### mesanu's blog

By mesanu, history, 2 months ago,

Thank you for participating!

1926A - Vlad and the Best of Five

Idea: flamestorm

Tutorial
Solution

Idea: mesanu

Tutorial
Solution

1926C - Vlad and a Sum of Sum of Digits

Idea: flamestorm

Tutorial
Solution

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
• +69

 » 2 months ago, # |   +74 For Problem-F, I have a bitmask dp solution: 247350477
•  » » 2 months ago, # ^ |   +15 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 →   +20 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: It is a good matrix (every black square doesn't have 4 diagonal black square) and $f(mask) = 0$: We can check it in $O(bit \ number \ of \ mask)$. 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, # ^ |   +1 Wow, amazing approach, thanks for sharing. It helped a lot.
•  » » 2 months ago, # ^ |   +4 great solution , can you share any resources from where I can learn it?
•  » » » 2 months ago, # ^ |   +1 Usaco Guide Bitmask DP Tutorial may be helpful.
•  » » » » 2 months ago, # ^ |   +1 Thanks
 » 2 months ago, # |   +26 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 →   0 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, # ^ |   0 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, # ^ |   0 Practice as more as possible. CodeForces is not the main OJ on which I solve problem.
•  » » » » » 3 weeks ago, # ^ |   0 Which platform you use for practice and whats the good place to practice ?
•  » » » » » » 2 weeks ago, # ^ |   0 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, # |   -13 problem f is so bad i cant solve it
•  » » 2 months ago, # ^ |   +21 it's hard for div4 I agree
•  » » 2 months ago, # ^ |   0 If you're looking for an easier bruteforce solution which worked!Have a look at this
 » 2 months ago, # |   +20 There's a much simpler solution for G. Root the tree. Do DFS 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. 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, # ^ |   0 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, # ^ |   0 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, # ^ |   0 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, # ^ |   0 you can try this hack 1 1 1 2 2 3 3 CCCPSPS anwser may be 2 or 3
•  » » » 2 months ago, # ^ | ← Rev. 2 →   0 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, # ^ |   +1 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, # ^ |   0 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, # ^ |   +1 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, # ^ |   0 Can you attach the submission link ? I tried the same thing although i am getting WA on Test 3
•  » » » 4 weeks ago, # ^ |   0
 » 2 months ago, # | ← Rev. 2 →   0 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, # ^ |   0 Bro why you did it in live stream?
•  » » » 2 months ago, # ^ |   0 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, # ^ |   +1 Sorry, i thought it was public during contest sorry for misunderstanding
•  » » 2 months ago, # ^ |   0 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, # |   0 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 constantI solved it other way: https://codeforces.com/contest/1926/submission/247402620 But still curious why first one is TLE
•  » » 2 months ago, # ^ |   0 Because of unordered_map blowing up. This blog tells more about it.
•  » » » 2 months ago, # ^ |   0 Goddamn. Thank you.
•  » » » 2 months ago, # ^ |   0 Thank you. i had no idea why my ans was getting TLE.
 » 2 months ago, # |   +3 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*2if it has a "010" then it is a triangle else is squareits an easier approach Solution#include #include #define ll long long using namespace std; int main() { ll t; cin>>t; while(t--){ ll n; cin>>n; string arr[n]; for(ll i = 0; i < n; i++) { cin >> arr[i]; } bool found = 0; for (ll i = 0; i < n; i++) { found = arr[i].find("010") != std::string::npos; if(found){ break; } } if (!found) cout << "SQUARE" << endl; else cout << "TRIANGLE" << endl; } }
 » 2 months ago, # | ← Rev. 10 →   +12 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 enoughhere's my solution
 » 2 months ago, # |   0 I solve problem C without precomputing in O(t*log10(n)).LOL 247319547
•  » » 2 months ago, # ^ |   0 Do you have the math proof of that?? I couldn't understand
•  » » 2 months ago, # ^ |   +1 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, # ^ |   0 same. I literally wasted 20 minutes trying to find a formula, the 0.5 second bugged me.
•  » » » » 2 months ago, # ^ |   0 me too,I didn't try to preprocess it
•  » » 2 months ago, # ^ |   0 Can you explain the mathematical proof behind this?
 » 2 months ago, # | ← Rev. 2 →   +19 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, # |   +8 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, # ^ |   +8 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, # ^ |   0 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, # |   0 The link to G in the tutorial is in russian for some reason.
 » 2 months ago, # | ← Rev. 2 →   0 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, # ^ |   0 8 * 1
•  » » 2 months ago, # ^ |   0 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, # ^ |   0 so that was And so on, until all cards are laid down.. part now I can see thank you
 » 2 months ago, # |   +27 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, # ^ |   +9 I tried modeling with network flow but failed too lol
•  » » 2 months ago, # ^ | ← Rev. 2 →   0 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 →   +8 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 →   +17 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 →   +16 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 →   0 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#include using namespace std; #define int long long int f(int i,int n){ int lo=1,hi=1e10; int ans=1; while(lo <= hi){ int mid = lo + (hi-lo)/2; if(pow(2,i) + (mid - 1)*(pow(2,i+1)) <= n){ ans=mid; lo = mid+1; } else{ hi=mid-1; } } return ans; } void solve(){ //find highest power of 2 which is less than n int n;cin>>n; int k;cin>>k; int h_p = log2(n); vectorv;//no of num we can take with 2^i for(int i=1;i<=h_p;i++){ v.push_back(f(i,n)); } // for(int i=0;itakes; takes.push_back((n+1)/2); for(int i=0;i>t; while(t--){ solve(); } }
 » 2 months ago, # |   0 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, # ^ |   0 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 →   0 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, # ^ |   0 use custom hash
•  » » » 2 months ago, # ^ |   0 I want to hack this not prevent it. Like I want understand how hack collision can be performed
•  » » » » 2 months ago, # ^ |   0 You can checkout usaco guide,they have a section about generating hack cases for hack collisions.
 » 2 months ago, # |   0 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, # |   +10 I don't get why dp is needed in G. Isn't it simple dfs? #include using namespace std; const int int_max = 1e5 + 10; char s[int_max]; vector 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 →   0 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, # ^ |   +3 https://codeforces.com/contest/1926/submission/247492595You need to use faster flow network
•  » » » » 2 months ago, # ^ | ← Rev. 2 →   0 Thanks. It got AC using Dinic's algorithm. After you said faster flow network I googled and found out it takes less time for unit capacity $O(M√M)$.Sorry, but since I didn't know about it before, I didn't really understand your implementation of it. Did you use the same?
 » 2 months ago, # |   +18 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 →   0 My solution in C was $O(N + t)$ 247266041. If $n \le 10^6$ my solution will still workBut it contains precomputation too
 » 2 months ago, # |   +3 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, # |   0 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 →   0 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, # ^ |   0 noooo I didnt get hacked. I got TLE in contest itself.
•  » » » » 2 months ago, # ^ |   0 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, # ^ |   0 ohhh thanks for the reply. Any idea why do they do this on purpose? And how do they achieve this exactly
•  » » » » » » 2 months ago, # ^ | ← Rev. 2 →   +8 They do this on purpose to make people come up with efficient algorithms that don't rely on hash tables. Of course, in C++ there's ordered map.For example, try solving:Subarray Sums I: https://cses.fi/problemset/task/1660Subarray Sums II: https://cses.fi/problemset/task/1661You will be surprised :DHere's a hack tutorial for Python and you can test it out yourself: https://codeforces.com/blog/entry/98994You exploit the hash function to create collisions and make any resizing attempts still maintain collisions.
•  » » » » » » » 2 months ago, # ^ |   0 holy hell thanks for the articles
•  » » 2 months ago, # ^ |   0 my c++ version also got TLE #include #include #include 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 counter; vector 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, # ^ |   0 Change unordered_map to map
•  » » » » 2 months ago, # ^ | ← Rev. 2 →   0 Could you please explain the reason behind this?
•  » » » » » 2 months ago, # ^ |   0 This because unordered_map in c++ has bad implementation
•  » » 2 months ago, # ^ |   0 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, # ^ |   0 https://codeforces.com/contest/1926/submission/247314271Here's slow Python IO passing.
•  » » » » 2 months ago, # ^ |   0 Hmm, this is interesting. I didn't come across such behaviour earlier. New learning for me.
•  » » » » » 2 months ago, # ^ | ← Rev. 2 →   0 Try this one: https://cses.fi/problemset/task/1661They have a test case that purposefully TLEs a Python hashmap and if you switch to a $O(n \log n)$ solution, with same IO, it passes.
 » 2 months ago, # |   0 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, # |   0 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, # ^ |   0 Your description sounds correct, your code doesn't quite do that. 1 3 110 110 000
•  » » » 2 months ago, # ^ |   0 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 →   0 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, # |   0 Why didn't my rating change ( my rating was 381 before the contest ) even though I solved the first three problems ?
•  » » 2 months ago, # ^ |   0 Have patience, it takes some time. Check now.
•  » » » 2 months ago, # ^ |   0 Alright thanks :)
 » 2 months ago, # |   0 how to solve F in polytime?
 » 2 months ago, # |   0 Why my solution gets TLE for Problem C? 247461107
 » 2 months ago, # |   0 This was my first Codeforces contest. Can anyone suggest study materials for dsa.
 » 2 months ago, # |   0 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, # |   +8 I think this Div4 is harder than previous Div4, because problem F and G was very hard for pupil
 » 2 months ago, # |   0 Could you please explain the approach of D ?
»
2 months ago, # |
Rev. 4   -8

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 →   0 first of all, put your code beetween   next time so its more readable, and I think your code fails on this test case:3557I'll leave it to you to figure it out as to why.
 » 2 months ago, # |   0 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, # |   0 Problem-D,Why does using unordered_map table time out?
 » 2 months ago, # |   +1 For problem C I've a O(n) solution which runs in about 15 ms (c++) 247602474
 » 2 months ago, # |   0 How to deal with Problem C if n<=1e18. (Chinese student,Englishi is very poor...)
•  » » 2 months ago, # ^ |   0 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 →   0 I don't sure but i think my solution here is O(log(n)) for each query #include 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, # ^ |   0 Thanks for your healping.Last night I solve Problem C with a O(log10 n * t) solution.This is my code: #include using namespace std; #define pii pair #define mp(x,y) make_pair(x,y) #define pqg priority_queue,greater> #define pql priority_queue,less> #define pqg_pii priority_queue,greater> #define pql_pii priority_queue,less> #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<>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
•  » » » » » 2 months ago, # ^ |   0 And it could solve even though n<=10^1000 if t<=1^4
•  » » » » » 2 months ago, # ^ |   0 Oh,I'm sorry.The time complexity of my code is $O(t \log_{10} n)$.
 » 2 months ago, # |   0 guys just to justify the tags for C, i came up with a digit DP solution. digit_dp
 » 2 months ago, # |   0 I wondered if there is a not hard brute-force approach for the problem F
 » 2 months ago, # |   0 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, # ^ |   0 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 submission248239875
 » 2 months ago, # |   0 For Problem D why my solution is giving TLE CODE-: int t; cin>>t; while(t--){ int n; cin>>n; vector v(n); for(int i = 0; i < n; i++){ cin>>v[i]; } int ans = 0; unordered_map 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<
 » 2 months ago, # |   +8 I made video editorial for problem G.
 » 2 months ago, # |   0 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#include using namespace std; #define int long long int f(int i,int n){ int lo=1,hi=1e10; int ans=1; while(lo <= hi){ int mid = lo + (hi-lo)/2; if(pow(2,i) + (mid - 1)*(pow(2,i+1)) <= n){ ans=mid; lo = mid+1; } else{ hi=mid-1; } } return ans; } void solve(){ //find highest power of 2 which is less than n int n;cin>>n; int k;cin>>k; int h_p = log2(n); vectorv;//no of num we can take with 2^i for(int i=1;i<=h_p;i++){ v.push_back(f(i,n)); } // for(int i=0;itakes; takes.push_back((n+1)/2); for(int i=0;i>t; while(t--){ solve(); } }
 » 2 months ago, # |   0 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, # |   0 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, # |   0 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 →   0 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   0

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 <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   0

// 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, # ^ |   0 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, # ^ |   0 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, # ^ |   0 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, # ^ |   0 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 = 24 0000 0000 0100 1110 2 11 11
»
4 weeks ago, # |
0

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   0

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<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, # |   0 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, # |   0 .ComplaintFrame { display: inline-block; position: absolute; top: 0; right: -1.4em; } .ComplaintFrame a { text-decoration: none; color: #ff8c00; opacity: 0.5; } .ComplaintFrame a:hover { opacity: 1; } ._ComplaintFrame_popup p, ._ComplaintFrame_popup button { margin-top: 1rem; } ._ComplaintFrame_popup input[type=submit] { padding: 0.25rem 2rem; } ._ComplaintFrame_popup ul { margin-top: 1em !important; margin-bottom: 1em !important; } New to c++ for problem c, can someone please explain why my logic is wrong , couldn't get the subject #include #include 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; } }