### JaySharma1048576's blog

By JaySharma1048576, 4 weeks ago,

Hello Codeforces,

I invite you to participate in Codeforces Round #785 (Div. 2) which will be held on 30.04.2022 17:35 (Московское время). The round will be rated for participants of Division 2 with rating strictly less than 2100. But as always, the participants of Division 1 are welcome to take part out of competition.

You will be given 6 problems and 2 hours to solve them. Moreover, there will be atleast $1$ and atmost $1048576$ interactive problems. So, you are advised to refer to the guide on interactive problems in case you are not familiar with them.

I would like to thank the following people for their contribution to the round:

This will be my second Codeforces round and based on the feedback by participants on Round #730, I have implemented the following changes:

• The round will have no theme. I have tried to keep the problem statements as short and to the point as possible.
• The pretests have been made stronger (specifically, more pretests have been included).
• I have tried to keep the samples and their explanation as useful as possible.
• No floating-point precision issues this time.
• The I/O constraints for the interactive problem(s) have been kept low to provide a fair ground to the languages with slower I/O.

The score distribution will be released strictly before the round and the editorial will be released strictly after the round.

Good luck and have fun!

Disclaimer

UPD: The score distribution is $500-750-1500-2000-2750-3250$.

UPD: Editorial

UPD: Congratulations to the winners

Div. 1 + Div. 2 -

Div. 2 -

First solves -

Problem Participant Time
A A_G 0:02
B Geothermal 0:05
C quailty 0:03
D BalintR 0:16
E zsxdcfv 0:27
F ksun48 0:11

• +369

 » 4 weeks ago, # |   +64 As a taster.. I want to assure the round is well balanced, sweet and spicy
•  » » 4 weeks ago, # ^ |   +68 By looking at your pfp, i can confirm that you are a taster :P
•  » » » 4 weeks ago, # ^ |   +44 Yes I do.. "easly morning dream" is always right!
 » 4 weeks ago, # |   +84 As a tester, problem statements are short and all have expected rating less than 1048576.
•  » » 3 weeks ago, # ^ |   0 Do you mean delta <= 1048576 but > 0?:)
 » 4 weeks ago, # | ← Rev. 2 →   +60 As a tester still waiting for jay to set a round without an interactive problem
 » 4 weeks ago, # |   +43 As a tester, I agree that no powers of 2 were harmed. JaySharma1048576 orz
 » 4 weeks ago, # |   +88 As a tester, I must say that it was quite hard to solve those 1048576 interactive problems.
 » 4 weeks ago, # |   +98 As a tester, I solved at most $1048576$ interactive problems.
 » 4 weeks ago, # |   +33 I care about tags :)
•  » » 3 weeks ago, # ^ |   +20 tags
 » 4 weeks ago, # | ← Rev. 2 →   -139 Oh no, mathforces again. Btw how did Mike let the "cheater" set a round again?
•  » » 4 weeks ago, # ^ |   -93 I don't understand why people are downvoting you. Though his contribution as a problem setter cannot be undermined, but he has cheated multiple times before. And allowing him to write more and more contests doesn't send a good message to the community.
•  » » » 4 weeks ago, # ^ |   +138 Initially I didn't want to feed the trolls but since this is happening repeatedly, I'll try to clarify this once and for all. Jay once submitted a problem from 2 accounts a long time ago. It was a moment's bad decision, he has accepted this since and also suffered a lot due to this. After which every time he was caught in plagiarism, it was in easy problems as he didn't use templates. I wish you could appreciate the amount of effort required to prepare a CF round and him trying to improve upon the flaws of his previous round.
 » 4 weeks ago, # |   +57 As a tester, good luck.
 » 4 weeks ago, # |   +61 No powers of 2 were harmed but testers were harmed with the powers of 2.The number 1048576 still haunts me :P
 » 3 weeks ago, # |   +14 Debugging interactive can be hard. Need this feature(google code jam has this) : https://codeforces.com/blog/entry/102256
•  » » 3 weeks ago, # ^ |   +15 This most probably will not be implemented on codeforces as their team is small and this feature is not very important. Better to setup your own environment. -is-this-fft- blog will be helpful for this.
 » 3 weeks ago, # |   +14 It seems to be interesting contest
 » 3 weeks ago, # |   +36 I really liked all your interactive problems. Looking forward to this round as well.
 » 3 weeks ago, # |   +12 Indian Round Let's Go!...
 » 3 weeks ago, # |   +27 Hey Jay! Good to see you back again with your interesting rounds!
 » 3 weeks ago, # | ← Rev. 2 →   +124 No powers of 2 were harmed during the making of this round.Not even $1$?
 » 3 weeks ago, # |   +25 Why is 1048576 occursed frequently?
 » 3 weeks ago, # |   -7 Excited for an Indian Round !
 » 3 weeks ago, # |   0 hope statements are short i don't know why but long statement make a feeling of scariness in itself.
 » 3 weeks ago, # |   -9 Hoping to retrieve my lost rating :)
 » 3 weeks ago, # |   -28 Hopefully I reach specialist today
•  » » 3 weeks ago, # ^ |   0 Hopefully I reach specialist today
•  » » » 3 weeks ago, # ^ |   +2 You did actually. Congrats :yayy:
 » 3 weeks ago, # | ← Rev. 2 →   -31 .
•  » » 3 weeks ago, # ^ | ← Rev. 4 →   -42 .
 » 3 weeks ago, # |   0 In general, A,B,C might not be interactive right?
•  » » 3 weeks ago, # ^ |   -6 96058983
•  » » 3 weeks ago, # ^ |   0 C may be interactive
 » 3 weeks ago, # |   +5 Super excited for this round !!
 » 3 weeks ago, # |   +3 good luck everyone
 » 3 weeks ago, # | ← Rev. 3 →   -21 _ _
 » 3 weeks ago, # |   0 I feel so sorry for missing out on the contest registration
 » 3 weeks ago, # |   +33 Problem D IS SUCH A PAIN
 » 3 weeks ago, # |   +154
•  » » 3 weeks ago, # ^ |   +1 Congrats for expert
 » 3 weeks ago, # |   +10 I really dislike Problem B. Problem B took me more than 1 hour, while I solved Problem C in less than 20 minutes. Proving that you algorithm works for Problem B is not straightforward, and while I got the main idea after a few minutes, I spent the entire hour trying to prove that it is true. I was extremely tempted to try proof by "accepted/try and see" vs mathematically proving a solution actually works, especially looking at the number of accepted submissions. At least I feel this shouldn't be encouraged for div2AB.One might argue that I just suck/am slow at proving things, which is fair, but I honestly don't think I'm that much worse than an average Div2 contestant in my proof skills.
•  » » 3 weeks ago, # ^ |   0 Tf were you proving?
•  » » » 3 weeks ago, # ^ |   +5 Let $prev(i)$ denote the maximum index $j •  » » 3 weeks ago, # ^ | 0 shit happens. i solved ab in 14 mins and couldnt handle with c lol •  » » 3 weeks ago, # ^ | +1 You can just check for all pairs of letters that condition is satisfied with prefix sums in n * 26^2, and there is nothing to prove there. •  » » » 3 weeks ago, # ^ | 0 Did that pass? The time limit was$1$second for$2*10^5*26*26 = 135200000$which I thought wouldn't pass. I ended up with$O(26n)$by noticing a counter example can always start and end with the same character. •  » » » » 3 weeks ago, # ^ | 0 C++ is very fast so it will probably pass. •  » » » » » 3 weeks ago, # ^ | 0 I'll try it once system tests are done. I thought of this idea literally 2 minutes into reading the problem, but since I thought the general rule was$10^8$ops/second, I thought it wouldn't pass.. •  » » » » » » 3 weeks ago, # ^ | 0 it passed i tried it •  » » » » » » » 3 weeks ago, # ^ | ← Rev. 2 → 0 Can I get a link to your submission? I'm getting TLEEdit: nvm, Got it AC by doing some low level optimisations. That's pretty surprising that it passed. •  » » » » » » » » 3 weeks ago, # ^ | 0 •  » » » 3 weeks ago, # ^ | 0 another way was calculating total number of distinct letters in the string (let it be) k, and for all 1 <= i, j <= n, s[i] == s[j], j > i you can check if j — i >= k. it must be true for all such pairs •  » » 3 weeks ago, # ^ | 0 my solution was to count the number of distinct characters. Now let's say the string s has 4 distinct characters. Then the first 4 characters must be some permutation of these 4 characters, and every set of 4 consecutive characters after must be the same permutation. EX: abdcabdcabd •  » » » 3 weeks ago, # ^ | 0 That passed the pretests? I thought of that and then realized the counterexample, what if the string doesn't have a permutation of the$d$distinct characters as a prefix of length$d$and instead has something like$abbabcabcabc$. •  » » » » 3 weeks ago, # ^ | 0 but for this case, the answer is NO, obviously, isn't it? •  » » » » 3 weeks ago, # ^ | 0 It's no because of the substring 'bb'. Here the difference between the number of b's and a's is 2 •  » » 3 weeks ago, # ^ | ← Rev. 2 → +3 only need to check if distance of two adjacent same characters is the number of distinct characters in s.because each of them should appear exactly once in this substring. •  » » » 3 weeks ago, # ^ | 0 Thanks, I didn't think of that. I ended up checking adjacent same characters, but brute-forcing the count for characters in the range using presum table. •  » » » 3 weeks ago, # ^ | 0 I approached problem B like this — the first n unique characters will be repeated maintaining their initial order until the end of the string. •  » » » » 3 weeks ago, # ^ | ← Rev. 2 → 0 How have you implemented that? Seems easy!!! •  » » » » » 3 weeks ago, # ^ | 0 https://codeforces.com/contest/1673/submission/155425009this is my solution. in the first loop i found out the substring formed by the first n unique characters. then for the rest of the string i just checked if they maintain the order or not. •  » » » » » » 3 weeks ago, # ^ | 0 Your implementation is really easy. I like it. I also had a similar approach using map. •  » » » 3 weeks ago, # ^ | 0 How to implement that in O(n)? •  » » » » 3 weeks ago, # ^ | ← Rev. 2 → 0 Check out my solution, if you want. I had the same approach. 155410182UPD: It is very bad implementation probably, but idc :D •  » » » » » 3 weeks ago, # ^ | 0 I like your implementation. Here is another way to do the last part of your code. while (sz(ans) < sz(s)) ans += ns; if (ans.substr(0,sz(s)) != s) cout << "NO"; else cout << "YES";  •  » » » » » » 3 weeks ago, # ^ | -6 Oh yeah... This looks much more easier, thanks :D •  » » 3 weeks ago, # ^ | 0 Probably, you choose a harder approach. My approach:Let's say the string has k distinct characters. Then the first k characters of a must contain all distinct characters, otherwise frequency of at least one letter in first k characters of string will be >= 2, making it imbalanced. Then, this substring of prefix must repeat, if it doesn't you can take [2...k+1], it will be imbalanced. Code of above approachvoid solve(){ int k,i, j, q, x, y; string a; cin >> a; n = sz(a); bool ok = true; map freq; each(c,a) freq[c-'a']++; k = sz(freq); forn(i,n) if(a[i] != a[i%k]) ok = false; pok(ok); }   » 3 weeks ago, # | +2 How can I solve problem D?  » 3 weeks ago, # | 0 How do you solve problem C. I tried to do a knapsack variationy sort of thing, but it counted the solutions 1 3 1 and 3 1 1 as different. •  » » 3 weeks ago, # ^ | +9 you can probably google it by searching "unbounded knapsack" •  » » 3 weeks ago, # ^ | +11 similar problem: CSES Coin Combinations II •  » » 3 weeks ago, # ^ | +9 This is simply solvable. Assume$p$is a set of weights. Then usual knapsack algorithm is for (int i = 0; i < n; i++) { for (int j : p) { if (i + j < n) ... } } Do instead for (int j : p) { for (int i = 0; i < n; i++) { if (i + j < n) ... } }  •  » » » 3 weeks ago, # ^ | 0 Could you please explain why this works though? •  » » » » 3 weeks ago, # ^ | ← Rev. 2 → 0 The idea is that once you pick a weight w[i], you make sure that you pick all possible amounts of it now so that you don't pick the w[i] again in the future. That way you wont have a case like 1,3,1. •  » » » 3 weeks ago, # ^ | +5 Also instead of for (int i = 0; i < n; i++) if (i + j < n) { ... }you can do for (int i = 0; i+j < n; i++) { ... }(not important for correctness or speed) •  » » 3 weeks ago, # ^ | 0 Backpack Counting.  » 3 weeks ago, # | ← Rev. 2 → +17 Perhaps I overcomplicated the casework in D but come on... This problem was just painful. On another note, does anyone know why mixing puts and cout results in a wrong answer? •  » » 3 weeks ago, # ^ | +1 Probably, you did not use flush in$cout$. Then they alternate flushing and produce mess in output. For$cout$it is$cout.flush()$and$cout « endl$.$puts$always flushes. •  » » » 3 weeks ago, # ^ | 0 That's probably it (since I'm not using endl). I'll be more careful next time (thank god I was able to suspect that puts was the problem). •  » » 3 weeks ago, # ^ | +4 always use same output method. •  » » 3 weeks ago, # ^ | +6 You are using ios_base::sync_with_stdio(false), which does not preserve the order of output streams.  » 3 weeks ago, # | +3 I literally submitted problem D 2 seconds before the contest ends but it said the contest is over... I forgot to take the answer mod 1e9 + 7. So sick of crap like this...  » 3 weeks ago, # | 0 ModuloForces  » 3 weeks ago, # | 0 In problem D, let say for a common difference d and first number c, how do you find smallest number in that sequence which exist in sequence B but not in sequence C?  » 3 weeks ago, # | +36 Trash Indian Round. Terrible problems : C , D . Maybe F is a good one , but generally a trash round. •  » » 3 weeks ago, # ^ | -25 C wasn't that bad actually, albeit a little standard. •  » » » 3 weeks ago, # ^ | +70 little completely •  » » » 3 weeks ago, # ^ | 0 Totallllly standard •  » » 3 weeks ago, # ^ | +9 I'm confused, are Indian rounds trash in general? or this particular round is trash among all the Indian rounds?  » 3 weeks ago, # | +8 how to solve E and F? E: I know from one index, at most 20 power could be extended. But don't get how to deal with K. •  » » 3 weeks ago, # ^ | +26 F: The goal is to find a good way to assign 0~1023 to all the cells, and edge connecting u and v is u^v. Gray code will do. •  » » » 3 weeks ago, # ^ | 0$2^5$x$2^5$gray code is yielding a sum of approximately 84k •  » » » » 3 weeks ago, # ^ | 0 You should use the 0, 2, 4, 6, 8-th bit for one dimension and 1, 3, 5, 7, 9 for the other dimension. •  » » 3 weeks ago, # ^ | +13 For problem E, consider for a certain$(i, j), i \le j$, how many times does$A_i ^ {A_{i + 1} \cdots A_j}$contribute to the final answer, when putting XOR both between$A_{i - 1}$and$A_i$, and between$A_j$and$A_{j + 1}$. •  » » 3 weeks ago, # ^ | +55 We just have to count the number of times$pow(l,l+1,\ldots,r)$appears in the final expression. As you said,$r-l \leq 20$holds since we take the answer modulo$2^{2^{20}}$, so we can brute force all pairs.The sequence will look something like$\ldots A_{l-2} ~?~ A_{l-1} \oplus A_l ~\hat{}~ \ldots ~\hat{}~ A_r \oplus A_{r+1} ~?~ A_{r+2} \ldots$. We just need to count the number of ways to fill in the$?$such that we satisfy the number of$\hat{}$being more than$k$. This is simple combinatorics and we get some expression of the form$\sum\limits_{k=0}^m \binom{n}{k}$. Remember we only need this expression under modulo$2$, also we will only require the value for about$20$values of$n$since$r-l \leq 20$, so just do prefix sums.Btw, you can find$\binom{n}{k}$under modulo$2$easily as$\binom{n}{k} \pmod 2= 1$iif$(k \& n)=k$. We can show this using Kummer's theorem, Lucas's theorem or just staring at an image of a Sierpiński triangle •  » » » 3 weeks ago, # ^ | +5 Found where I was stuck...Thanks very much.  » 3 weeks ago, # | 0 Problem C is similar to this problem  » 3 weeks ago, # | +5 Tests for problem B is very weak. My submission https://codeforces.com/contest/1673/submission/155406728 fails on test 1 abaabba of which the answer should be NO, but this code prints YES.  » 3 weeks ago, # | 0 Good Questions and Questions were arranged according to its level of difficulty  » 3 weeks ago, # | +4 The pretests have been made stronger (specifically, more pretests have been included).Are you sure for problem B?:( •  » » 3 weeks ago, # ^ | 0 :(  » 3 weeks ago, # | +39 So much 2 in problem E...  » 3 weeks ago, # | ← Rev. 2 → +5 Can somebody please tell me what's wrong with my approach in C: I made an array of all palindroms from 1 to 1e5 (v) (v[j] — j-th palindrome starting from 1) I also made a 2-dimensional array dp, when for each i<=4*1e4 and j that j-th palindrome (lets say v[j]) (starting from 1) is no more than 1e5, dp[i][j] = number of different sums of palindromes which are equal i and in each sum all the palindroms used are equal or less than v[j]. I filled this array for i from 1 to 40000for each i I found certain l that v[l] (l-th palindrome) <=i, but v[l+1]>i;Let's notice that dp[i][j] (v[j]<=i) = dp[i][j-1]+dp[i-v[j]][j], because for each sum of palindromes equal i, and each palindrome <=v[j], either in this sum is included v[j] as a palindrome, and all the other palindromes are <=v[j], (dp[i-v[j]][j] sums in total), or v[j] is not included => all the other palindromes are <=v[j-1] (dp[i][j-1] sums in total).since dp[i][0] = 0, then for each 1 <= j <= l dp[i][j]=dp[i][j-1]+dp[i-v[j]][j]. if j>l, dp[i][j] = dp[i][l], because i < v[l+1] <= v[j] (therefore in each possible sum of palindromes equal i every palindrome is <=v[l], and dp[i][j] = amount of all possible sums of palindromes equal i = dp[i][l].)After filling our array for each 1 <= i <= n, we can print dp[n][r], if n < v[r] <=1e5 and it will be our answer, (dp[n][r] = number of sums of palindromes equal n, and therefore each palindrome <= n < v[r], so dp[n][r] = amount of all possible sums).I will be glad if someone can explain what's wrong with my approach. Or is it correct and my realisation is wrong? •  » » 3 weeks ago, # ^ | +3 I looked at your solution. Your dp is correct, but you output the dp[n][i], such that it has the maximum value (since it's in modulo 1e9+7 that would be wrong). Just output dp[n][590] and that would pass the testcases •  » » » 3 weeks ago, # ^ | +3 Thank you so much, AC now)(OMG i can't believe I made this mistake, i'm such a moron. No words for that)  » 3 weeks ago, # | 0 what it means by tester( everyone in the comment is saying that he/she is a tester)? •  » » 3 weeks ago, # ^ | 0 if you read the blog carefully you will why they say their testers and not everyone is a tester .  » 3 weeks ago, # | +14 Technical question: The API was unavailable while contest. I this something that will happen now more often intentionally, or a coincidence?I use some tooling that uses the API to parse problem statements, tests etc. •  » » 3 weeks ago, # ^ | +3 I met the same situation: cf-tool warns Cannot find csrf, and I can neither login nor parse problem statements.  » 3 weeks ago, # | +1 What's an upper bound on the number of divisors of$n$?. I mean, is it something like$2$to the power of [number of smallest prime numbers with product up to$n$]? For example if$n = 10^9$I can take up to$2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29$so is$2^{10}$a safe upper bound? I was a little bit confused about iterating divisors in problem D, so I didn't even try, but I guess that's what I had to do. •  » » 3 weeks ago, # ^ | +9 It's$2 \sqrt n$but on average it's much lower •  » » » 3 weeks ago, # ^ | 0 Ok, I get it, for some reason I was trying to do better unnecessarily, we just iterate from$1$to$\sqrt{n}$and check if it's a divisor, that's enough here. I feel a bit dumb now but happens. Thank you. •  » » » 3 weeks ago, # ^ | 0 Cubic root is also okay: https://codeforces.com/blog/entry/651 •  » » » 3 weeks ago, # ^ | +3 It's$O(log(n))$on average. •  » » 3 weeks ago, # ^ | 0 look up OEIS  » 3 weeks ago, # | 0 Just asking: what is the point of making F interactive? I don't think there exist offline solutions since you can just ask all cells anyway. •  » » 3 weeks ago, # ^ | +31 Even if I ask the queries offline, the grader still needs the length of the roads to generate the queries which itself needs the problem to be interactive.  » 3 weeks ago, # | 0 I found problem C very hard to solve. At first, I tried to formulate the number of ways to create the integer using only 1-9, but got stuck there. Can anyone please tell me how to approach this kind of problem and solve it? •  » » 3 weeks ago, # ^ | 0 You have to realize that there are only ~500 palindromes less than 4e4. So you can simply do dp on this where dp[i] is the number of ways to get sum i. The dp method is mentioned somewhere above.My submission: 155444840 •  » » 3 weeks ago, # ^ | +1 My thought process (and I think pretty much everyone else's) was, 'This$n \le 4 \cdot 10^4$constraint is a bit weird, surely there must not be too many palindromic numbers anyway', so I quickly generated the palindromes and saw there were around$500\$, so the constraints were enough for a knapsack, and I just had to be careful to first iterate the palindromes in the knapsack to count only ordered sums.
•  » » » 3 weeks ago, # ^ |   0 Wow. i also tried to count the number of palindrome integers, and they were not many. But i didn't realize it can be approached as a knapsack problem.
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 You can find all the palindromes less than 4e4 with a simple python script. There are less than 500. after that, Visit the land of CSES Problemset, and find the one named Coin Combinations II. It will guide you.
•  » » » 3 weeks ago, # ^ |   0 Thanks a lot
 » 3 weeks ago, # |   -20 Generating all palindromic numbers for C was available online on GFG https://www.geeksforgeeks.org/generate-palindromic-numbers-less-n/
•  » » 3 weeks ago, # ^ |   +13 I'm pretty sure that you can use google, also it's not even that hard to generate palindromes
 » 3 weeks ago, # | ← Rev. 2 →   0 oh
•  » » 3 weeks ago, # ^ |   +1 Input1 aaabbbaaabbabbabaaababbac  Expected OutputNO  Your OutputNO Although your submission outputs the correct answer, it has a non-zero exit code, with stack smashing errors.
 » 3 weeks ago, # |   +18 Ratings updated preliminarily. We will remove cheaters and update the ratings again soon!
 » 3 weeks ago, # |   +5 Glad to be green again :D Thanks for this amazing contest!
»
3 weeks ago, # |
-41

Can Somebody help me why my solution for B problem is wrong? Thanks

# define lcm(a, b) ((a)*1ll * (b)) / gcd(a, b)

using namespace std; //vector sml={'a','b','c','d','e','f','g','h','i','j','k'}; //vector cap={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'}; bool prime(ll a) { for(int i=2;i*i<=a;i++) { if(a%i==0) return false;

}
return true;

} ll ceel(ll a,ll b) { if(a%b==0) return a/b; else return a/b+1;

} bool comparator(string a,string b) { return a<b; }

void sol() { string s; cin>>s; ll n=s.size(); ll i; priority_queuep; map<ll,ll>mp; vectorv(26,-1);

for(i=0;i<n;i++) { mp[s[i]]++;

} for(i=0;i<n;i++) { if(v[s[i]-97]==-1) { v[s[i]-97]=mp[s[i]]; }

} // for(i=0;i<26;i++) // { // cout<<v[i]<<" "; // } i=0; ll j=n-1; ll flag=0; while(i<j) { ll m1=INT_MIN; ll m2=INT_MAX; //cout<<"I: "<<i<<" "<<"J: "<<j<<endl; for(ll k=0;k<26;k++) { if(v[k]!=-1) { m1=max(m1,v[k]); m2=min(m2,v[k]); } } // cout<<m1<<" "<<m2<<endl; if(m1-m2>1) { flag=1; break; } if(v[s[i]-97]<v[s[j]-97]) {

v[s[i]-97]--;
i++;
}
else
{

v[s[j]-97]--;
j--;
}

} if(flag==0) { cout<<"YES"; } else { cout<<"NO"; }

cout<<endl;

}

int main() { ios_base::sync_with_stdio(false); cin.tie(0);
cout.tie(0); long long int t,i; cin>>t; for(i=0;i<t;i++) { sol(); }

return 0;


}

 » 3 weeks ago, # |   +3 If you prefer command line, checkout CF Doctor for more flexibility and faster feedback.If you are/were getting a WA/RE verdict on problems from this contest, you can get the smallest possible counter example for your submission on cfstress.com. To do that, click on the relevant problem's link below, add your submission ID, and edit the table (or edit compressed parameters) to increase/decrease the constraints. If you are not able to find a counter example even after changing the parameters, reply to this thread (only till the next 7 days), with links to your submission and ticket(s).
 » 3 weeks ago, # | ← Rev. 2 →   0 problem c : dont know why im getting wa , https://codeforces.com/contest/1673/submission/155442169updated : ac
 » 3 weeks ago, # |   -30 The problems were so interesting and fun. Unlike other contests, AB was really good. I like them
 » 3 weeks ago, # | ← Rev. 2 →   -7 good contest
 » 3 weeks ago, # |   0 When you find out that all div2 winners are alt accounts
 » 3 weeks ago, # |   0 I wish it be GREAT
 » 3 weeks ago, # |   0 Really? just 2 seconds? Humans can do that?
•  » » 3 weeks ago, # ^ |   +22 That's what she said :(
 » 3 weeks ago, # |   0 in this submittion: https://codeforces.com/contest/1673/submission/155426364. I write vector cnt(26,0); in line 23. then I have Runtime error on test 11 but I write int cnt[26]; for(int k=0; k<=25;k++) cnt[k]=0; then there is no error. I want to know WHY???
•  » » 3 weeks ago, # ^ |   0 x[s[i]-‘a’]++
•  » » 3 weeks ago, # ^ |   0 cnt[k+'a']
 » 3 weeks ago, # |   0 problem B looks similar to this problem https://codeforces.com/gym/307122/problem/A
 » 3 weeks ago, # |   0 I am avinash204, I am writing about the C problem of the contest. It happens that my solution coincides with another solution. The C problem was similar to a problem that I have done before on CSES (https://usaco.guide/problems/cses-1636-coin-combinations-ii-ordered/solution) and I happen to use the some part of the solution which I submitted last time on CSES. The AI has detected my solution to be coinciding with another user. Please look into the matter.
 » 3 weeks ago, # |   0 I am .DJ., I am writing about the B problem of the contest. It happens that my solution coincides with another solution. I don't know why the AI has detected my sol to be similar to other users. I have done it on my own, and it's just that I tested my code on an online compiler called programiz.com, and it could have my code leaked from there. You can also see my prev solution of B, which I got wrong and was some same approach. It's my original solution. I don't know how to prove can you help me with it. Please look into the matter.
 » 3 weeks ago, # | ← Rev. 2 →   0 Unfortunately, I did not know the rules, so I solved problems with my friends at school "binary129478" , "sameh_tarek1". but now i read the rules and understand. **I'm the one who solved the two problems