### BledDest's blog

By BledDest, history, 10 days ago,

1535A - Fair Playoff

Idea: BledDest

Tutorial
Solution (Neon)

1535B - Array Reodering

Idea: BledDest

Tutorial
Solution (Neon)

1535C - Unstable String

Idea: BledDest

Tutorial
Solution (Neon)

1535D - Playoff Tournament

Idea: BledDest

Tutorial
Solution (Neon)

1535E - Gold Transfer

Tutorial

1535F - String Distance

Idea: BledDest

Tutorial
Solution (BledDest)

• +83

 » 10 days ago, # |   +42 The round was really educational. thanks!
 » 10 days ago, # |   -13 Solution to problem c was much simpler . Just use basic DP . Here's my solution -> ll dp[n+1][2]; dp[0][0] = 0; dp[0][1] = 0; ll ans = 0; for(ll i = 1 ; i <= n; ++i){ if(s[i-1] == '?'){ dp[i][0] = dp[i-1][1] + 1LL; dp[i][1] = dp[i-1][0] + 1LL; }else if(s[i-1] == '0'){ dp[i][0] = dp[i-1][1] + 1LL; dp[i][1] = 0; }else if(s[i-1] == '1'){ dp[i][1] = dp[i-1][0] + 1LL; dp[i][0] = 0; } ans += max(dp[i][0] , dp[i][1]); } cout << ans << endl;
•  » » 10 days ago, # ^ |   0 Can you explain the logic behind this Solution?
•  » » 10 days ago, # ^ | ← Rev. 3 →   +18 Without DP Time Complexity — O(N) Space Complexity — O(1)int main() { int tc; cin>>tc; while(tc--) { ll i,c1,c2,f,ans; string s; cin>>s; c1=0; c2=0; f=0; ans=0; for(i=0;i
•  » » » 10 days ago, # ^ |   0 Auxiliary Space, not Space Complexity.
•  » » » » 9 days ago, # ^ |   0 difference ?
•  » » » » » 9 days ago, # ^ |   0 Google.
•  » » » 10 days ago, # ^ | ← Rev. 2 →   0 Logic Please
•  » » » 7 days ago, # ^ |   0 Somehow I understood this task "the number of beautiful substrings of the string s" as follows: For string 0101, the beautiful substrings are: 0 1 01 10 010 101 0101 i.e. 7 substringsIt turns out that the solution expects 10 for this example: 0 1 0 1 01 10 01 010 101 0101 i.e. 10 substrings (with duplicates at different indexes)
•  » » » 4 days ago, # ^ |   0 Can you explain the logic behind this solution?
•  » » 10 days ago, # ^ |   0 I am trying to solve the problem C with dp ,but unable to solve. Can you explain how you came up with the solution. What is the logic for dp relation?
•  » » » 10 days ago, # ^ |   0 The logic for dp is suchIf character at ith index is '0' you can't do anything but take transition from previous index's '1' that is dp[i-1][1] where dp[i][j] stores the number of valid substrings if the ith index ends with j (0 or 1). Similarly the same has to be done if the ith index is '1'Now if the ith index is '?' you would want the the alternating string ending at '?' to be as huge as possible therefore u pick dp[i — 1][0] or dp[i — 1][1] depending which is bigger.
•  » » 10 days ago, # ^ | ← Rev. 9 →   0 That's what I call simpler :) char s[7 << 15]; long long r; int p, i, j, k; int main() { for(scanf("%*d"); ~scanf("%s",s + 1); r=p=i=j=k=0) { for(; s[++i]; r += i - k) if(s[i] != '?') { int x = s[i] ^ i & 1; if(p != x) k = j, p = x; j = i; } printf("%lld\n", r); } } 
•  » » 6 days ago, # ^ |   0 Why is this downvoted so much? the solution is much simpler than the official one.
 » 10 days ago, # | ← Rev. 2 →   +42 I accidentally solved a harder version of problem E where you're given not a path from $v_i$ to the root in the second query type but ANY vertical path. I think it might be fun for you to solve if you're interested.UPD: Oh yeah, btw, I can modify this solution a little bit so that it works not only for vertical but for arbitrary paths.
•  » » 10 days ago, # ^ |   0 Can u provide the link to the referred problem?
•  » » » 10 days ago, # ^ | ← Rev. 2 →   +1 There's no such problem. I was just solving problem E from this contest and forgot that the path goes from the root.
•  » » 10 days ago, # ^ |   +8 I think SecondThread also did something similar by bashing it with LCT with path aggregates: https://codeforces.com/contest/1535/submission/118434117He explains it in https://youtu.be/1x4mkj1kE1E?t=7051
•  » » » 10 days ago, # ^ |   +8 Omg. LCT. No, my submission 118419626 is just a little bit harder than the one from the editorial. Yeah, I also thought about LCA but it's too much for problem E I guess.
 » 10 days ago, # | ← Rev. 2 →   -13 I Loved problem D, the first time I got a chance to use Segment tree in a CF contest and the editorial also good. Thanks to the team behind the contest.
•  » » 10 days ago, # ^ |   0 You already commented the same before ¯_(ツ)_/¯
•  » » » 10 days ago, # ^ |   0 Both posts are regarding the same contest and the same problem so I think it's fair to post the same comment.
 » 10 days ago, # |   +1 Thanks for such an interesting contest! I really enjoyed the problems, especially B and C. Actually, I think problem A was a little bit too easy in comparison with the last contest. I hope the next contest will be more and more exciting! :)
 » 10 days ago, # |   +109 I have a solution to problem F in $O (n * m * log n)$ (where $m$ is the length of the string). Similarly to the author's, we split strings into equivalence classes. We only need to quickly calculate the number of pairs of strings with a distance equal to 1. Divide each string into the minimum number of blocks so that the characters in each block are sorted. Note that for the string $s[i]$ this partition is unique and is specified only by those positions $j$, where $s[i][j] < s[i][j - 1]$. Suppose we have two strings $s1 < s2$. Note that we can sort in $s2$ only a segment nested in some block in $s1$. Then, in fact, we can expand this segment to the size of the block. That is, for each block $[l, r]$ in $s1$, we need to find the number of such $s2$ so that if we sort the segment $[l, r]$ in $s2$, then $s2$ becomes equal to $s1$. Note that in this case $s1[:l-1]$ = $s2[:l-1]$ and $s1[r + 1:]$ = $s2[r + 1:]$. Let's iterate over $l$ for the equivalence class and split into new equivalence classes by prefixes of length $l$. Now, note that if some block $[l, r]$ of string $s[i]$ begins in $l$, then we need to find out the number of such $s[j] > s[i]$ in the new equivalence class $s[i]$ that the length of the common suffix $s[i]$ and $s[j]$ $> = m - r$. This is easily done by simply sorting all the reversed strings and doing a binsearch using a sparse table on the lcp array. Further, going through the strings in descending order, add 1 to their position in the new array, for beginning of the block find out the amount on the segment using any convenient data structure.My submission — 118431979.P.S. If you don't understand something, ask questions. Sometimes even I don't understand myself.
•  » » 9 days ago, # ^ |   +8 "Let's iterate over l for the equivalence class and split into new equivalence classes by prefixes of length l"How do you split into new equivalence classes ?
•  » » » 8 days ago, # ^ |   0 Strings $s1$ and $s2$ in the same new class, if the length of their common prefix is ​​at least $l$ and their multisets of characters are the same. Therefore, when incrementing $l$ by 1, you only need to create subclasses using the character at position $l + 1$ (numbering from 1).
•  » » 8 days ago, # ^ |   +8 Can you explain about vector> to(m, vector (sz)) in your code?
•  » » » 8 days ago, # ^ |   0 $to[j][i]$ — the number of the equivalence class of $s[i]$, if $l$ (which I described earlier) is equal to $j$. It is not hard to see that the equivalence classes form segments in the sorted array of strings.
•  » » 6 days ago, # ^ |   +8 vector> bup(sz, vector (lg, -2)); for(int i = 1; i < sz; ++i) { int a = ord2[i]; int b = ord2[i - 1]; while(rs[a][bup[i][0]] == rs[b][bup[i][0]]) { bup[i][0]++; } I wonder what this initiation means. Isn't this -2 out of bound?
•  » » » 6 days ago, # ^ |   +13 lol really. You are right, this is out of bound. :) I was lucky that there was no RE.
 » 10 days ago, # |   +1 I think I found an easier solution to problem B (I am not saying it is faster — just easier). There's simply no need to move the even numbers to the beginning, just check for each pair (i < j) if __gcd(a[i], 2 * a[j]) > 1 or __gcd(a[j], 2 * a[i] > 1.Here is my submission: 118428074.
 » 10 days ago, # |   0 editorial for c is more complicated to understand then the solution itself
 » 10 days ago, # |   0 simple (not) solution for C, no Dp, just string compression and index checking and then straight up calculating, also if an array of len n is beautiful then all its substrings will be beautiful, and an array of len n will have n*(n+1) substrings, I have used this concept.solution
 » 10 days ago, # |   0 Can someone please explain the solution to B in python ?
•  » » 10 days ago, # ^ |   0 Bro just think about this, put all positive number in the beginning , so for each positive number you’ll have n-i (where i is it’s one based index ) pairs , then put all odd numbers in sorted order and manually calculate the pairs using two for loops, simple as that. Try to prove why this is the best ordering yourself ;)
•  » » 4 days ago, # ^ |   0 from math import gcd def solve(): n = int(input()) # List of elements created with even and odd element are separated arr = sorted(list(map(int,input().split())),key =lambda x:x%2) # finding good indexes pairs count = 0 for i in range(n): for j in range(i+1,n): if(gcd(arr[i],2*arr[j])>1): count +=1 print(count) if __name__=="__main__": t = int(input()) while t>0: solve() t-=1 Main idea behind this code is that when we calling gcd(a,2*b) then case when "a" is odd we get large number of solution as gcd()<=1 which we don't want. So we want STOP this type of case gcd(odd,2*b). so that is why if we make array like this even,even, odd... in which all odd elements are at last. We get our solution. Order of element doesn't matter. Note: For learning purpose after you code refer solution of this profile he use such elegant code in python. You definite like his work...Happy coding.
 » 10 days ago, # |   0 my submission = https://codeforces.com/contest/1535/submission/118565398pls help me. I can't find the reason of runtime error
•  » » 9 days ago, # ^ | ← Rev. 4 →   +8 try to rewrite your cmp function like this Linkthis is a common RTE mistake you can search on CF,some blogs has been written about this
•  » » » 8 days ago, # ^ |   0 thank you
 » 9 days ago, # |   0 MikeMirzayanov, I just got a message that my solution Kal-El/118413234 for 1535C coincides with low_profile/118412998,K0000/118413793, shakeitbaby/118414205, O_BhosDiwale_ChaCha/118414232, madarakaguya1234/118414304, XENOX_GRIX/118417732, codeforcesalt11/118418351, yash_agarl_/118423400I think this is either coincidence(I used a simple 2 pointer approach for it) or the people mentioned above are indulging in cheating. I have never indulged in leaking my solution or copying someone else code (you can have a look at my profile to confirm it), and looking at the timestamps it is clear that I did not copy paste someone else code.If u look at template of my other submissions on Codeforces it is similar to my submission for 1535C but for the people mentioned above their code style is not same as their submission for 1535C. I do not know how they got access to my code or was it just a mere coincidence.I sincerely participated in the contest and it is a humble request to you to not skip my submissions for the contest.
 » 9 days ago, # |   -13 Can anyone tell why my solution for problem B giving TLEll gcd(ll a, ll b){ if (b == 0) return a; return gcd(b, a % b); }int main(){ ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); int t; cin >> t; while(t--){ ll n; cin >>n; vector v; ll a[n],h=n-1,ans=0; for(ll i=0;i>a[i]; if(a[i]%2==0){ ans+=h; h--; } else{ v.push_back(a[i]); } } for(ll i=0;i 1){ ans++; } } } cout<
•  » » 9 days ago, # ^ |   0 Store v.size() in some variable and then pass it into loops.
•  » » » 9 days ago, # ^ |   0 Thanks but then how did it passed the pretest-1
•  » » » 9 days ago, # ^ |   0 But why it does work? Any reason ?
•  » » 9 days ago, # ^ |   0
•  » » » 9 days ago, # ^ |   0 Thanks but then how did it passed the pretest-1
•  » » » » 9 days ago, # ^ |   0 Pretest-1 doesn't have a case where the number of odd numbers is zero.
•  » » 9 days ago, # ^ |   0 for(ll i=0;i
 » 9 days ago, # |   0 What data structure has the model solution used in the solve_short() function in the last problem? Is it Aho Corasick or something similar?
 » 9 days ago, # |   0 Can Anyone explain the dp solution for problem C in a little more detailed manner .
 » 9 days ago, # |   0 my submission = https://codeforces.com/contest/1535/submission/118565398pls help me. I can't find the reason of runtime error
 » 9 days ago, # |   -10 Hello guys. I have a little question. Does this python code equivalent to this C++ code?Python: res = [] for _ in range(int(input())): a = input().split() if (min(a[0], a[1]) > max(a[2],a[3])) or (max(a[0], a[1]) < min(a[2],a[3])): res.append("NO") else: res.append("YES") for i in res: print(i)C++: #include using namespace std; int main() { int t; cin >> t; while (t--) { vector s(4); for (int& x : s) cin >> x; if (min(s[0], s[1]) > max(s[2], s[3]) || max(s[0], s[1]) < min(s[2], s[3])) cout << "NO\n"; else cout << "YES\n"; } }
 » 9 days ago, # |   0 Can somebody explain how sort function is written in the code of question B?
•  » » 4 days ago, # ^ |   0 For std::sort(first,last,comparatorFunction) this is the syntax of sort function for more you can refer to this Link. In editorial 3rd parameter in sort function is lambda expression which might confused you I guess. You can refer this Link.
 » 9 days ago, # | ← Rev. 2 →   -34 I have seen others' solutions on the same logic in Problem B , which I have used,But I am getting TLE,but others didn't get Can someone tell me , why I am getting TLE?My Code:(getting TLE :( ) #include #define ll long long using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); ll t; cin>>t; while(t--) { ll n; cin>>n; ll arr[n]; ll i; ll odd=0; ll eve=0; vectoro; for(i=0;i>arr[i]; if(arr[i]&1) { odd++; o.push_back(arr[i]); } else eve++; } ll ans=eve*odd+(eve*(eve-1))/2; for(i=0;i1) ans++; } } cout< using namespace std; #define pb push_back #define ppb pop_back #define make make_pair #define FAST ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); typedef long long int ll; typedef double dd; #define test int t;cin>>t;while(t--) #define f(i,a,b) for(ll i=a;i=b;j--) #define ln cout<<"\n"; #define se second #define fi first #define vll vector using namespace std; const ll mod=1e9+7; const ll MAX=LLONG_MAX; const ll MIN=LLONG_MIN; int main() {FAST; test { ll n; cin>>n; vll v,w; ll c=0; ll a[n]; f(i,0,n) {cin>>a[i]; if(a[i]%2==0) { v.pb(a[i]); } else { w.pb(a[i]); } } for(ll i=0;i
•  » » 9 days ago, # ^ | ← Rev. 3 →   0 If o is empty, then o.size() - 1 will evaluate to 2^64 − 1 since size_t is unsigned.
 » 9 days ago, # | ← Rev. 2 →   0 For Problem D: Can someone tell me why i am getting TLE in tc 11?code : 118688525
 » 8 days ago, # |   +16 Trying to implement something "smart" in F I accidentally got Accepted with "stupid" $O(n*n)$ (or even $O(n*n*m)$) solution. After removing all unnecessary trash it has only several lines of code: 118716718In particular it uses the following code:  rep(i, n) rep(j, i) ans += ...; `where $n$ — the size of a current equivalence class.It seems C++ is too fast :) and, probably, the tests do not cover the worst case for this solution. Also please note, that all string are different, so the maximum size of an equivalence class is $200000/8 = 25000$.
 » 8 days ago, # |   0 Problem C: Are tutorial and solution the same?
 » 3 days ago, # |   0 In python 3.8 i used my own implementation for gcd (mod euler) and it got time limit exceeded using the math.gcd method it got accepted.
•  » » 28 hours ago, # ^ |   0 Have you used Euclid's algorithm?