mesanu's blog

By mesanu, history, 2 months ago, In English

Thank you for participating!

1926A - Vlad and the Best of Five

Idea: flamestorm

Tutorial
Solution

1926B - Vlad and Shapes

Idea: mesanu

Tutorial
Solution

1926C - Vlad and a Sum of Sum of Digits

Idea: flamestorm

Tutorial
Solution

1926D - Vlad and Division

Idea: mesanu

Tutorial
Solution

1926E - Vlad and an Odd Ordering

Idea: flamestorm

Tutorial
Solution

1926F - Vlad and Avoiding X

Idea: flamestorm

Tutorial
Solution

1926G - Vlad and Trouble at MIT

Idea: mesanu

Tutorial

Solution coded by Dominater069, thanks!

Solution
  • Vote: I like it
  • +69
  • Vote: I do not like it

»
2 months ago, # |
  Vote: I like it +74 Vote: I do not like it

For Problem-F, I have a bitmask dp solution: 247350477

  • »
    »
    2 months ago, # ^ |
      Vote: I like it +15 Vote: I do not like it

    I am interested to know how you formed intitution for bitmask dp when it looks like a graph based problem at first.

    • »
      »
      »
      2 months ago, # ^ |
      Rev. 3   Vote: I like it +20 Vote: I do not like it

      Firstly, I thought if we precalculate answers for all matrixes at the beginning, then we can solve queries in $$$O(n^2)$$$ which equals reading matrix.

      But how can we precalculate answers for all matrixes. Like in the editorial, we can separate matrix into 2 sets; which one of them has 25 elements, other's 24. How can we calculate answers for all matrix designs in one set.

      And then I reduced problem to a bitmask DP problem. And I found a classicial DP solution:

      Try to show possible matrixes as a masks, which black square is 1, else 0. Then $$$f(mask)$$$ will show answer for $$$mask$$$.

      We have 2 easy cases:

      1. It is a good matrix (every black square doesn't have 4 diagonal black square) and $$$f(mask) = 0$$$: We can check it in $$$O(bit \ number \ of \ mask)$$$.

      2. It isn't a good matrix, and suppose we transform $$$k.$$$ black square to white: $$$f(mask) = \min_{\forall k \in mask} f(mask \otimes 2^k) + 1$$$ And our case will has a same complexity as 1. case.

      Then we will have $$$2^{25}$$$ mask for first set, and $$$2^{24}$$$ for second set. And final complexity will be $$$O((2^{25} + 2^{24}) \times 25 + (T \times 49))$$$ or $$$O(2^{\frac{N^2}{2}} \times N/2 + T)$$$. Nearly $$$2 \times 10^9$$$ which is in TL. But it is too close to TL so it can get TLE if your constant is big.

      Note: You can fast your solution with magical bitwise operations.

      Also sorry for my poor English.

      • »
        »
        »
        »
        2 months ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        Wow, amazing approach, thanks for sharing. It helped a lot.

  • »
    »
    2 months ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    great solution , can you share any resources from where I can learn it?

»
2 months ago, # |
  Vote: I like it +26 Vote: I do not like it

I tried hard to solve F using bipartite matching but I failed. Does F can't be solved by bipartite matching? I think it can be solved by find minimum vertex cover with good network modeling. Plz help me.

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

    I think F is very difficult to be solved by flow (network modeling).

    I tried to find a modeling of minimum cut, but I found that it's difficult to produce by flow that for an illegal black cell, we need only to change one cell's color to make it legal. In other words, for a black cell that is not legal, we need to manipulate at least one other cells to make it legal. The "at least one" operation is difficult for network flows.

    But perhaps there is some subtle transformation in this problem that eliminates the "at least" condition. I'm looking forward the subtle way to solve it expectantly.

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

      hey how and where do you practice < I saw you're doing really good on contests.

      Any tips !!!

      My current level is solved 400+ on leetcode.

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

        Practice as more as possible. CodeForces is not the main OJ on which I solve problem.

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

          Which platform you use for practice and whats the good place to practice ?

          • »
            »
            »
            »
            »
            »
            2 weeks ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            The same as many Chinese CPer, I use Luogu mainly. You can wait for the international version of the site to use...

»
2 months ago, # |
  Vote: I like it -13 Vote: I do not like it

problem f is so bad i cant solve it

»
2 months ago, # |
  Vote: I like it +20 Vote: I do not like it

There's a much simpler solution for G.

  1. Root the tree.
  2. Do DFS
  3. if your current node $$$u$$$ is P, install a wall between $$$u$$$ and every child node $$$v$$$ that has S. The same goes for S.
  4. if your current node $$$u$$$ is C:
    • let $$$cnt_s$$$ be the number of childs have S.
    • let $$$cnt_p$$$ be the number of childs have P.
    • if $$$cnt_s < cnt_p$$$, add $$$cnt_s$$$ to the answer and change the letter on $$$u$$$ to P (Changing the letter to P means that the parent of $$$u$$$ should never be S since we have some unwalled edges contain P).
    • if $$$cnt_s > cnt_p$$$, the opposite of the previous step.
    • if $$$cnt_s = cnt_p$$$, It doesn't matter whether we should change $$$u$$$ to S or P (We should change it to be the same as the parent) so we left it to be C as it's.
  • »
    »
    2 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Well done, I attempted the same DFS traversal, but never thought of changing C to P. What I did instead was to keep a count of P and S during traversal, and set either P or S to 0 once the wall is put. But haven't quite figured out how to make it AC. Although I have the same $$$cnt_s$$$ $$$cnt_p$$$ comparison :D

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

      I did the same and got WA2, but I got lucky and found the bug 3 minutes before the contest ended :D

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

    Could you explain why: when $$$cnt_s$$$ = $$$cnt_p$$$,We should change it to be the same as the parent.

    I didn't do this, so I got WA2.But I don't understand why we have to do this.

    thank you

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

      you can try this hack 1 1 1 2 2 3 3 CCCPSPS anwser may be 2 or 3

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

      Because if the parent is S and we changed the current node to P, this would take an extra operation.

      For example :

      6
      1 2 2 2 2
      SCPPSS
      

      We should change node 2 to S (total cost = 2), but if we change it to P, the total cost will be 3.

  • »
    »
    2 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Amazing, I also tried same approach but couldn't figure out last condition. This last condition cnts==cntp is what I was missing out and getting WA on test case 2.

    Thanks it helped me getting AC. My AC Solution 247695620

  • »
    »
    7 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Can you explain if node == c why you gave hasS = 1 and hasP = 1 instead of giving hasS = 0 and hasP = 0?

    • »
      »
      »
      7 weeks ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      I didn't do that. They are initially 0, and I change them based on which gives the minimum answer.

          if (s[node] == 'C')
          {
              cnt += min(cs, cp);
              if (cs > cp)
                  hasS[node] = 1;
              else if (cs < cp)
                  hasP[node] = 1;
          }
      
  • »
    »
    4 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Can you attach the submission link ? I tried the same thing although i am getting WA on Test 3

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

I recorded myself live while solving A,B,C,D,E and thinking G. Hope it helps someone: https://www.youtube.com/watch?v=Zr2JbyTwq0A

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

    Bro why you did it in live stream?

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

      The stream was private during the contest. After the contest, I made it public. idk why there are so many downvotes. Many YouTube creators upload codeforces screencasts.

      • »
        »
        »
        »
        2 months ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        Sorry, i thought it was public during contest sorry for misunderstanding

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

    By the way why so many downvotes its just private live and made public once the contest is over and do consider checking it out its actually helpful.

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

Why this solution get TLE? https://codeforces.com/contest/1926/submission/247402006 I can't think of a single reason, every operation in loops are constant

I solved it other way: https://codeforces.com/contest/1926/submission/247402620 But still curious why first one is TLE

»
2 months ago, # |
  Vote: I like it +3 Vote: I do not like it

I wrote this solution which just check each line if it has substring of "010" since k>1 that means a square is at least 2*2

if it has a "010" then it is a triangle else is square

its an easier approach

Solution
»
2 months ago, # |
Rev. 10   Vote: I like it +12 Vote: I do not like it

I have a bitmask dp solution for problem F.

first assume that B = 1 and W = 0, and change the grid accordingly.

Let's say the bit on cell (i, j) is bad iff grid[i][j] = 1 and grid[i - 1][j - 1] = 1 and grid[i - 1][j + 1] = 1.

With that, let's define dp[i][j][k] to be the minimum number of flips considering the first i lines, the current mask on the i'th line is j, and k is a bitmask saying which bits from j are bad. Clearly, k is a submask of j, since the x'th bit from j can only be bad if grid[i][x] = 1, this is important because now we can iterate on j and k in $$$O(3^N)$$$. Now we need to transition to i + 1. How? simple, just iterate on all possible masks, and check if they are valid.

A mask is valid iff for all x, if x is bad than mask CANNOT have both (x — 1) and (x + 1) BOTH on. otherwise we would have that grid[i - 1][x - 1] = 1 and grid[i - 1][x + 1] = 1 and grid[i + 1][x - 1] = 1 and grid[i + 1][x + 1] = 1, which we cannot have.

With some clever preprocessing of the masks, this dp can be achieved with a complexity of $$$O(N * 2^N * 3^N)$$$, with N = 7 and 200 testcases, this is fast enough

here's my solution

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

I solve problem C without precomputing in O(t*log10(n)).LOL 247319547

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

    Do you have the math proof of that?? I couldn't understand

  • »
    »
    2 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    I was literally trying to formulate a thing like this but failed on it badly and messed up with my rating. going back to grey again :(

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

      same. I literally wasted 20 minutes trying to find a formula, the 0.5 second bugged me.

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

        me too,I didn't try to preprocess it

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

    Can you explain the mathematical proof behind this?

»
2 months ago, # |
Rev. 2   Vote: I like it +19 Vote: I do not like it

I have a different approach with dp for problem F. First consider the following dp: dp[i][msk1][msk2].
i = current line
msk1 = what is the state of the last line
msk2 = what are the positions that, if we put two black cells in the diagonal in this line will make a X.
The transition for this dp will be: try every mask, see if the mask is possible and, if possible, calculate the cost of the mask.
The complexity of this would be O(7*2^7*2^7*2^7*7), and this is too slow for 200 test cases. But we can formulate the masks with a different way, take a number in base 3.
if the digit at pos j =
0: the pos j at last line was a W.
1: the pos j at last line was a B, and I can put two black cells in the diagonal.
2: the pos j at last line was a B, and I can't put two black cells in the diagonal.
This dp will be O(7*3^7*2^7*7), this will be sufficient for 200 test cases.
https://codeforces.com/contest/1926/submission/247403810

»
2 months ago, # |
  Vote: I like it +8 Vote: I do not like it

For problem D, rather than using XOR, I simply added the two numbers and checked to see if they were equal to 2^31 - 1. Here was my submission: 247301401. I then sorted the array and used two pointers. Is the editorial solution superior to mine? If so, why?

  • »
    »
    2 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    It's not superior. You also have a hard time accomplishing it in Python (dict can be TLE hacked). Sorting and 2 pointers is the safest option.

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

    Your solution probably will get better runtime than the editorial if implemented in C++, since the constant factor of map is bigger than traditional sorting.

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

The link to G in the tutorial is in russian for some reason.

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

have problem with reading problem statement E. wouldn't all cards with number from 1 to n appears? for example 8, it's clearly not odd, it can't be 2 * (any odd number) number, it can't be 3 * (any odd number), it can't be 4 * (any odd number). 8 never appears, thus constraint on k, n is not valid

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

    8 * 1

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

    The process does not stop at the step 4 -- in the step 8, you will have $$$8 \cdot 1 = 8$$$, because $$$1$$$ is an odd number.

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

      so that was And so on, until all cards are laid down.. part now I can see thank you

»
2 months ago, # |
  Vote: I like it +27 Vote: I do not like it

So is there a polynomial solution for F? I tried modeling with network flow but failed, and I think such approach probably won't work.

  • »
    »
    2 months ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    I tried modeling with network flow but failed too lol

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

    How can greedy approach ( 247465029 ) result in more deletions that optimal?

    upd: think I understood why: it's a knapsack, in case anyone wondering

  • »
    »
    2 months ago, # ^ |
    Rev. 4   Vote: I like it +8 Vote: I do not like it

    Okay, I've got it. Since we know from editorial that at most 4 squares are needed in each parity, we can replace knapsack with 4-sum (1-2-3-sums in fact) with total complexity $$$O(n^3)$$$. Slightly trollish but polynomial

    • »
      »
      »
      2 months ago, # ^ |
      Rev. 2   Vote: I like it +17 Vote: I do not like it

      Sorry, but this solution is nothing different from the one in the editorial. Your solution requires the fact that "at most 4 squares needed in each parity", however, this won't work for $$$n\ge 8$$$. So I had to say that this is not a true polynomial solution. It's indeed $$$\mathcal{O}(n^{\lfloor \frac{n}{2}\rfloor})$$$.

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

        Indeed, should have checked it before. Thanks for pointing out.

        While researching I found that low frequency set can be approximated, maybe that's the intended polynomial solution.

»
2 months ago, # |
Rev. 4   Vote: I like it 0 Vote: I do not like it
Can anyone tell me why this is giving tle for E [submission](https://codeforces.com/contest/1926/submission/247361946)
i did the same as in the editorial but i used binary search to fing no of elements we can take at each step
am i missing something 
i tried to add link for my submission but i'm unabel to do that
code
»
2 months ago, # |
  Vote: I like it 0 Vote: I do not like it

For F,my submission 247420824, why i add the dfs branch (the //add part),the result become worse. For example,

1
BBBWBBB
BBBBBBB
BBBBBBB
BBBBBBB
BBBBBBB
BWBWBBB
BBWBBWB

If I remove the comment symbols in the code and dfs by five cases, I get a result of 7.While following the code inside my link to dfs with three cases gives a result of 6.

Besides , What is the problem with the code inside my link and why can't AC.

Sorry for my poor English, can anyone help me ?

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

    I found the problem, I forgot to add "return" in this place during dfs

    if (check(x, y) == 0)
            "return" dfs2(u + 1, cnt);
    

    This can lead to otherwise legal points that I still go to enumerate to modify its diagonal, resulting in a large answer. And the more cases of dfs, the more likely it is to lead to a large answer, which is what led to my question above.

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

How can I do hash collision on this solution?

from collections import defaultdict

t = lol()[0]
int_max = 2**31-1
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split())
    c = defaultdict(int)
    count = 0
    for e in a:
        if c[int_max ^ e]:
            count += 1
            c[int_max ^ e] -= 1
        else:
            c[e] += 1
    print(count + sum(c.values()))
  • »
    »
    2 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    use custom hash

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

      I want to hack this not prevent it. Like I want understand how hack collision can be performed

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

        You can checkout usaco guide,they have a section about generating hack cases for hack collisions.

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

Can somebody share the flow networks approach and solution for Problem G. I want to know how maxflow/mincut can find the solution. Thanks in Advance.

»
2 months ago, # |
  Vote: I like it +10 Vote: I do not like it

I don't get why dp is needed in G. Isn't it simple dfs?

#include <bits/stdc++.h>
using namespace std;
const int int_max = 1e5 + 10;
char s[int_max];
vector<int> children[int_max];
int n;
 
int dfs(int i){
    int sm = 0;
    for (int j = 0; j < children[i].size(); j++){
        sm += dfs(children[i][j]);
    }
    int count_p = 0;
    int count_s = 0;
    for (int j = 0; j < children[i].size(); j++){
        if (s[children[i][j]] == 'P'){
            count_p++;
        }
        else if (s[children[i][j]] == 'S'){
            count_s++;
        }
    }
    if (s[i] == 'P'){
        return sm + count_s;
    }
    else if (s[i] == 'S'){
        return sm + count_p;
    }
    else{
        if (count_p > count_s){
            s[i] = 'P';
            return sm + count_s;
        }
        else if (count_p < count_s){
            s[i] = 'S';
            return sm + count_p;
        }
        else{
            return sm + count_p;
        }
    }
}
 
void solve(){
    cin >> n;
    for (int i = 0; i <= n; i++){
        children[i].clear();
    }
    for (int i = 2; i <= n; i++){
        int e;
        cin >> e;
        children[e].push_back(i);
    }
    for (int i = 1; i <= n; i++){
        cin >> s[i];
    }
    cout << dfs(1) << '\n';
}
 
 
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin >> t;
    while(t--){
        solve();
    }
    return 0;
}
  • »
    »
    2 months ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    I was thinking if we can solve G using min-cut.

    Basically my idea was to combine all the 'P' to a node 0 in a new_graph and all the 'S' to a node n+1, which are basically acting as the source and sink respectively, and keeping the configuration of the 'C' vertices constant, which are acting as intermediate nodes. There can be multiple edges.

    Then we calculate the max-flow, where each edge has a capacity of 1 which will be our final answer. But using Ford-Fulkerson gives $$$O(nm^2)$$$, thus giving TLE.

    Can we use the fact that the capacity of each edge is 1 thus somehow reducing it to linear time ?

»
2 months ago, # |
  Vote: I like it +18 Vote: I do not like it

I would like to mention my solution for F.

Observation 1: Grid can be divided into 2 independent sets of cells based on whether (i+j) is odd or even, with 25 and 24 cells in each.

Considering only black cells, we can brute force all possible flips in O(2^25 * 25). But this is slow.

Observation 2: Flipping cells from the grid boundary is not necessarily needed, we will still find an optimal flipping.

This reduces our cell count to 13 and 12. And the brute force is fast enough now.

Code

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

My solution in C was $$$O(N + t)$$$ 247266041. If $$$n \le 10^6$$$ my solution will still work

But it contains precomputation too

»
2 months ago, # |
  Vote: I like it +3 Vote: I do not like it

A very simple solution for B. find the first instance of 1 in the grid. then check if the next cell is 1 and if the bottom cell is 1. if so then its square otherwise triangle.

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

for problem D i got TLE. I believe its o(n) can someone correct

from collections import Counter

for i in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    c = Counter(a)
    count = 0
    for e in a:
        k = 2147483647-e
        if c[k] and c[e]:
            count+=1
            c[e]-=1
            c[k]-=1
    print(n-(count))
  • »
    »
    2 months ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    Counter behaves similarly to unordered_map in C++. It can be TLE hacked where c[k] or c[e] can be $$$O(n)$$$.

    What you can do is a random.shuffle(a) to make the hackers work harder :D

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

      noooo I didnt get hacked. I got TLE in contest itself.

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

        They made the test case that will TLE if you use a dict or a similar hash based data type. They created a "hack" before the hacking phase.

        Try adding a import random and random.shuffle(a) and you will see you will get AC (the test case that TLEs python will no longer work).

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

    my c++ version also got TLE

    #include <iostream>
    #include <unordered_map>
    #include <vector>
     
    using namespace std;
     
    int main() {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
     
        int t;
        cin >> t;
     
        while (t--) {
            int n;
            cin >> n;
     
            unordered_map<int, int> counter;
            vector<int> a(n);
     
            for (int i = 0; i < n; ++i) {
                cin >> a[i];
                counter[a[i]]++;
            }
     
            int count = 0;
            for (int i = 0; i < n; ++i) {
                int k = 2147483647 - a[i];
                if (counter[k] > 0 && counter[a[i]] > 0) {
                    count++;
                    counter[a[i]]--;
                    counter[k]--;
                }
            }
     
            cout << n - count << "\n";
        }
     
        return 0;
    }
    
  • »
    »
    2 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Did you try using Fast IO? Python's Hash table uses Sip Hash. Not exactly similar to CPP's unordered_map but very strong. IMO hash table is not the cause for TLE.

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

As a nooooooooooob,even I can't solve D and E lol Next I will record the results of each of my races. I believe I can be LGM before 2026!

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

For problem B, I have a solution in which I get the numbers of 1s in the first line the number 1 starts appearing.

If the numbers of 1s in the next line that has 1 is different (for example, line 1 has 3 ones, line 2 has 1 one) then it is a triangle, otherwise it is a square. However this solution doesn't work on test case 2 and I've not been able to find any edgecase in my mind that defies this logic: 247338503

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

    Your description sounds correct, your code doesn't quite do that.

    1
    3
    110
    110
    000
    
    • »
      »
      »
      2 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thank you, my code didn't check for the case in which there're lines full of 0s below the shape, and it compared 0 number of 1s to the number of expected 1s to make a square (which was already completed) and flagged it as a triangle because these numbers are different.

      Thank you for finding the test case!

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

Amazing work setters!! My Java Binary Trie Solution 247338014 with complexity O(30*N) TLEd on System Test which ideally should not. In contest it showed a runtime of 1091 ms , since it had a buffer of almost 1 sec , so i thought it was feasible. And again no surprise exact same code in C++ passes. 247463968 UPD : Got my AC back :)

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

Why didn't my rating change ( my rating was 381 before the contest ) even though I solved the first three problems ?

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

how to solve F in polytime?

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

Why my solution gets TLE for Problem C? 247461107

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

This was my first Codeforces contest. Can anyone suggest study materials for dsa.

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

Can anyone explain this line of the editorial solution of Problem G:Vlad and Trouble at MIT?

dp[u][0] = min({tot, dp[u][1] + 1, dp[u][2] + 1});

I can not understand it. Please give some explanation.

»
2 months ago, # |
  Vote: I like it +8 Vote: I do not like it

I think this Div4 is harder than previous Div4, because problem F and G was very hard for pupil

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

Could you please explain the approach of D ?

»
2 months ago, # |
Rev. 4   Vote: I like it -8 Vote: I do not like it

Problem C: Why does it fail on Test #8 ? : Submission

`

include <bits/stdc++.h>

using namespace std;

int sumDigits(int n) { int sum = 0; while(n > 0) { sum += n%10; n /= 10; }

return sum;

}

void solve(vector v, int n, vector &copy) { map <int, int> mp; int ans = 0; int p = 0; for(int i = 1; i <= n; i++) { if(i == v[p]) { ans += sumDigits(i); mp[i] = ans; p++; continue; } else ans += sumDigits(i); }

for(int i = 0; i < copy.size(); i++) {
    cout << mp[copy[i]] << endl;
}

}

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

int t; cin >> t;
int maxN = 0;
vector<int> v;
while(t--) {
    int n;
    cin >> n;
    maxN = max(maxN, n);
    v.push_back(n);
}

vector<int> copy = v;
sort(v.begin(), v.end());
solve(v, maxN, copy);

return 0;

} `

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

    first of all, put your code beetween ` ` next time so its more readable, and I think your code fails on this test case:

    3

    5

    5

    7

    I'll leave it to you to figure it out as to why.

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

For Problem-F, we just need to check the cells marked as '@' below:

. . . . . . .
. . @ @ @ . .
. @ @ @ @ @ .
. @ @ @ @ @ .
. @ @ @ @ @ .
. . @ @ @ . .
. . . . . . .

For the blue part, we need to check these cells:

. . . . . . .
. . @ . @ . .
. @ . @ . @ .
. . @ . @ . .
. @ . @ . @ .
. . @ . @ . .
. . . . . . .

So we just brute force from 0 to 4095.

For the red part, we need to check these cells:

. . . . . . .
. . . @ . . .
. . @ . @ . .
. @ . @ . @ .
. . @ . @ . .
. . . @ . . .
. . . . . . .

So we just brute force from 0 to 511.

Here is my solution: 247559944.

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

Problem-D,Why does using unordered_map table time out?

»
2 months ago, # |
  Vote: I like it +1 Vote: I do not like it

For problem C I've a O(n) solution which runs in about 15 ms (c++) 247602474

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

How to deal with Problem C if n<=1e18. (Chinese student,Englishi is very poor...)

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

    That means how to deal it with a O(t) solution. I think there is a solution which used math.

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

      I don't sure but i think my solution here is O(log(n)) for each query

      #include <bits/stdc++.h>
      using namespace std;
      #define ll long long
      
      ll sumOfDigits(ll n) {
          if (n < 10)
              return n * (n + 1) / 2;
      
          ll d = log10(n);
          ll *a = new ll[d+1];
          a[0] = 0, a[1] = 45;
          for (ll i = 2; i <= d; i++)
              a[i] = a[i-1] * 10 + 45 * ceil(pow(10, i-1));
      
          ll p = ceil(pow(10, d));
          ll msd = n / p;
      
          return msd * a[d] + (msd * (msd - 1) / 2) * p + msd * (1 + n % p) + sumOfDigits(n % p);
      }
      
      int main() {
          ios_base::sync_with_stdio(false);
          cin.tie(0); cout.tie(0);
      
          int t;
          cin >> t;
          while(t--) {
              ll n;
              cin >> n;
              cout << sumOfDigits(n) << "\n";
          }
      
          return 0;
      }
      
      • »
        »
        »
        »
        2 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Thanks for your healping.Last night I solve Problem C with a O(log10 n * t) solution.This is my code:

        #include<bits/stdc++.h>
        using namespace std;
        #define pii pair<int,int>
        #define mp(x,y) make_pair(x,y)
        #define pqg priority_queue<int,vector<int>,greater<int>>
        #define pql priority_queue<int,vector<int>,less<int>>
        #define pqg_pii priority_queue<pii,vector<pii>,greater<pii>>
        #define pql_pii priority_queue<pii,vector<pii>,less<pii>>
        #define scnaf scanf
        #define rt register int
        #define int long long
        int n,T,a[6],x[6],k[6];
        string s;
        signed main() {
        //	freopen(".in","r",stdin);
        //	freopen(".out","w",stdout);
        	cin>>T;
        	k[0]=45;
        	for(int i=1; i<5; i++)
        		k[i]=k[i-1]*10+45*pow(10,i);
        	cout<<endl;
        	while(T--) {
        		cin>>s;
        		int ans=0;
        		a[0]=a[1]=a[2]=a[3]=a[4]=0;
        		x[0]=x[1]=x[2]=x[3]=x[4]=0;
        		n=s.size();
        		for(int i=0; i<n; ++i)
        			a[i]=s[i]-'0';
        		for(int i=1; i<n; ++i)
        			x[i]=x[i-1]+a[i-1];
        		for(int i=0; i<n/2; ++i)
        			swap(a[i],a[n-i-1]),swap(x[i],x[n-i-1]);
        		if(a[0]!=0)
        			ans=(a[0]+1)*a[0]/2+a[0]*x[0];
        		for(int i=1; i<n; i++)
        				ans+=a[i]*pow(10,i)*x[i]+k[i-1]*a[i]+a[i]+(a[i]-1)*a[i]*pow(10,i)/2;
        		cout<<ans<<endl;
        	}
        	return 0;
        }
        
        • »
          »
          »
          »
          »
          2 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          And it could solve even though n<=10^1000 if t<=1^4

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

          Oh,I'm sorry.The time complexity of my code is $$$O(t \log_{10} n)$$$.

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

guys just to justify the tags for C, i came up with a digit DP solution.

digit_dp

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

I wondered if there is a not hard brute-force approach for the problem F

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

I think problem F can further be optimized by considering only the inner 5x5 square (the same idea, just considering less squares for backtrack). This approach should be more than 10x faster.

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

    I have upsolved problem F based on your idea by just checking the 5x5 square but using bit 1 in the generated binary numbers for painting (the 13 cells and 12 cells in the 5x5 square) rather than DFS.

    Haskell submission
»
2 months ago, # |
  Vote: I like it 0 Vote: I do not like it

For Problem D why my solution is giving TLE

CODE-:

int t;
cin>>t;
while(t--){
    int n;
    cin>>n;
    vector<int> v(n);
    for(int i = 0; i < n; i++){
        cin>>v[i];
    }

    int ans = 0;
    unordered_map<int, int> mpp;
    for(int i = 0; i < n; i++){
        if(mpp.count(INT_MAX - v[i])){
            mpp[INT_MAX - v[i]]--;
            if(mpp[INT_MAX - v[i]] == 0)
                mpp.erase(INT_MAX - v[i]);
            ans++;
        }else{
            mpp[v[i]]++;
        }
    }

    for(auto it : mpp){
        ans += (it.second)
    }

    cout<<ans<<endl;
}
»
2 months ago, # |
  Vote: I like it +8 Vote: I do not like it

I made video editorial for problem G.

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

Can anyone tell me why this is giving tle for E i did the same as in the editorial but i used binary search to fing no of elements we can take at each step am i missing something

code
»
2 months ago, # |
  Vote: I like it 0 Vote: I do not like it

For problem d:

for i in range(int(input())): n = int(input()) l = list(map(int,input().split()))

my_dict = {}

for num in l:
    if(num not in my_dict):
        my_dict[num] = 1
    else:
        my_dict[num] += 1

count = 0

for num in l:
    if(my_dict[num]!=0):
        count+=1
        xor_val = ((1<<31)-1)^num
        if(xor_val in my_dict and my_dict[xor_val]!=0):
            my_dict[xor_val]-=1
        my_dict[num]-=1

print(count)

i am getting tle... how can i optimise it??

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

I found this solution helpful as a beginner programmer. However, I believe attaching a video solution would greatly benefit learners like me who might need a bit more visual guidance

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

Can anyone give test case for the case 1(no type of water flowing up) in case of Question G: Vlad and Trouble at MIT? I feel that the latter two cases are enough.

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

For Problem-C, I have a O(1) time complexity solution: 247367957. O(#digits(n))...better than the one in tutorial?

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

In the problem C, you don't need to use the sum of digits function at all. You can also use dp to find the answer. This removes the logn factor.See my code;

include <stdio.h>

include <stdlib.h>

int main() { int* a = (int*) malloc(sizeof(int) * (2000001)); for (int i = 0; i <= 9; ++i) { a[i] = i; } int c = 1; for (int i = 10; i <= 2000000; i++) { if(i / c == 10) { c = 10; } a[i] = a[i — c] + 1; } int b = (int*) malloc(sizeof(int) * (2000001)); b[0] = 0; for (int i = 1; i <= 2000000; ++i) { b[i] += b[i — 1] + a[i]; } int t; scanf("%d",&t); for (int i = 0; i < t; ++i) { int n; scanf("%d", &n); printf("%d\n", b[n]); } return 0; } If you have any doubt, feel free to ask.https://codeforces.com/contest/1926/submission/249616774

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

// Could someone try to identify the logic mistake for the B problem? I am just checking if any //row's sum is equal to 1 if yes it is a triangle else square.

include <bits/stdc++.h>

using namespace std;

int main() { ios_base::sync_with_stdio(false); cin.tie(0);

int t, n, flag, sum;
char c;
cin>>t;
while(t--){
    cin>>n;
    flag = 0;
    for(int i=0; i<n; i++){
        sum = 0;
        for(int j=0; j<n; j++){
            cin>>c;
            if(c=='1')
                sum++;
        }
        if (sum==1){
            flag = 1; 
            cout<<"TRIANGLE"<<"\n";
            break;
        }
    }
    if(!flag) cout<<"SQUARE"<<"\n";
}
return 0;

}

  • »
    »
    4 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    first you have to take ALL the inputs otherwise inputs of other tests may overlap, second what if there is only one 1 in the whole grid? this is square not triangle!

    • »
      »
      »
      4 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      first you have to take ALL the inputs otherwise inputs of other tests may overlap could you please explain the above statement because I have used break so it shouldn't overlap right. Also, the input grid contains atleast 2*2 sqaure.

      • »
        »
        »
        »
        4 weeks ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        my bad, then you just have to take all the inputs, imagine with me that the shape it triangle, so in the first row of this triangle there is only one 1, your code just after getting that whole row prints triangle then break BEFORE you continue to take the coming rows for the triangle so those will go the next testcase

        • »
          »
          »
          »
          »
          4 weeks ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Thanks for the hint. My solution was accepted after I removed the break statement.

          Reason: If we use break, it prints "Triangle" but after that it breaks. So instead of getting the "n" it just inputs the next line as "n".

          For example, in the below case, it should go and input next "n" and not next line i.e. it inputs n = 1110 instead of n = 2

          4 0000 0000 0100 1110 2 11 11

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

For Problem-E, I have written this code but I am not able to understand the problem in my logic , and it gives wrong result for n=10^ and k=10^9. Please help me with this.

include <bits/stdc++.h>

using namespace std;

int main() { int t; cin >> t; while (t) { long long int n, k, m = 1, j = 0; cin >> n >> k; long long int r = 0; long long int x = 0; x = n / 2 + n % 2; while (1) { if (k <= x) { while (k) { r = m * (2 * j + 1); j++; k--; } break; } else { m = m * 2; k = k — x; x = x / 2 + x % 2; } } cout << r << endl; t--; } return 0; }

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

For problem C, I guess there is a solution with complexity O(t). I am just a beginner, so I can be mistaken. Someone more experienced can check this. Here is my solution :

include

include<math.h>

using namespace std;

int main()

{

ios_base::sync_with_stdio(0);

cin.tie(0);

int t;

cin>>t;

for(int tests=0;tests<t;tests++)

{

int n;

cin>>n;

int e=n%10;

n/=10;

int d=n%10;

n/=10;

int c=n%10;

n/=10;

int b=n%10;

n/=10;

int a=n%10;

n/=10;

int p=n;

bint res=(e+1)*(a+b+c+d+p+e/2)+(d>0)*d*(5*(d-1)+45+10*(p+a+b+c))+(c>0)*c*(50*(c-1)+100*(a+b+p)+900)+(b>0)*b*(500*(b-1)+1000*(a+p)+13500)+(a>0)*a*(5000*(a-1)+10000*p+180000)+(p>0)*p*(50000*(p-1)+2250000);

if(e%2==1)

{

   res+=(e+1)/2;

}

cout<<res<<endl;

}

return 0;

}

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it
g[vec[idx].first][vec[idx].second] ^= 1;

In the solution of Problem G,this option is only need when g[][] is 1.

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

New to c++ for problem c, can someone please explain why my logic is wrong , couldn't get the subject

#include <iostream>
#include <string>

int main()
{
	int testcase ;
	std::cin >> testcase;
	while (--testcase >= 0)
	{
		int a;
		int b = 0;
		int i = 1;
		std::cin >> a;
		while (a != 0)
		{
			if (i == 10)
				i = 1;
			b += i;
			i++;
			a--;
		}
		std::cout << b << std::endl;
	}
}