### Ahmed_Hosssam's blog

By Ahmed_Hosssam, history, 11 days ago,

Here is the editorial. Please provide your feedback on each problem so that we can improve upon them the next time.

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### 1670F - Jee, You See?

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• +69

 » 11 days ago, # |   -48 Feed back for C: While giving problems which require answer modulo p. Its good to provide 1 sample test where answer is greater than p so that we can understand that mod was wrong.
•  » » 11 days ago, # ^ |   +68 In this particular problem the size of the array should have been at least 30 in order to have an answer that exceeds 1e9 + 7, this seems too large for samples.
 » 11 days ago, # |   +15 can you provide c++ solutions?
•  » » 11 days ago, # ^ | ← Rev. 2 →   -6 Yeah they should provide code in atleast in 2 languages , just saying . Btw they have explained the editorial very clearly ... Thanks a lot guys.
•  » » 9 days ago, # ^ |   0 W
•  » » » 9 days ago, # ^ |   0 Cute doggy
 » 11 days ago, # |   +5 My solution for Problem C shouldn't this give TLE because of memset, if t = 1e5?
•  » » 10 days ago, # ^ |   +28 Hacked :)
•  » » » 10 days ago, # ^ |   0 I won't lose rating right? :'(
 » 11 days ago, # |   +11 Please provide your feedback on each problem so that we can improve upon them the next time. My feedback on each problem
 » 11 days ago, # |   0 Apart from my stupidity on getting unnecessary wrong answers. The contest was interesting :).
 » 11 days ago, # |   +5 A simple solution for D problem without binary search, exploiting the fact that for 3 types of lines the numbers should be as close as possible (+-1 difference)
 » 11 days ago, # |   0 Can someone please explain to me that why my solution is getting "time limit exceeded"? 156095877
•  » » 10 days ago, # ^ |   0 i think you can do something faster by doing bool isspecial[27] and check for them in O(1) instead of this "if s[i] in a:"
•  » » » 10 days ago, # ^ |   0 It's still exceeding time limit : 156179910 Or did I do something wrong?
•  » » » » 10 days ago, # ^ |   0 try fast io
•  » » » » » 10 days ago, # ^ |   0 Thanks it worked with fast io with pypy
•  » » 10 days ago, # ^ |   0 I was facing the exact same issue. TLE on pretest 3. In fact, changing lookup to O(1) didn't fix it either.
•  » » 10 days ago, # ^ | ← Rev. 3 →   0 This problem has quite a huge input (3e5 lines) and output (1e5 lines) and output (1e5 lines)It seems it won't pass without optimizing on this levelLookup fast io in python or pypy
•  » » » 10 days ago, # ^ |   0 Thanks for the help , it worked with fast io with pypy
 » 10 days ago, # |   +16 For D somehow this works ans=ceil(sqrt(n/6.0)*3); 
•  » » 10 days ago, # ^ |   -8 I was lucky to have found a similar formula and submit during contest. I am surprised this is not the solution used in the editorial.
 » 10 days ago, # |   +11 Video Solutions for C,D,E
•  » » 10 days ago, # ^ |   +4 Watched your previous videos, they are really good
•  » » » 10 days ago, # ^ |   +3 Thanks :) I really appreciate the feedback.
 » 10 days ago, # |   +5 A great contest, thanks to the writers :)I was really close to solving problem E during the contest. At first, I tried several approaches to choosing the root node, for instance, I have considered the node which is on the diameter of the tree, or the one which has the largest degree. But neither of them work.I completely fell into the trap of this problem, since it asks to select one node as the root, and I believe that the root node must have some special properties, which should be the "key" to the final solution. It is like 10 minutes before the end, I realized that we could always construct the tree to obtain "n" as the answer. I noticed that, except for "100..00", the other values could be divided into (n-1) pairs, (1xxx,0xxx), (1yyy,0yyy), and so on. We could always make the prefix XOR be less than or equal to "10000".It is a pity that I didn't finish the codes before the end, but this problem is really exciting, especially that "trap". Hope that next time I could do better.
 » 10 days ago, # |   0 In editorial of D: In second para "Number of intersections" should be "number of intersections at center of some hexagon" as intersection on some vertex won't create any triangle
•  » » 10 days ago, # ^ |   0 yes it is number of intersections in the middle of hexagon.
•  » » 10 days ago, # ^ |   0 You're right, fixed.
 » 10 days ago, # |   0 My solution for C without using dsu https://codeforces.cc/contest/1670/submission/156098315
•  » » 10 days ago, # ^ |   0 C++ solution for C (also without DSU): 156119537
 » 10 days ago, # | ← Rev. 2 →   +13 Can someone pls explain F a bit more (last line of third para "we only have to know the difference between the previous bits of X (add 1 if the current bit is on) and the sum of the generated bits.") I am unable to get it from the editorial.thanks in advance.
 » 10 days ago, # |   +5 I don't like the editorial for AFirst it gives a claim without proof Then it has a solution that doesn't directly use this claim at all"So let's say the number of negative elements is k. Then we must check that the first k elements are non-increasing and the remaining elements are non-decreasing." Why is it true? How does the solution use this fact?
•  » » 10 days ago, # ^ | ← Rev. 2 →   +1 Notice that applying the operations doesn't change number of negative elements. When sorted all the negative elements should be before all the positive elements. So if there are k negatives then we need to make the first k elements negative. For negative numbers to be sorted the absolute values must be non increasing. For positive numbers it should be non decreasing.
 » 10 days ago, # | ← Rev. 5 →   0 can anyone explain why the code for the C question is getting WA on test case 8?https://codeforces.cc/contest/1670/submission/156185766approach: if c[i] != 0 then c[i] = a[i] or b[i], but in front, we should not count a cycle that has a[i] or b[i], because we don't have any option, for this I am storing the values that cannot count in cycles in a set container.if(a[i] == b[i]) then also I inserted a[i] in the set. and while checking for no. of connected components I am checking whether the parent of i from 1 to n does not belong in the set.Yes I know, I made my approach messy using a set, I can use a visited array instead, but in the contest, this approach got stuck in my mind, and cannot able to change it...but I wanted to know the reason why my approach is wrong...
•  » » 10 days ago, # ^ | ← Rev. 2 →   0 1 3 2 1 3 3 2 1 0 0 3 
•  » » 10 days ago, # ^ |   0 You're only checking the root is not in s. You need to make sure none of the children are in s as well.
•  » » » 9 days ago, # ^ |   0 ya, you are correct, thank u !!
 » 10 days ago, # |   0 Problem C is so much like https://codeforces.cc/contest/1534/problem/C.
»
10 days ago, # |
0

what's wrong with my solution for (B). It says time limit exceeded on test 35.

# include <bits/stdc++.h>

using namespace std; int main() { long long t; cin>>t; while(t--) { char ch; long long arr[26]={0},n,val,max=0,ind,k,i,flg=0; cin>>n; string s; cin>>s>>k; while(k--){ cin>>ch; arr[ch-'a']=1; } for(i=0;i<n;i++) { if(arr[s[i]-'a']==1) { if(flg==0) {max=i;flg=1;} else if(i-ind>max) max=(i-ind); ind=i; } } cout<<max<<"\n"; } }

•  » » 10 days ago, # ^ |   0
•  » » » 10 days ago, # ^ |   0 thanks
 » 10 days ago, # |   0 Can someone explain how https://codeforces.cc/contest/1670/submission/156198034 this solution can not pass ?
•  » » 10 days ago, # ^ |   +3 Reading input is taking lot of time. Add these lines at beginning of main.  std::ios::sync_with_stdio(false); cin.tie(NULL); 
•  » » » 10 days ago, # ^ |   0 oh I see now... thanks a lot!
•  » » » 10 days ago, # ^ |   0 For understanding what this actually does, see -is-this-fft-'s comment.
 » 10 days ago, # |   +11 To get the sum of the first X groups it will be 3X^2. Again, such claims have to be proven
 » 10 days ago, # | ← Rev. 2 →   0 NVM. Figured it out
•  » » 10 days ago, # ^ |   0 You can’t use the parent array directly, use the find function instead.
•  » » » 10 days ago, # ^ |   0 Yeah, just figured it out. Thanks :)
 » 10 days ago, # |   0 Nice round brothers, you are a source of pride <3
 » 10 days ago, # |   0 Can someone help me to figure out the what mistake I have done https://codeforces.cc/contest/1670/submission/156213481My logic First I clearify the string i.e if the special characters occurs more then 1 time adjacently I remove the it like "kmjjjffj" if j is special characters then my function clearify the string as kmjffj. After that I find out the max distance between the special characters.
•  » » 10 days ago, # ^ |   0 Failing testcase: Ticket 6718
 » 9 days ago, # |   0 1670B — Dorms War I still didn't understand this problem editorial. (can't understand the solution as I don't know java) can somebody please explain that in detail.
 » 9 days ago, # | ← Rev. 2 →   0 can u please provide more explanation about (rem and nextRem) in dp function i can't see what is benefit, and thank u for the great contest Ahmed_Hosssam
•  » » 9 days ago, # ^ | ← Rev. 3 →   0 problem F ,,, the real magic happens here, [ Your code here... int currentBitSum = (val & 1L << idx) == 0 ? 0 : 1;int nextRem = 2 * (rem + currentBitSum — i); Your code here... ]can any one explain it Ahmed_Hosssam
 » 8 days ago, # |   0 can any one explain solution of problem d ?? i didn't understand the editorial!
•  » » 8 days ago, # ^ |   0 Which part you did not understand? Instead of using pen and paper, You can download a hexagonal graph paper and use sketch.io for better clarity
 » 7 days ago, # | ← Rev. 2 →   0 can someone check my solution for problem B? Im getting a TLE at 35th test case.
•  » » 7 days ago, # ^ |   0 here is your accepted code just add this line ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); it make your io operations fast as in this problem t<=10**5 which is large.
 » 7 days ago, # |   0 can someone help me with my solution to problem C, I'm getting TLE 156607712
•  » » 7 days ago, # ^ | ← Rev. 5 →   0 here is your accepted code the modifications are the following: max size of arrays vis and adj should be 10**5+1 not 10**5 only since you use 1-based index don't print pow(2,k) without modulus ,instead do this Spoiler if(ans>1 && j==d[i]){ k=(k*2)%mod; } 3.the cause of TLE is that you clear adj at every test case which is very time consuming so clear only n size portion from adj Spoiler for(int u=0;u
•  » » » 6 days ago, # ^ |   0 Now I see, thank you, I really appreciate