### feecIe6418's blog

By feecIe6418, 2 months ago, Thanks for participation!

If your solution to D involves any data structures and is not $O(n)$ -- please read the "solution 1". I believe it is very interesting, but to make the difficulty suitable for D we allowed not $O(n)$ solutions.

1696A - NIT orz!

Hint 1
Solution

1696B - NIT Destroys the Universe

Hint 1
Hint 2
Solution

1696C - Fishingprince Plays With Array

Hint 1
Hint 2
Solution

1696D - Permutation Graph

This problem has two different solutions. The first one is more beautiful, but less straight-forward.

Hint 1 for solution 1
Hint 2 for solution 1
Hint 3 for solution 1
Solution 1
Hint 1 for solution 2
Hint 2 for solution 2
Solution 2

1696E - Placing Jinas

Hint 1
Hint 2
Hint 3
Solution

1696F - Tree Recovery

Hint 1
Hint 2
Solution

1696G - Fishingprince Plays With Array Again

Hint 1
Hint 2
Hint 3
Solution

1696H - Maximum Product?

Hint 1
Hint 2
Solution Tutorial of Codeforces Global Round 21 Comments (157)
 » Thank you for fast editorial!
 » fast editorial, good problemsgreat round!
 » What a speed!
 » Editorials with hints. Appreciable efforts..Thanks!
 » I hope I'm not the only one who got WA on question B for not realizing that the answer is at most 2... XD
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   .
•  » » got 4 times wa then realised .
•  » » Lol Same...
•  » » LOL ! I also did not realize it.
•  » » got WA 3 times not realizing answer will be at most 2, one more WA for a mental typo using max(ans,2) instead of min(ans,2).
 » Thanks for so much early tutorial
 » Best way to present editorial is with hints Thanks for fast and crisp editorial
•  » » Yeah, totally agreed! I love it when editorials have Hints as it allows many of us to solve problems (sometimes we just need that train of thought and a little nudge). Honestly, it's better than looking at the solution and then doing it, as there's still that feeling of 'doing it ourselves'
 » Tq for fast Editorial
 » Thank you for the fast editorial feecIe6418 Congrats on your first Global Round. It's hard to imagine that you single-handedly created all except for one problems for this round, that's amazing! Great work! FeedbackI really liked the problems. Personally I felt that the problem statements were not very lengthy (which is good) and not too hard to understand for low-rated players. Honestly, I only checked out the first three problems, but I found them really interesting. I was able to work out the solutions for the first two, sample test cases were also good in my opinion, no problems with that.Unfortunately due to a power cut in my locality, I didn't have internet access for the first hour of the contest and this decided not to participate. Looking forward to upsolving in virtual participation.Also, many congratulations once again on your first Global Round. It's hard to imagine the amount of work that goes into making these problem statements and testing them. Kudos to the team and to all testers and problem setters on the platform!!
 » Thanks for the fast editorial. By the way,E
 » Thank you for Editrorial!
 » 6 weeks ago, # | ← Rev. 2 →   Feeling sorry for Tourist. Dude is facing a lot of negative deltas in recent times. I remember reading a blog few days back, when someone predicted Tourist's fall. It's becoming real now :(
•  » » Just out of curiosity, could you link that blogpost? I'd like to read it too
•  » » Brother Legends never dies Tourist is Legendary No one takes his place
•  » » » There is nothing called legendary. If you consistently put in the effort for over a decade(He has been into mathematics and programming since age 10), you could become like him(may be better than him).
•  » » » » And you're still newbie after 3 years.
•  » » » » » Because I am not consistent :( I am trying to be consistent.
•  » » » » » And you are still unrated after 2 years
•  » » » » » I am saving the link to this comment because i want to comment here when i get a decent level of rating in CF.
•  » » » » » » and what's your definition of "decent"?
•  » » » » » » » RED
•  » » » » » » » » lol, okay.
•  » » » » » » » » » I want to give the same reply to u one day
•  » » A Lion takes its one step back before making its bigger attack
•  » » Normal competition when everyone is too good.
•  » » Share the blog . I want to see what happen to our king !!.
•  » » i Dont think its a downfall if you get -56 for global rank 7.
 » Thank you for Editrorial!
 » Solution for E without using the identity given in the editorial: Let $D_i$ be the $i$'th diagonal, the set of $(x, y)$ such that $x+y=i$. Calculate the answer by summing each diagonal. When moving from $D_i$ to $D_{i+1}$, the current number of moves is simply multiplied by $2$, except in the case where some cell in $D_i$ is a "boundary" (the cell down or to the right is black). But there are only $O(n)$ such boundary cells, so we can simply subtract their contribution as we go along.
 » Thank you for such a fast editorial , But I would like to say , talk is cheap , show me the code
 » Plz show codes for the questions
 » Are you sure memory limit will not exceed if we expand them? If not why did it happen to me?
•  » » because you do not need to store entire expanded array. you can simply store counts of repeating factors.
•  » » You shouldn't make the whole array just you can store the occurences
•  » » » can you pls explain how to store the occurrences such that they appear in the correct order as well?
•  » » » » I did it just storing a pair of values: the factors and its number of ocurrences. For example, for M = 3 and A = {6 3 4} -> {[2,3] [1,3] [4,1]} Did it for both vectors and compare them
•  » » » » 161770389 You can check here. I've just stored it as pair like if A={4,2,6} and M=2 then it will be {{2,3},{3,2}}.
•  » » » This is how I rewrote my solution after hitting the memory limit.https://codeforces.com/contest/1696/submission/161775684
 » Thanks for the fast Editorial!
 » Can someone explain why the following code for Problem C is getting RTE on Pretest 3? Codevector breakdown(ll t, ll m) { ll recur = 1; while(t%m==0) {t/=m; recur*=m;} return vector(recur, t); } int main() { setIO(); TestCase() { int n, m; cin >> n >> m; vector a; for(int i = 0; i < n; i++) { ll x; cin >> x; vector bx = breakdown(x, m); for(auto& z: bx) a.push_back(z); } int o; cin >> o; vector b; for(int i = 0; i < o; i++) { ll x; cin >> x; vector bx = breakdown(x, m); for(auto& z: bx) b.push_back(z); } cout << (a==b ? "Yes" : "No") << endl; } } 
•  » » The maximum size of the array that can be created is of order 1e5, you are pushing more values than the max size, thus giving RTE. Instead of pushing all values, just store it in pair of vectors {value, frequency}.
•  » » » Makes sense, thank you so much!
•  » » » Excuse me, why is it giving RTE and not memory error?
•  » » » » As far as I know, segmentation fault = RTE in online judges
•  » » » » » I meant how it's a segmentation fault? Like the code doesn't set any array size constraints as such
•  » » » » » » Because, after the array is completely filled, he is trying to insert more elements in it...
 » Flash like Editorial!!!
 » "Find the maximum possible value of the maximum value in a after any number (possibly zero) of operations."In Problem A this line mislead me into thinking that we have to maximize the maximum value of the original array, so I took the maximum value say "V" in the array and expected V|Z to be the answer, but after spending like an hour and half I realized that the answer would be the maximum attainable value of any value in the array, I would like to request the authors to write clearer statements to avoid such confusion, like the line mentioned above could be replaced by, "Find the maximum possible value in a after any number (possibly zero) of operations." Those who faced the same difficulty can upvote the post so that it reaches the authors. Peace.
•  » » Exact same thing happened with me!!! wasted 24 mins on a simple problem.
 » The problem F is great !
 » Easier solution for F:Firstly, notice that if you know a single edge of the tree, you can figure out all the edges. (If you know an edge u,v first all w such that dist[u][v] = dist[u][w] and add edge u,w. Then continue this process). In the end, you can verify the existence in O(n^3). Now, realize that out of (1,2), (1,3) ... (1,n), one of the edges must be present. So, try all of these n-1 possible edges and check if you get a valid tree.
 » fast editorial, good problemsgreat round!
 » D was really great, I have solved it with stack, segment tree, and bfs :).
 » Was it really a good idea to give n+1 number in array for task E ? I didn't solve it only because of this. The tasks mustn't be with such a potential trap I think.
•  » » lmao came here to comment this
 » D is really cool, I solved it with mono stack and binary search
•  » » I cant understand the second approach given in editorial . Need help
 » could someone please find the mistake in my submission for C?
•  » » seems integer overflow problem.
•  » » Int overflow
 » For Question 2 it says that it will give you the smallest non-negative integer (**zero or positive number**) which does not appear in S. For test case 0 2 3 0 1 2 0 . if i take the entire array and put it in the mex function . how will it return any non-negative number that does not appear in the array ?
•  » » if you take entire array, it will return 4. return value of mex doesn't have to appear in array.
•  » » it will give 4. so the array will be 4 4 4 4 4 4 4. then do mex again for the entire array, it will give 0 this time. hence all 0.
 » thanks for the editrorial!
 » pls give some proofs for math formulas you are using in E solution otherwise it doesnt tutorial, because your solution based only on this math fact so you have to prove it
•  » » It's just hockey stick identity. Nothing more, nothing less. I don't think one has to prove a well-known, simple identity in combinatorics in their editorial
•  » » I'll leave a link to the solution of last problem from last Starters since it also uses this identity and gives a 'proof' based on another (maybe more well-known) identity.
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   For row i : the summation is iCi + (i+1)Ci + (i+2)Ci ... + (i+v[i]-1)Ci.Now if we know iCi = 1 = (i+1)C(i+1)Replacing this in the above formula we get the summation for row i as : (i+1)C(i+1) + (i+1)Ci + (i+2)Ci ... + (i+v[i]-1)Ci. Combine the first 2 terms : (i+1)C(i+1) + (i+1)Ci = (i+2)C(i+1) [ Using the simple property nCr = (n-1)C(r-1) + (n-1)Cr ]Replacing this in the equation we get : (i+2)C(i+1) + (i+2)Ci ... + (i+v[i]-1)CiAgain now, we can combine the next 2 terms and so on. So finally we get (i+v[i])C(i+1)
•  » » » 6 weeks ago, # ^ | ← Rev. 4 →   Thinking on how to come up with a visual approach, I have found the following one. It does nothing different than what 123thunderbuddie did, but supports the idea visually.We want to find the sum of the number of ways to reach every cell in the $i^{th}$ row. Assuming there are $c$ cells, the sum is equal to the following (just what 123thunderbuddie wrote): $\displaystyle \binom{i}{i} + \binom{i+1}{i} + \cdots + \binom{i+c-1}{i}$Suppose there is a row below that has the same number of cells $c$ (not the $(i+1)^{th}$ row but a temporary one). This sum is actually equal to number of ways to reach the last cell in that row, which is $\binom{i+c}{i+1}$ (number of ways to reach cell $(i+1, c-1)$ from $(0, 0)$). Here is a visualization: » Why is this submission with long long giving WA, but this isnt.
•  » » The variable is z_cnt is not initialized in both of them resulting in UB. You got lucky with the second solution.
•  » » » Thanks :) me rn » Kudos to the authors for an amazing round. Finally reached Purple!
 » I've noticed that in G there are no small tests before the first big one, maybe that's the case in other problems too, keep in mind that it's bad.
 » I am not sure of the time complexity of this solution for D can someone help out https://codeforces.com/contest/1696/submission/161799451I feel it should be $O(n)$ because any next greater or next smaller chain is travelled once.
 » 6 weeks ago, # | ← Rev. 2 →   We can do D in O(N). By maintaining premax,premin,sufmax,sufmin. Spoilerll n; cin>>n; vector a(n); cin>>a; vector loc(n+1); for(ll i=0;i> adj(n+1); vector sufmax(n+2),sufmin(n+2); sufmax[n-1] = a[n-1]; sufmin[n-1] = a[n-1]; for(ll i=n-2;i>=0;i--){ sufmax[i] = max(sufmax[i+1],a[i]); sufmin[i] = min(sufmin[i+1],a[i]); } vector premin(n+1),premax(n+1); premin = a; premax = a; for(ll i=1;i a[i] and a[i] == premin[i]){ adj[i+1].push_back(loc[premax[i]]); adj[loc[premax[i]]].push_back(i+1); } if(sufmax[i] > a[i] and a[i] == sufmin[i]){ adj[i+1].push_back(loc[sufmax[i]]); adj[loc[sufmax[i]]].push_back(i+1); } if(sufmin[i] < a[i] and a[i] == sufmax[i]){ adj[i+1].push_back(loc[sufmin[i]]); adj[loc[sufmin[i]]].push_back(i+1); } } debug(adj); auto bfs = [&](ll x){ vector vis(n+1),dist(n+1); vis[x] = 1; dist[x] = 0; queue q; q.push(x); while(q.size()){ auto node = q.front(); q.pop(); for(auto child: adj[node]){ if(vis[child] == 0){ vis[child] = 1; dist[child] = 1 + dist[node]; q.push(child); } } } return dist[n]; }; cout << bfs(1) << endl; 
•  » » thanks for the code
•  » » » 6 weeks ago, # ^ | ← Rev. 2 →   you're welcome. Some lines are redundant you can remove them.
•  » » Doesn't this code will add multiple edges between two nodes? Though it won't affect our result.
•  » » » Yup it will i thought of making vector but i ran out of time. see my submission during contest. I clutched this question after being brutally harrassed by Problem C(nice problem) in last 3 min.
 » Thank you very much for the fast editorial! I learned a lot from the problems I didn't solve.
 » 6 weeks ago, # | ← Rev. 2 →   I solved problem D using greedy approach :) 161785869 here it is
 » Can someone explain C please
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   Just maintain vector of pairs where pairs.first = actual value and pair.second = frequency . Make vector of both a and b . And check at last a == b or not. Since we can divide every no m times even in the worst case it's log(n), which means atmost 30times for each element. which is still O(N). Code-- Spoiler#include #include #include using namespace std; #define mem(a, b) memset(a, (b), sizeof(a)) #define all(x) begin(x), end(x) #define rall(x) rbegin(x), rend(x) #define mp make_pair #define print2d(mat,n,m){for(int i=0;i void __print(const pair &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';} template void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";} void _print() {cerr << "]\n";} template void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);} #ifndef ONLINE_JUDGE #define debug(x...) cerr << "[" << #x << "] = ["; _print(x) #else #define debug(x...) #endif template istream& operator>>(istream& is, vector& v){for (auto& i : v) is >> i;return is;} template ostream& operator<<(ostream& os, vector& v){for (auto& i : v) os << i << ' ';return os;} const int N = 2e5 + 100; const ll inf = 2e18; ll mod1 = 998244353; ll mod = 1e9 + 7; const int imax = 1 << 30; // C:\mingw64\include\c++\12.1.0\bits g++ -std=c++20 stdc++.h void loop(){ ll n,m; cin>>n>>m; vector a(n); cin>>a; ll k; cin>>k; vector b(k); cin>>b; auto get = [&](vector &a) -> vector> { vector> res; ll sz = a.size(); for(ll i=0;i T; ll cnt = 1; ll cur = a[i]; while(cur % m == 0 and cur){ cnt *= m; cur /= m; } if(res.empty() or res.back().first != cur){ res.push_back({cur,cnt}); } else{ res.back().second += cnt; } } return res; }; vector> A = get(a); vector> B = get(b); cout << (A == B ? "YES" : "NO") << "\n"; } int main(){ ios_base::sync_with_stdio(false);cin.tie(nullptr); cout.tie(nullptr); ll t = 1; cin>>t; for(ll i=0;i
•  » » » Nice username name, totally relatable ;)
 » Another solution for F: Let $S(u)$ be a set of pairs $(x, y)$ such that $d(x, u) = d(y, u)$. Notice that $u$ with minimal number of elements in $S(u)$ is a leaf and the vertex connected to it is a vertex $v$ such that $S(v)$ contains $S(u)$ and for every pair $(x, y)$ such that $(x, y)$ belongs to $S(v)$ and doesn't belong to $S(u)$ either $x = u$ or $y = u$. We can remove that leaf and continue the process.
•  » » Notice that u with minimal number of elements in S(u) is a leaf. Do you know how to prove this? It does make sense intuitively.
 » 6 weeks ago, # | ← Rev. 2 →   Is it just me or are the codes not attached?
 » https://codeforces.com/contest/1696/submission/161805451 Can someone help me figure out why my problem C code is giving WA on test 9?
 » Hope I'm not the only one who overcomplicated D with sparse tables, sets and binary searches.
 » 6 weeks ago, # | ← Rev. 3 →   I was dazzled when I found the "more beautiful but less straightforward" solution to problem D. It was perfectly splendid! orz
 » Can any one tell why in Problem C Why my Map based solution does not work but it work by using vector>
•  » » order matters, in map is ordered by key in increasing order so.
•  » » » Okk Thanks for this
 » Any other guy who got wrong answer on problem B untill realising answer can be atmost 2 !!
 » "EDITRORIAL" of Codeforces Global Round 21. Lol :D
 » 6 weeks ago, # | ← Rev. 3 →   I have another solution for F. It was very painful to implement, though, so I couldn't finish it during the contest. Maybe some of you had a similar thought.First of all, if x and y are even distance apart, then there must be a valid z that's equidistant from them (center of their path). Also, if x and y are an odd distance apart, then no valid z will occur, clearly. Therefore we can create a graph where there is an edge between x and y if there is a valid such z. This graph must be bipartite, so we will be able to split vertices into 2 categories.Now we will solve for the adjacent vertices of every vertex. After this, it will be pretty easy to construct a solution (and check if it is false) as we can just add in all the apparent edges. Consider the vertex v. Now, for all other vertices, we can form equivalence groups by their distance to v. Then, for each of these groups, we can check if every 2 vertices of that group are equidistant. If so, then clearly every one of these vertices is in a different component of the "graph without v". Now the size of the largest of those groups must be the degree of v, as each outgoing edge of v signifies one connected component on the "graph without v". Of course, we still are not able to pinpoint which one of these groups contains the vertices distance 1 around v. Let's say there are m such groups (groups where every 2 vertices are equidistant from each other and every vertex is equidistant from v with the largest possible size). Then they must be "circles" around v from distance 1 to m. We are trying to find the group of distance 1. This is where the bipartite property comes in: consider any point p from a group k distance away from v. If p and v are in different bipartite categories, then they are at an odd distance away from each other. Otherwise, they are even distanced away, and there will be exactly one single vertex z that is equidistant from them both, which will be in the middle of the path between p and v. This z will clearly be in the equivalence group with distance k/2 away from v. let's call this operation (obtaining a new vertex with k/2 distance) splitting. Now consider the equivalence group that splits the most. It is clear that if $k=2^ba$( a odd), then b will be the times it splits. Then, it is obvious that the largest power of 2 less than or equal to k will satisfy this property. The important part is that a vertex with a power-of-2 distance will eventually split into a vertex of distance 1 from v, and so we can obtain all the other vertices of distance 1 around v, and we are done.I believe this method as described takes $O(n^3\log{n})$ time, and it could be simplified to $O(n^3)$
 » In Q2) Won't a case like this 0 2 3 0 4 0 5 0 return 3 as answer since there are three subarrays which need to be turned into zero?
•  » » In this Optimal Move would be to take whole subarray and calculate its MEX and replace it with all elements in the array then second move would be converting all array value zeroIn your Example 0 2 3 0 4 0 5 0 We first find out the mex of whole array which is 1 so array will be 1 1 1 1 1 1 1 1 then in second operation we will convert the above array to 0 0 0 0 0 0 0 0 so no more than two ways are possible
•  » » » Thanks now I get it
•  » » » For 0 2 3 0 4 0 5 0 ans is 1 how ?
•  » » » » No ans is not one in this case beacause we will take min(2,count of total continuous non zero element) we need maximum two operation only min(2,3) will be 2
 » Here is my solution to problem D:I have a recursive function that takes two-parameter $(l, r)$ positions of the starting and ending index in an array and calculates the optimal answer for this range.For the optimal answer, I first find the index of the minimum element and maximum element in the range $(l, r)$ and store them in $s$, $e$. $(s < e)$. I am using a segment tree and a map to find the index of the minimum and maximum in the range $(l, r)$. It is optimal to have an edge between $(s, e)$, so we add one edge.So, The answer for the range $(l, r)$ is $1$ + the answer for $(l, s)$ + the answer for $(e, r)$, and the base case is that if $l$ == $r$, there is no need to add an edge, so return $0$.
 » Here is another solution for F:Assume that for some two nodes $u$ and $v$, we know there exist an edge between them in the solution.From this, we can find all vertices $w$ such that $dist(u, v) = dist(v, w)$, then the solution must also contain the edge $(v, w)$.Repeating that process, we can find the solution (if exists) in a bfs-like way.Therefore, for each $x$ from $2$ to $n$ we can suppose that the edge $(1, x)$ exists in the solution, try to find that solution, and see if it is valid.
 » Great round. I especially liked that nothing between A and E is too implementation-heavy.
 » Damn my $O(n \log^2n)$ solution for D managed to pass. :)
 » 6 weeks ago, # | ← Rev. 2 →   Minor issue: blog title has a typo (Editorial not editrorial)feecIe6418
•  » » There r quite a lot of typos .
 » Here is $O(n^3)$ solution to F. First we can find the farthest distance from any vertex $x$ to other. So we know the diameter of the tree. If we root the tree at the center, we can know the depth of every vertex. So we can determine at least one edge in the tree (node with depth 1 and the root), then use the method in the editorial. It also shows there is at most one valid answer.
 » That's quick.
 » I think the contest has too much observation.
•  » » As is literally every codeforces round including div 3, div.2, educational, div1+div2 combined, div1, olympiad-based rounds?(yes I know that div.4 exists but I don't have a very fond experience in div.4)
 » a bit of pity for Dpass pretest make me think it works anywaybut didn't come up with cases like 1 100000 2 99999 3 99998 4 99997 ...then got fsttry to consider more next time. or maybe always need to try to achieve as low complexity as i canit's actually a great problem though, and solution 1 is awesome.
•  » » why so many downvotes
 » For D, what if the maximum element is 1 or n?
 » This time i screwed up my time on Div2C using Map to store the counts and matching them (which is not possible) because it leads to a very bigger value which cannot be stored. Using two pointers is a saviour
 » Could someone please help me to find me the error in my code of problem D I am using approach 2 https://codeforces.com/contest/1696/submission/161834104
•  » » Take a look at Ticket 14387 from CF Stress for a counter example.
•  » » » Ohh yes Got it Thanks!!
 » 6 weeks ago, # | ← Rev. 2 →   In problem G, when I set const double INF = 1.0 / 0.0; It got TLE on test 76 (or Pretest Passed 5880ms, and TLE on test 7 if I unroll the loops in matrix multiplication myself) with GNU C++14 (32bit), but ~850ms with GNU C++20 (64bit, winlib).However, when I set const double INF = 1e18; Its speed remains unchanged with GNU C++20 (64bit, winlib), but when compiled by GNU C++14 (32bit), it works in <500ms and becomes the fastest submission.Is there any logic behind this (about inf in double)?
•  » » 6 weeks ago, # ^ | ← Rev. 3 →   An explanation for the first (but not very informative) is that diving by zero (even with floating numbers) is undefined behavior, (according to [expr.mul.4], or discussions in StackOverflow), and anything can happen with undefined behavior.
•  » » 6 weeks ago, # ^ | ← Rev. 2 →   a better suggestion might be using std::numeric_limits::infinity() as the value for inf. this is not undefined behaviour and still compares bigger than the maximum value representable with double.
 » 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).
 » great tutorial
 » Can you please brief me, if the test case would 5 2 0 3 0 5 how the answer would be 2 or less explaining the l and r for each operation?I didn't understand...sorry :(
•  » » in first operation,you can take l=1 and r=6(basically the entire array),so the new array will be 1,1,1,1,1,1. In second operation,again select l=1 and r=6,the array will now be 0,0,0,0,0,0.
•  » » » For a collection of integers S, define mex(S) as the smallest non-negative integer that does not appear in S....Then it means I can take any smallest non negative integer? it's not necessary that it should be less than all the elements of the array right?
•  » » » » No it's not necessary.But you cannot just take "any" smallest non negative integer. for example-Mex(5,2,0,3,0,5) will be 1 because 1 is the smallest non negative integer which does not appear in this array.
•  » » » » » thank you so much... <3
•  » » » » » But how number of operation for (5,2,0,3,0,5) is 1.
 » can anyone please explain the proof for problem c with intution by taking an example? i couldnt understand editorial
•  » » Try expanding $a$ and $b$ into $a^\prime$ and $b^\prime$ in the example test cases.
•  » » » i have expanded both the arrays upto maximum extent as possible but im not getting proof/intution.
•  » » » » Now notice that it is possible to obtain $a^\prime$ from $a$ after some operations, and it is possible to obtain $b$ from $b^\prime$ after some operations, so if $a^\prime = b^\prime$, we can obtain $b$ from $a$.We still need to prove that if $a^\prime \ne b^\prime$, it is impossible to obtain $b$ from $a$ after any number of operations.To do so, note that the fully expanded form of $a$ does not change when you perform one of the operations on it.
 » shit contest. downvoted
 » What does it mean by "z will always become a submask of itself after any number of operations, so ai will always be a submask of (ai or z) after any number of operations"? Please explain.
•  » » Basically, $z$ will only lose bits.So $(a_i\mathrm{\ or\ } z^\prime) \le (a_i\mathrm{\ or\ } z)$, where $z^\prime$ is $z$ after some operations.
•  » » » Okay, but what is the meaning of submask here specifically. Is it like changing the set bits of z after any number of operations so the resulting bit representation i.e. bit mask obtained is called submask?
•  » » » » $z^\prime$ is a submask of the bitmask $z$, since every bit that is set in $z^\prime$ is also set in $z$, but not necessarily vice versa.
•  » » » » » It means new bitmask obtained after changing only the set bits of a bitmask is known as submask of that bitmask. Am I right?
•  » » » » » » 6 weeks ago, # ^ | ← Rev. 2 →   Sounds right
 » the explanation for problem c is very good.
 » Find Solve problems for C ,orz,%%%，thans
 » 6 weeks ago, # | ← Rev. 3 →   Why this code is working fine for question A? for(int i = 0; i < a.size(); i++){ cin >> a[i]; a[i] = a[i] | z; z = a[i] & z; ans = max(ans, a[i]); } Here I am performing both the operations — bitwise or and bitwise and. Also why it is not needed to perform second operation i.e bitwise as given in tutorial, any proof?
•  » » You are doing:  x = (a[i] | z) y = x & z z = y Every bit in z that is 1 is also 1 in x, so (x & z) equals z. In other words, you aren't changing the value of z and the second operation is doing nothing.
 » 6 weeks ago, # | ← Rev. 2 →   Have a solution different from the tutorial for problem D which does not use segment tree or stacks, very difficult for me to explain how it works but the following ideas might help-$1.$ $Its$ $always$ $optimal$ $to$ $jump$ $to$ $the$ $farthest$ $index$ $from$ $current$ $index.$$2.$ $We$ $always$ $jump$ $from$ $a$ $local$ $maxima$ $to$ $a$ $local$ $minima,$ $and$ $then$ $to$ $a$ $local$ $maxima$ $and$ $so$ $on.$161975222
 » 6 weeks ago, # | ← Rev. 2 →   For F, for each $i$, one can assume that edge $(1, i)$ is in the tree, and then try to construct the rest of the tree, as per hint 2. We can then check if the tree is valid.At the start, for each node $v$, we can do a $O(n^2 \alpha(n))$ precomputation of the equivalence classes of nodes equidistant from $v$. Then, we can do each tree reconstruction and check in $O(n^2)$, for a total complexity of $O(n^3 \alpha(n))$.
 » 6 weeks ago, # | ← Rev. 2 →   -
 » Can someone explain the equation given in Problem E?
 » Hi, could someone take a look at why I'm getting WA on test case 3 of F? My submission is here.Specifically, I create equivalence classes for each node; each equivalence class represents equal distance from said node. This is done using DSU's, which should make the whole thing have complexity $O(n^3)$. Thanks!
 » I got a solution for F which solves it in $O(n^3)$Well-known Theorem: there is atmost two centres in a tree (a center is a vertex with minimum value of maximum distance of a vertex from the given vertex: this value is called eccentricity).Note that using the given information, we can find the eccentricity of a vertex: this is just the number of equivalence classes of vertices wrt any given vertex (two vertices belong to the same equivalence class wrt a given vertex if they have the same distance from that given vertex).So we can easily find the centre(s) of the tree. If there are two centers, there is a theorem that they must be connected, so we find an edge immediately and proceed to construct the tree.If there is one center, then we can prove that all adjacent vertices have eccentricity of exactly one greater than the center, and all other non-adjacent vertices have eccentricity $\geq e+2$, where $e$ is eccentricity of the center. So in this case, we can find any node with $e^\prime = e+1$ and hence, these two must be connected, and we have found an edge.