| # | User | Rating |
|---|---|---|
| 1 | jiangly | 3810 |
| 2 | Benq | 3676 |
| 3 | Kevin114514 | 3655 |
| 4 | maroonrk | 3463 |
| 5 | strapple | 3447 |
| 6 | Um_nik | 3387 |
| 7 | heuristica | 3322 |
| 8 | turmax | 3317 |
| 9 | tourist | 3307 |
| 10 | jiangbowen | 3291 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 156 |
| 2 | nik_exists | 150 |
| 2 | maspy | 150 |
| 4 | Um_nik | 143 |
| 5 | Errichto | 139 |
| 6 | adamant | 137 |
| 7 | AmShZ | 135 |
| 8 | maroonrk | 133 |
| 9 | BledDest | 132 |
| 10 | qwexd | 129 |
|
0
Host every contest offline, registering for a context would mean you would have to travel offsite to give the contest. |
|
0
In problem B solution, the extra — in the beginning is confusing. |
|
+56
Because |
|
+75
Okay, let me enter depression mode! |
|
0
How would why handle this case — -3 -3 -3 -3 -3 -3 -3 -3 -3 -3 -3 -3 -4 and k = 1000 I am not sure that taking only 11 elements would work. |
|
+3
Divide the array into two sets let call them A and B, where all the positive elements will be in A, and negative in B, find the top K sum in both the set individually, and then use a priority queue to find the top K from both A and B. Top K for each set can be found by sorting and some observations. |
|
0
what was your dp state for 250 points problem? codebuster_10 UPD: got it dp[i][j] -> # of ways to get j vowels odd number of times with i as the string length |
|
0
Where is it given that directions changes between neighbors |
|
0
Why is the output for Problem E test 2 'No'? |
|
+11
I Don't think Artificial intelligence is enough for CP. |
|
On
chokudai →
TOYOTA SYSTEMS Programming Contest 2021(AtCoder Beginner Contest 228) Announcement, 5 years ago
0
In problem E this Binary Exponentiation code gives WA it gives the correct answer when I add n %= mod why is this line needed? |
|
+3
Game Theory Problem for sure! |
|
0
Are the contests going to be updated manually by you or by automation? |
|
+3
Here we go again... |
|
+10
For authentication, they are using firebase. |
|
0
No, I was not. |
|
+1
Okay, then I can stop regretting not being able to solve this problem. |
|
0
Consider all the cells needs to be cleaned how your bfs is going to work? |
|
+3
I Don't think this will work, as the robot can clean cells while returning to (0,0) |
|
+4
Mike_Mirzayanov_Is_Pussy is your solution for problem 3 takes O(4 ^ K) if not then how did you solve it? UPD: 4 ^ K = 16777216 as max k is 12 we can slightly optimize it by observing that the robot can't go beyond a distance of k / 2 from (0,0) as he also has to return to the cell (0,0). I think this should work. |
|
+3
why are you checking only till sqrt(maxvalue), how will it work for the case 21, 14 |
|
+3
ig it will give wrong answer for test 6 14 3 7 the answer is 2 (3 and 7), but according to your approach answer is 3 (2,3,7) |
|
+10
Solution Good Sleep. |
|
+23
Binary Search on the answer, Now how to check if we can make M teams of K size each Note that we need a total of M * K members now from each country, we can take a maximum of M members, so if the summation of Min(A[i],M) overall "i" is greater than or equal to M * K then we can form M teams of K size each, otherwise not. |
|
+2
What is goc33? |
|
+3
Spoiler ig this is a google OA problem |
|
0
was this contest for full time or intern hiring? |
|
0
Were this challenge was for full time or intern hiring? |
|
0
I used it in 2 contest work so good thanks, man. |
|
+3
How are you inspired from SecondThread ? |
|
+3
Completely agreed, that's why I wrote "etc". |
|
+3
Companies like Uber, Codenation, Sprinklr, Google(Maybe) you need to have good experience in CP, but other companies including Amazon, Microsoft, etc leetcode/InterviewBit/GFG is the best resource you can use. You can still keep on giving cf contest. |
|
0
|
|
0
so that we don't replace two adjacent elements with the same prime number. consider the case 3 3 3 |
|
0
All the best bro. |
|
0
Thanks a lot. |
|
+1
Can anyone help can't figure out what's wrong with my problem D solution. WA on Test 5 114051343 |
|
0
Thanks. |
|
+4
VScode forever. |
|
0
oh Nice. |
|
0
Also, likes and dislikse of each problem like leetcode are also good. |
|
+6
Cool App. |
|
0
real data analyst |
|
0
Thanks really helped alot. |
|
+8
It takes time and effort to make a contest all we can do is wait. |
|
0
Practice more. |
|
0
Cool. |
|
0
Practice. |
|
0
I am not getting problem D. |
|
0
I am not getting problem D. |
|
On
MikeMirzayanov →
Technocup 2020 — Elimination Round 4 + Codeforces Round 606: Editorial, 5 years ago
0
Why is the answer of sample test 1(test case 2) is 0 what about the path 1->2->3->4 upd: My bad understood now |
|
+4
Why I am not able to see the editorial? zscoder |
|
0
arham_doshi I have a different approach for the problem "Flight Discount" But getting wrong answer on 3 TC can you help me please,I can't find any TC. |
|
0
Can E be solved with Matrix exponentiation?? |
|
On
-is-this-fft- →
[Tutorial] The DFS tree and its applications: how I found out I really didn't understand bridges, 6 years ago
+8
The sad part is that I can only give you upvote for this amazing article. p.s this is the best blog I have read on cf. |
|
On
-is-this-fft- →
[Tutorial] The DFS tree and its applications: how I found out I really didn't understand bridges, 6 years ago
0
Got it. This is because the back edge vertex can be an articulation point. So, if we do low[v] = min(low[v],low[to]) then it will connect both of the components. see Example 7 8 1 2 1 3 2 4 3 4 4 5 4 6 5 7 6 7 |
|
On
-is-this-fft- →
[Tutorial] The DFS tree and its applications: how I found out I really didn't understand bridges, 6 years ago
0
-is-this-fft- I feel dumb asking this. In the implementation to find Bridges. // in dfs if (visited[to]) { low[v] = min(low[v], tin[to]); } Why can't we do low[v] = min(low[v],low[to]) or both things are different? |
|
0
. |
|
0
Okay thanks! |
|
+1
I have a small doubt in problem statement For n=6 m=2 k=2 2 2 2 1 2 2 Is this a valid combination? |
|
0
Brute force on the persuade people by taking two-pointer one starting from 0 index(say left) and one at n-min(k,m-1)-1 (say right), then for each pair of left and right, find the minimum element you can get by brute-forcing from left to right with the same approach of taking two-pointer one starting from left(say l) and another at n-m+left(say r), then find minimum from max(a[l], a[r], r till right. Do the same for all the values of left and right. And the maximum of all the values will be the answer. |
|
0
Can you help me to find the time complexity of my submission with explanation for Div2 C problem 70354436 |
| Name |
|---|


