation. But this formula gives the wrong answer for the 4th example test. I haven't known what is wrong with my solution yet so I hope u guys will help me with it. Thanks in advance!

  • »
    »
    16 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I'm struggling with the same sitaution have you figured out what's wrong with the approach @Dyln

    • »
      »
      »
      16 months ago, # ^ |
      Rev. 2   Vote: I like it 0 Vote: I do not like it

      So when we use $$$C_n^2$$$ and multiply it with $$$(n-2)!$$$ it will count some cases multiple times because the positions of the second max elements are not fixed (ie. when the array is 1 1 2). Therefore, we have to fixed $$$k+1$$$ elements first and the max element will be placed anywhere but the last position, so our formula will be $$$A_n^{n-k-1}k!k$$$, which will simplify to $$$\frac{n!}{k+1}k$$$.

      • »
        »
        »
        »
        16 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Hey its still doubtful as by doing nC2 we ensure that the second (atleast) one is behind the max one. So how can it count multiple cases

        • »
          »
          »
          »
          »
          16 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          So let take a look at the array i mentioned: 1 1 2. First we take the first 1 (1st index) and the max value, then pick two random positions for them, let assume they are 1 and 2, so our permutation will be 3 1 2. But when we consider the second 1 (2nd index) and the max value, if we take position 1 and 3 then our permutation will also be 3 1 2. This permutation is counted twice because we didnt fix the position of the 1s. Therefore, the original formula is wrong although intuitively, it seems right XD.

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

There is a typo in the last line of paragraph 5, 1569D. strret->street :D @awoo

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

In C, this line of code if (cmx == 1) ans = (ans - sub + MOD) % MOD; Why do we use mod like this: (.. + MOD % MOD)? We already calculated 'ans' and 'sub' under MOD, why do we use it again in the if statement?

  • »
    »
    3 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    you need to +MOD because ans-sub may be a negative number , you need to %MOD because if ans-sub is non-negative , the former +MOD will give a value greater than ( or equal to ) MOD so you have to modulo again