flamestorm's blog

By flamestorm, 12 months ago, In English

We hope you enjoyed the contest!

1829A - Love Story

Idea: SlavicG

Tutorial
Solution

1829B - Blank Space

Idea: mesanu

Tutorial
Solution

1829C - Mr. Perfectly Fine

Idea: SlavicG

Tutorial
Solution

1829D - Gold Rush

Idea: flamestorm

Tutorial
Solution

1829E - The Lakes

Idea: mesanu

Tutorial
Solution

1829F - Forever Winter

Idea: flamestorm

Tutorial
Solution

1829G - Hits Different

Idea: flamestorm

Tutorial
Solution

1829H - Don't Blame Me

Idea: SlavicG

Tutorial
Solution
  • Vote: I like it
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12 months ago, # |
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thanks for the speedy editorial.

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12 months ago, # |
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wow what speedy editorial!

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12 months ago, # |
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In F , according to the constraints , if m = 1 , then how x and y could be both greater than 1 such that x*(y+1) = m = 1 ?

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    12 months ago, # ^ |
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    It is guaranteed that this graph is a snowflake graph.

    So m=x+xy>2

    m>=1 doesn't means that m can be 1.

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    12 months ago, # ^ |
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    In fact, $$$m=n-1$$$

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12 months ago, # |
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Problem H can be solved in O(k * 2^k + n) using AND Convolution, as described here: https://mirror.codeforces.com/blog/entry/115438

Implementation: https://mirror.codeforces.com/contest/1829/submission/204840516

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    12 months ago, # ^ |
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    Wow, dude, this is really nice! Because of commutativity of AND, this is possible! I thought that I could solve it as well with convolutions, but didn't see the commutativity and ended up solved it using the standard n * 2^k dp.

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    12 months ago, # ^ |
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    This is very interesting! However, I have a major doubt: How did you realize that applying the inverse SoS on the amount of subsets whose AND contain mask is the same as the amount of subsets whose AND is EXACTLY mask? In the linked blog the example is given for pairs, but I fail to realize how to connect these ideas formally. Did you prove it? Can you share your insight?

    Maybe I would understand it if I understood why it works for pairs, lol.

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12 months ago, # |
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Thanks for the round!!
I liked problem G(only problem which I couldn't AC :sadge:) and the intended solution idea.

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12 months ago, # |
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Wow. Loved G. Though was not able to solve it. But didn't even notice that just rotating it will make it a standard 2-D prefix sum problem. Lol struggled the whole 1hr during the contest.

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    12 months ago, # ^ |
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    Can you please explain it?

    I am not able to get the tutorial.

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      12 months ago, # ^ |
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      Maybe it's because you don't know that technique called 2-D prefix sum. Google it and study it.

      The pattern was [1,3,6..][2,5,9..].

      Imagine this as the 2-D grid.

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    12 months ago, # ^ |
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    G was a really good problem, it reminded me of http://www.usaco.org/index.php?page=viewproblem2&cpid=416.

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    12 months ago, # ^ |
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    Why I always get tle?

    long long f[1000010];
    int main() {
    	f[1] = 1;
    	long long cnt = 2;
    	for (int i = 2; i < 1420; i++) {
    		f[cnt] = f[cnt - i + 1] + cnt * cnt;
    		cnt++;
    		for (int j = 2; j < i; j++) {
    			f[cnt] = f[cnt - i] + f[cnt - i + 1] + cnt * cnt - f[cnt - i + 1 - i + 1];
    			cnt++;
    		}
    		f[cnt] = f[cnt - i] + cnt * cnt;
    		cnt++;
    	}
    	int t;
    	scanf("%d", &t);
    	while (t--) {
    		int n;
    		scanf("%d", &n);
    		printf("%lld\n", f[n]);
    	}
    	return 0;
    }
    
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12 months ago, # |
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Didn't get H

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12 months ago, # |
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G is possible by simply using sum of squares of N natural numbers for each row and is easy implementation too.

https://mirror.codeforces.com/contest/1829/submission/204823035

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    6 months ago, # ^ |
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    can you explain your approach a bit?

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12 months ago, # |
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I think problem D amounts to checking whether the equation 2^a * n == 3^b * m has a solution where b >= a. I used two-pointers to search for a and b.

Implementation: https://mirror.codeforces.com/contest/1829/submission/205007790

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    12 months ago, # ^ |
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    Yup, I noticed that too.

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    12 months ago, # ^ |
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    can you explain more about your approach?

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      12 months ago, # ^ |
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      think like that any number a can be divided to the form a/3 , 2a/3 which further divide into a/9 ,2a/9 and 4a/9 and so on so the number m should be of latter form . so find how much is the differences of maximum power of 3 say y and divide the first number by that . and then try checking for all powers of 2 if u can get to number m before finishing the tries u have which is y

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12 months ago, # |
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in problem F what is the output of this case :

1
2 1
1 2
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12 months ago, # |
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Tonight's problems' interesting and not typical, thank the writer and codeforces for such a great contest!

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12 months ago, # |
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Have anyone solved E using BFS?

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i solved 5/8. Great contest everyone!!

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12 months ago, # |
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Alt solution to G:

From the given $$$n$$$, try to go upwards to the left block $$$l_n$$$, and to the right block $$$r_n$$$.

If we can visit both ways, there some blocks will be taken twice. So we find the common blocks between $$$l_n$$$ and $$$r_n$$$ and subtract them from the answer. If the common block exists, it's either $$$l_{r_n}$$$ or $$$r_{l_n}$$$.

There is a recursive relation, until the blocks exist. It can be either of:

$$$ans_n = n*n + ans_{l_n} + ans_{r_n} - ans_{l_{r_n}}$$$
$$$ans_n = n*n + ans_{l_n} + ans_{r_n} - ans_{r_{l_n}}$$$

Implementation: 204825154

G

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    12 months ago, # ^ |
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    Do you mean to say like this: My Submission I guess it's easier this way.Not sure!

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    12 months ago, # ^ |
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    I also coded it this way, but if you look closely, this is actually just the 2d prefix sum solution!

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      12 months ago, # ^ |
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      It is!

      This is just a different way of looking at it, without all the modifications. I feel this is more intuitive.

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    12 months ago, # ^ |
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    this is 2023 per test case right?

    originally, this solution was right on the edge of passing :( and was supposed to not pass. But setters remembered its a div4.

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      12 months ago, # ^ |
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      It's actually $$$10^6$$$ for all test cases since you can do memoization.

      Also you do not need all of the $$$2023$$$ rows.

      $$$x*(x+1)/2 = 10^6$$$
      $$$x \approx 1414$$$
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    11 months ago, # ^ |
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    My Submission my logic is to go upwards in the form of an inverted tree until the base of the inverted tree is greater than the number of cans in that level, after which all the cans above this level will fall, and if the can lies on the left or right boundary then only the boundary cans fall, but it seems to fail for large testcases can you please explain why?

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12 months ago, # |
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For problem F, if you see: Wrong answer on test 2 wrong answer 65th numbers differ

Then it is the x == y+1 edge case mentioned in editorial.

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12 months ago, # |
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Someone please help me spot the error in my solution to D I spent almost the entire contest trying to fix it:

def solve(n, m):
    if n == m:
        return(1)

    elif n % 3 != 0 or m > n:
        return(0)

    elif n // 3 >= m:
        return(solve(n // 3, m))

    else:
        return(solve(n-(n//3), m))

for tt in range(int(input())):
    n, m = map(int, input().split())
    if solve(n, m):
        print("YES")
    else:
        print("NO")



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    12 months ago, # ^ |
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    Here is a failing test case:

    Input:
    1
    27 8
    
    Correct Output:
    YES
    
    Your Output:
    NO
    

    Can you see why this happens?

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      12 months ago, # ^ |
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      Ohhhhhhhhh, thank you so much.

      Yeah I see now.

      27 // 3 is 9 -> 9 > 8, so my solution tries to split up 9 to get 8 -> outputs NO.

      And a fix be to use return (solve(n//3, m) or solve(n-n//3, m)) right?

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        12 months ago, # ^ |
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        Yeah, this should fix the problem.

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        12 months ago, # ^ |
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        did u solved it?

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          12 months ago, # ^ |
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          yup, here is final submission:

          def solve(n, m):
              if n == m:
                  return(1)
          
              if n % 3 != 0 or m > n:
                  return(0)
          
              return solve(n//3, m) or solve(n-n//3, m)
          
          for tt in range(int(input())):
              n, m = map(int, input().split())
              if solve(n, m):
                  print("YES")
              else:
                  print("NO")
          
          
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12 months ago, # |
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There is a no brainer dp solution to C.
DP
basically if we have S as a set of skills we took then xor(s)=3 ATP.
we have to find a set such that cost is minimum

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12 months ago, # |
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204848665

my solution for (E) The lakes is clearly an O(mn) solution, but was exceeding limit, can anyone explain where it can be optimised?

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    12 months ago, # ^ |
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    204858437 is logically same as my solution 204848665 , but the first solution was accepted but mine got TLE, both are written in python

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    12 months ago, # ^ |
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    Currently you are adding the same squares into the queue multiple times: when adding a new square to the queue you're only checking if you've been to the square before. Importantly, you're not checking if the square has already been added to the queue. This starts actually growing exponentially and the time complexity is definitely not $$$O(nm)$$$. It can be fixed with a small modification — changing the place where some operations are done. This ensures that no square will be added to the queue more than once and the complexity is now truly $$$O(nm)$$$.

    204888301

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      12 months ago, # ^ |
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      understood, thanks for the explanation

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        12 months ago, # ^ |
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        Actually an even simpler modification gets AC:

        Before adding the next squares into the queue, just make sure that grid[i][j] > 0. Now no exponential growth will happen and each square will get added to the queue at most 4 times.

        204889157

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    12 months ago, # ^ |
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    Use visited array

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12 months ago, # |
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I tried Solving E using Standard DFS in python Python Code. But I was continuously getting Runtime Error at TC 6. C ++ did the trick though (for the same) C++ code. Can someone tell me why I was getting Error in Python for the same Implementation ?

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    12 months ago, # ^ |
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    Python's default stack limit is 1000 which can be changed with sys.setrecursionlimit(big number), on cases where you are forced to recurse deeper then 1000 layers (imagine a spiral pattern of values separated by 0's) you will run time error. (Be careful python recursion is very slow so you may run into issues with time limit)

    Changing to c++ fixed the issue because c++ does not have this issue with low default stack limit.

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      12 months ago, # ^ |
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      Thanks a lot. That solve a lot of my problems with python. Been getting these kinda errors frequently.

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      12 months ago, # ^ |
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      for beginners, how large can the "big number" can be in-case of O(N) required space complexity ? Will 10^4 do?

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      12 months ago, # ^ |
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      Ps Update: Just tried using it. It didn't make any difference in the outcome of the submission (RE Again ;_;). But thanks again, It was Informative.

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        12 months ago, # ^ |
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        Since the grid can be of size 1000 * 1000, you would need the stack limit to be that big as well (ie 1e6+ a bit). Though you might lead to TLE since recursion is quite slow in python. I would suggest if you do get tle (and can't switch languages) to try programming the flood fill as an iterative bfs as although the time complexity is the same the constant factor should be significantly faster.

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12 months ago, # |
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204849479
Why am I getting runtime error in this code?

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    12 months ago, # ^ |
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    Python's default stack limit is 1000 which can be changed with sys.setrecursionlimit(big number), on cases where you are forced to recurse deeper then 1000 layers (imagine a spiral pattern of values separated by 0's) you will run time error.

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12 months ago, # |
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WoW Fast Editorial
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12 months ago, # |
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wow. so speedy

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12 months ago, # |
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JUST CAME BACK TO VISIT A TAYLOR SWIFT ROUND. OMG!!

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    12 months ago, # ^ |
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    come back...be here (another Taylor's song)

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      12 months ago, # ^ |
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      message in a bottle
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12 months ago, # |
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Can anybody hack this solution for TLE ?? https://mirror.codeforces.com/contest/1829/submission/204874235

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Problem F: Observe that the starting vertex is the ONLY vertex that is not a leaf AND also has no leaf neighbours. 204824064

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    12 months ago, # ^ |
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    Or you can go with that whenever there is a level greater than 2 then that is surely not our central node of snowflake.204843315

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12 months ago, # |
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for problem D, why we used this formula for master theorem: T(n) = 2T(n/3) + O(1) , but it's really T(n/3)+T((2*n)/3)+O(1) why we can Ignore that coeff (2*) ? thank you

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    12 months ago, # ^ |
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    Usually for time complexity we don't consider coefficients so we just say it is 2 * T(n/3), since the order is the same.

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      12 months ago, # ^ |
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      Thank you!

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        11 months ago, # ^ |
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        I thik this is wrong. Recurrence T(n)=T(n/3)+T(2n/3) +c has an O(n) solution for T(n). In fact, T(n) is Omega(n). This can be proved by induction easily. That is, you can prove that T(n) >= kn for some k and for every n sufficiently large.

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          4 weeks ago, # ^ |
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          I agree with juanxo_gu the solution using the master theorem given in solution list lower bound of T(n).

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12 months ago, # |
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12 months ago, # |
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Have anybody solved problem E without using bfs and DFS???

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12 months ago, # |
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$$$H$$$ can be solved in $$$(maxa_i^2 * t + ∑n)$$$ where you can just keep track of how many ways each number between $$$0-63$$$ inclusive can be constructed. Submission: 204886301 (binpower is not needed, we can just pre-calculate powers of $$$2$$$).

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12 months ago, # |
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task E is the best

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12 months ago, # |
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wow, fast editorial <3

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that was a good contest, thanks :)

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Amazing

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12 months ago, # |
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There is a much faster solution to D

Each time you either multiply by 1/3 or 2/3, so the final multiplier must be 2^a / 3^b where a <= b.

Then just check whether the ratio m/n can be expressed in that form.

Code (gets AC):

t=int(input())
import math
def ispow(a, b):
    if a==1:
        return True
    else:
        return a%b==0 and ispow(a//b, b)

for i in range(t):
    line = [int(a) for a in input().split()]
    n = line[0]
    m = line[1]
    s = m // (math.gcd(n,m))
    r = n // (math.gcd(n,m))
    if ispow(s,2) and ispow(r,3) and math.log2(s) <= math.log(r,3):
        print('YES')
    else:
        print('NO')

204904968

There is an even more cheesy solution, where once you realize the 2^a / 3^b and a<=b part, you precompute a list of all such fractions of that form within 10^7, and for each test case, just check whether m/n is equal to some fraction in that list:

t=int(input())
def equal(a,b,c,d):
    return a*d==b*c
fractions = []
for b in range(15):
    for a in range(b+1):
        fractions.append([2**a, 3**b])
for i in range(t):
    line = [int(a) for a in input().split()]
    n = line[0]
    m = line[1]
    ans = False
    for frac in fractions:
        if equal(m, n, frac[0], frac[1]):
            ans = True
    if ans:
        print('YES')
    else:
        print('NO')

(also AC)

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12 months ago, # |
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One thing I noticed on F:

Each node is in the 2nd layer of nodes if and only if its degree is $$$1$$$. That means we can perform the following:

  1. Find a node with degree $$$1$$$. (Second layer)
  2. Get its neighbor, which will have $$$y + 1$$$ edges. (First layer)
  3. Since $$$x*y=m$$$, we can divide $$$m$$$ by $$$y$$$ to get $$$x$$$.

my submission

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12 months ago, # |
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my strat for F was to take a node with degree 1, and go to the next node. the degree of this node is y + 1. then go to a node which degree is greater than one from this node, and the degree of this node is x.

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12 months ago, # |
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question E was interesting. loved solving it.

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Can someone explain in problem c why did we use >1e6 for the -1 case.

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    11 months ago, # ^ |
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    Doesn't really matter you just have to take a number which is larger than any input number which are given less than or equal to 10^5 so any number greater than 10^5 will work.

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12 months ago, # |
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One interesting Observation in F is that all leaf nodes have degree 1 so count all degree 1 lets say total and then other are degree >1 which means they would be x+1 as one node with degree >1 would be central vertex so answer would be x and total/x.

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For E, my dfs solution gives TLE which has the same logic as the solution given. When I changed the visited array and input array to global variables, it got accepted. Why is that so?

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12 months ago, # |
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Can someone tell me why my H problem is giving TLE with $$$dp[n][64]$$$ 204872960 but accepted with space optimization 204874966?

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12 months ago, # |
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Hi, I wonder is this round rated for all who has a rating < 1400? If so, why my ratings didn't change? I'm new to codeforces so please forgive me for asking these questions. Thanks!

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12 months ago, # |
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I submitted the question (C) 2 min before the contest end , last time it was showing it green now its showing it red ?? WHY?

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12 months ago, # |
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Some of the hacks are very... strange looking.

First

Second

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12 months ago, # |
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I solved G in a similar way. Let $$$dp_i$$$ be the answer for $$$i$$$ and $$$x_i$$$ be the row for $$$i$$$. Now we know that $$$dp_1 = 1$$$ , and $$$dp_0 = 0$$$, and for each $$$1 \leq i \leq N$$$ we have two cases:

$$$i - x_i$$$ only lies on the previous row $$$x_i - 1$$$ , then $$$dp_i = i^2 + dp_{i-x_i}$$$

$$$i - x_i + 1$$$ only lies on the previous row $$$x_i - 1$$$ , then $$$dp_i = i^2 + dp_{i-x_i+1}$$$ both $$$i - x_i$$$ and $$$i - x_i + 1$$$ lie on the previous row $$$x_i - 1$$$. Here we can't simply do $$$dp_i = i^2 + dp_{i-x_i} + dp_{i-x_i+1}$$$ because $$$dp_{i-x_i}$$$ and $$$dp_{i-x_i+1}$$$ have common numbers that were added to both of them. We can see that they have different answers except the one they took from $$$dp_{i-2(x+1)}$$$ so it becomes $$$dp_i = i^2 + dp_{i-x_i} + dp_{i-x_i+1} - dp_{i-2(x+1)}$$$

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12 months ago, # |
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I approached F the following: let be the number of outermost nodes l.

One can easily find this "l" from the input. It turns out that l = x*y, also, the total number of nodes is n = 1 + x + x*y. So we have due independent equations which yields to x and y values

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12 months ago, # |
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alt for D, we can just check wether we can get from m to n if we can do the operations below

m := 3*m

m := 3*m/2 (if m mod 2 = 0)

So we need to check if there exist integer pair x and y (x >= y and m mod 2^y = 0) s.t. m * 3^x / 2^y = n

we can do it in O(log2(n))

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12 months ago, # |
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How could H be solved with larger values? for example (a[i] <= 1e6 or 1e9) and of course a corresponding K.

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12 months ago, # |
Rev. 3   Vote: I like it +11 Vote: I do not like it

I am Expert now thanks to this round, SpeedForces forever xD

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12 months ago, # |
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What about this?

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12 months ago, # |
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I dont know why my submission 204795848 in the contest using C++ 17 giving the TLE but it's Accepted if I submit it with C++20 204978167

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12 months ago, # |
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I set the recursion limit and I still get Runtime Error: 204981121 Can anyone help?

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12 months ago, # |
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Thanks for a great round! I solved 5/8 problems, which is the highest ever (usually no more than 3). I solved the first 4 problems in 30 minutes, and then an hour later I solved F, which seemed to me easier than E, the idea of which I did not understand.

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12 months ago, # |
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/* this question is basically of precomputation + dp in this question we will first precompute the matrix then make a loop from 1 to 1e6 and calculate the answer for all the elements and then for each query we will give answer in O(1) / / dp of current element will be the x^2 (dp[k]=(vec[i][j]*vec[i][j])) + dp of the two elements above it if exists i.e. { if(i-1>=0 && j-1>=0) dp[k]+=dp[vec[i-1][j-1]]; if(i-1>=0 && j<vec[i-1].size()) dp[k]+=dp[vec[i-1][j]]; }

and then subtracting the element which is at the top of those two elements
because that will be the intersection and it will be added twice so we have to minus it
if exist

i.e. if(i-2>=0 && j-1>=0 && j-1<vec[i-2].size()) dp[k]-=dp[vec[i-2][j-1]];

-> inclusion / exclusion principle 

as in given example for calculating the answer for 9
dp[9] = 9*9 + dp[6] + dp[5] - dp[3]

*/ // do a dry run you will get by yourself

/* <--------------------------------- THANK YOU -----------------------------> */

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

My Python implementation of the editorial solution passes only first 5 tests, then gets Runtime Error on test 6. https://mirror.codeforces.com/contest/1829/submission/205011680

Can someone help find the problem?

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

    I got the same problem. After some trials (to me) it seems impossible to pass the problem in python with recursion. I used bfs to solve this. (dfs is also not ok for it will cause MLE)

    Here's my submission: 204988219

    Tip: check each cell if it is visited or is 0 before bfs to optimize the time

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12 months ago, # |
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C hack has been evil U_U

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12 months ago, # |
Rev. 4   Vote: I like it 0 Vote: I do not like it

Just an implementation detail about the tutorial solution for problem 1829H - Don't Blame Me: if you are using GNU C++20 compiler, then you do not have to use the built-in function __builtin_popcount. You can use the standard library function std::popcount instead. Also, as the next-state at any index $$$i$$$ depends only on the previous state at index $$$i-1$$$, it is sufficient to compute the answer using two vectors only instead of using a matrix of $$$n+1$$$ vectors.

Accepted Solution
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12 months ago, # |
  Vote: I like it +4 Vote: I do not like it

Nice problems ,i have got +156 points (:

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12 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

I had different idea for G. Keep start and end positions of cups in a line then go up and use math formula for the sum of n squares: 205064632 O(1) space and O(n) run

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12 months ago, # |
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Hello! Does anyone know how to solve H using combinatorics?

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12 months ago, # |
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Can someone post DP solution of G?

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12 months ago, # |
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This is my solution, I just calculate how many numbers appear once. Then i subtract that number from number of nodes and i subtract -1 to get rid of the first node and from there i get x.

void solve() { int n,m; cin >> n >> m; map<int,int> mp;

for(int i = 0; i < m; i++)
{
    int prvi,drugi;cin >> prvi >>drugi;
    mp[prvi]++;
    mp[drugi]++;
}
int ans = 0;
for(auto x : mp)
    if(x.second == 1)
        ans++;
cout << n-ans-1 << ' ' << ans/(n-ans-1) << endl;

}

int main() { ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

int tt;cin >> tt;
while(tt--)solve();

}

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11 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Can anyone please explain the tutorial for F (Forever Winter)? I am unable to understand the logic behind it. I only coded a TLE solution which was accepted at the time of the contest, then it changed to TLE.

EDIT:- Nevermind. I understood with the help of other's submissions.

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11 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

UPD : solved

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11 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Nice tutoriel

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11 months ago, # |
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include

include

include

using namespace std; typedef long long ll;

bool solve1(ll n, ll x, vector& dp) { if (n == x) { return true; } if (x > n || n % 3 != 0) { return false; } if (dp[n] != -1) { return dp[n]; } bool ans = solve1(n / 3, x, dp) || solve1(2 * n / 3, x, dp); dp[n] = ans ? 1 : 0; return dp[n]; }

void solve() { ll n, m; cin >> n >> m; vector dp(n + 1, -1); if (solve1(n, m, dp)) { cout << "YES" << endl; } else { cout << "NO" << endl; } }

int main() { cin.tie(0); cin.sync_with_stdio(0); cout.tie(0); cout.sync_with_stdio(0);

int t;
cin >> t;
while (t--) {
    solve();
}

return 0;

} Why this code is giving Time complexity in Problem D, ;Gold Rush?

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11 months ago, # |
  Vote: I like it +5 Vote: I do not like it

Editor seems to be fan of Taylor Swift

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11 months ago, # |
  Vote: I like it 0 Vote: I do not like it

In problem G, I was using bfs for solution but it didnt seem to give any answer for the last test of test case 1. Can anyone tell why.

include<bits/stdc++.h>

include <ext/pb_ds/assoc_container.hpp>

include <ext/pb_ds/tree_policy.hpp>

using namespace std; using namespace __gnu_pbds;

define ll long long

define pb push_back

define msort(v) sort(v.begin(),v.end())

define loop(ii,n) for(long long ii = 0; ii < n; ++ ii)

define pii pair<int,int>

define ff first

define ss second

// #define ordered_set tree<int, null_type,less, rb_tree_tag,tree_order_statistics_node_update>

define ordered_set tree<pair<int,int>, null_type,less<pair<int,int>>, rb_tree_tag,tree_order_statistics_node_update>

define o_key order_of_key

define o_set ordered_set

bool cmp(pair<int,int>& i1, pair<int,int>& i2) { return (i1.first < i2.first); } bool prime[1000005]; void sieve() { memset(prime, true, sizeof(prime));

for (int p = 2; p * p < 1000005; p++) {

    if (prime[p] == true) {

        for (int i = p * p; i < 1000005; i += p)
            prime[i] = false;
    }
}

} ll len(ll a) { int ans=0; while(a>0){a/=10; ans++;} return ans; } map<int,int> mprime; void pf(int n) {

while (n % 2 == 0)
{
    mprime[2]++;
    n = n/2;
}
for (int i = 3; i <= sqrt(n); i = i + 2)
{

    while (n % i == 0)
    {
        mprime[i]++;
        n = n/i;
    }
}
if (n > 2) mprime[n]++;

} long long binpow(long long a, long long b) { long long res = 1; while (b > 0) { if (b & 1) res = res * a; a = a * a; b >>= 1; } return res; } ll cnt(ll i) { ll ans=0; while(i>0) { ans+=(i%10); i/=10; } return ans; } /////////////////////////////////////////////////////////////////////////// //////////////////// DO NOT TOUCH BEFORE THIS LINE //////////////////////// ///////////////////////////////////////////////////////////////////////////

int main() { ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

ll t;
cin>>t;
while(t--)
{
    // cout<<"ds";
    ll n;
    cin>>n;

    vector<vector<ll>> adj(1000005);
    ll k=1;
    for(int i=1;i<=n;i+=(k-1))
    {
        for(int j=0;j<k;j++)
        {
            adj[j+i+k].pb(i+j);
            adj[j+i+k+1].pb(i+j);
        }
        k++;
    }
    vector<ll> vis(n+5,0);
    ll ans=0;
    queue<ll> q;
    q.push(n);
    // ans+=q.front();
    // cout<<"Ds";
    while(!q.empty())
    {
        // cout<<n<<endl;
        ll tmp=q.front();
         q.pop();
         ans+=tmp*tmp;
         vis[tmp]=1;
         for(auto i:adj[tmp])
         {
             if(vis[i]==0)
             {
                 vis[i]=1;
                 q.push(i);
             }
         }
    }
    cout<<ans<<endl;
}

return 0;

}

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11 months ago, # |
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Alternate approach for G: https://mirror.codeforces.com/contest/1829/submission/207013133

Basically whenever we move up the tower, we make the left top and right top fall down. But the sum of a single element is not simply the sum of its parents and itself, this solution deals with that by choosing the right upper part only for the right parent, hence minimizing any overlap.

Examples: If we were to calculate for n=13, it would go the following way-


1 / 2 3 / / 4 5 6 \/ / 7 8 9 10 \/ 11 12 13 14 15 --
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11 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Problem G is just wow, it couldn't be dissected more beautifully, could someone clear out what i and j mean in the solution mentioned in the editorial. i is the row number and j is the diagonal number?

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11 months ago, # |
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Could someone explain what did I did wrong for the fourth question. When I submitted by recursive code it gets accepted, but when I memoized the code, I am getting TLE. Thanks for explanation in advance. Below is the code, RECURSIVE

bool solve(int n, int m) {
	if(n == m) return true;
	if(m > n || n % 3 != 0) return false;
	
	int ind = n / 3;
	bool left = solve(ind, m);
	bool right = solve(n - ind, m);
	return left || right;
}
// Invoking function
bool res = solve(n, m);

MEMOIZED CODE

bool solve(int n, int m, vector<int> &dp) {
	if(n == m) return true;
	if(m > n || n % 3 != 0) return false;
	if(dp[n] != -1) return dp[n];
	
	int ind = n / 3;
	bool left = solve(ind, m, dp);
	bool right = solve(n - ind, m, dp);
	return dp[n] = (left || right);
}

//Invoking function
vector<int> dp(n + 1, -1);
bool res = solve(n, m, dp);
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10 months ago, # |
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can anyone tell that in question G how to decide what should be size of 2D matrix....

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10 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

C. Mr. Perfectly Fine

//Solution without using map....

#include<bits/stdc++.h>

using namespace std;

int main(){

int t;
cin >> t;

while(t-- > 0){
    int n;
    cin >> n;
    int min0 = INT_MAX;
    int min1 = INT_MAX;
    int min2 = INT_MAX;

    for(int i=0;i<n;i++){
        int m;
        cin >> m;
        string s;
        cin >> s;

        if(s[0] == '1' && s[1] == '1' && m < min2){
            min2 = m;
        }
        if(s[0] == '1' && m < min0){
            min0 = m;
        }
        if(s[1] == '1' && m < min1){
            min1 = m;
        }
    }
    if(min0 == INT_MAX || min1 == INT_MAX){
        cout << -1 << endl;
    }else{
        cout << min(min2,(min0 + min1)) << endl;
    }
}

}

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9 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

I did the exact same solution with python and somehow exceeded the time limit on test 8

here is my solution :

def is_possible(n, m):
    if n == m:
        return True
    elif n % 3 != 0:
        return False
    else:
        return is_possible(n/3, m) or is_possible(2*n/3, m)

t = int(input())
for i in range(t):
    n, m = map(int, input().split())
    if is_possible(n, m):
        print("yes")
    else:
        print("No")
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9 months ago, # |
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still unclear why for problem D, why we used T(n) = 2T(n/3) + O(1) , but it's really T(n/3)+T((2*n)/3)+O(1). you could say the 2/3 will dominate so b should be 1.5 but then time complexity is n^(log1.5(1)) which is O(n).

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7 months ago, # |
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I solved problem D using the same recursive method as the problem solution, but I saw dp in the problem label. I want to know how to solve it with dp? Can you help me? Thank you very much

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6 months ago, # |
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Can anyone explain how the time complexity of solution D(Gold Rush).

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3 months ago, # |
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I used topological sorting in F and passed :)

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3 months ago, # |
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wow!强大!

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4 weeks ago, # |
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I used dfs to solve the G, just search the (i — 1, j) and (i — 1, j — 1)!

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6 days ago, # |
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In problem G. How do you come up with the number of rows and columns as 1500. Isn't is supposed to be 2024?