By Vladosiya, history, 11 months ago,

1851A - Escalator Conversations

Tutorial
Solution

1851B - Parity Sort

Tutorial
Solution

1851C - Tiles Comeback

Tutorial
Solution

1851D - Prefix Permutation Sums

Idea: Gornak40, prepared: senjougaharin

Tutorial
Solution

1851E - Nastya and Potions

Tutorial
Solution

1851F - Lisa and the Martians

Idea: Gornak40, prepared: Gornak40

Tutorial
Solution

1851G - Vlad and the Mountains

Tutorial
Solution
• +96

 » 11 months ago, # |   +9 Very interesting competition! Thanks to Vladosiya and your team!
 » 11 months ago, # |   0 I am stuck at the first question itself for a while
•  » » 11 months ago, # ^ |   0 It was quite hard to read, but the idea at least was simple
 » 11 months ago, # | ← Rev. 2 →   +5 I have an idea to solve G in another way (a bit complicated). Height of vertex = value of vertex The best way from point $a$ to $b$ by traversing vertexex with minimum values. So, we can make a tree/forest (if graph is not connected) on our graph and use only simple paths to move between vertexes. We sort our vertexes by value. Maintain DSU. Iterate every edge of every vertex and add edge if vertex are in diffrerent unions.Our query is reformulated: check if maximum of value of every vertex on simple path between $a$ and $b$ is smaller or equal to $value[a]$ + $energy$. If $a$ and $b$ are in different unions then path does not exists. It could be done by finding $c = lca(a,b)$. We divided our task to find maximum on path between $a$ and $c$, and on path between $b$ and $c$. And resulting maximum will be maximum from these. To find LCA and maximum on path between $a,b$ and $lca(a,b)$ we can use binary lifting.
•  » » 11 months ago, # ^ |   0 I just got TLE in test 9 (5 sec) time complexity should be O(nlogn+qlogn) Not sure what's going wrong
•  » » » 10 months ago, # ^ |   0 Me too. Any idea?
•  » » » 10 months ago, # ^ |   0 Hey, Try adding  #pragma GCC optimize("O3,unroll-loops") #pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt") I got it accepted thru this.
 » 11 months ago, # | ← Rev. 2 →   0 I nearly got F. I know there is something to do with sorting but don't have enough time to implement it
 » 11 months ago, # |   0 Why problem E doesn't have cycles? I read E description for very long time but still don't know why :(
•  » » 11 months ago, # ^ |   +20 It is guaranteed that no potion can be obtained from itself through one or more mixing processes.
•  » » » 11 months ago, # ^ |   -17 How you take know? The reason wey I dey ask be sey most of Una dey like to talk things wey be sey no dey correct at times
•  » » » 11 months ago, # ^ |   0 For problem E, Is it possible to have the following criteria1 needs 4 and 5 for mixing & 2 needs 1 and 5 for mixing.For 1, we already have 5 included, so how will that affect the answer for 2 which has both 1 and 5. Is such a case even possible , if yes , how will the answer be calculated (adding both 1 and 5 or only 1)?
•  » » » » 11 months ago, # ^ |   +8 That is valid.Find cost of 4 and 5, now you have the cost for 1. Use the cost of 1 and 5 to get the cost for 2.
•  » » 11 months ago, # ^ |   +10 "Moreover, no potion can be obtained from itself through one or more mixing processes." here it is
•  » » 11 months ago, # ^ |   0 "It is guaranteed that no potion can be obtained from itself through one or more mixing processes." This statement means that agent i cannot be mixed directly or indirectly by Agent i(that is, itself)
•  » » » 11 months ago, # ^ |   +1 thank you, seems like I need to improve my English reading skill...
•  » » » 11 months ago, # ^ |   0 Oh man... This "one or more mixing processes" got me.I definitely need to get used to problem statements on Codeforces.
 » 11 months ago, # | ← Rev. 2 →   0 bruh my solution for C would have worked if i didnt put nums2 instead of nums1 at https://mirror.codeforces.com/contest/1851/submission/215591370working solution: https://mirror.codeforces.com/contest/1851/submission/215744363wasted :(is alright tho learnt that i always have to double check my code
 » 11 months ago, # | ← Rev. 5 →   +11 In problem F can also be understood using equation$a+b = a ⊕ b+ 2*(a$&$b)$$a = a_{i} ⊕x$$b = a_{j} ⊕x$$\large \frac{(a_{i} ⊕x) + (a_{j} ⊕x) - (a_{i} ⊕x ⊕a_{j} ⊕x)}{2} = \frac{(a_{i} ⊕x) + (a_{j} ⊕x) - (a_{i} ⊕a_{j} )}{2}$So we need to get min pair xor and $x$ which increases positive term.
•  » » 11 months ago, # ^ |   0 this is cool, but I dont really know how to find x. Then I saw a lot of people do like ((1 << k) — 1) ^ (a[i] | a[i-1]); after sorting. can anyone explain this a bit?
•  » » » 11 months ago, # ^ |   0 We know that all given numbers are strictly less than $2^k$, so the biggest possible number with the most bits (exactly k bits) set to 1 is equal to $2^k - 1$, to maximize our expression (a[i] ^ x)&(a[i-1] ^ x) we intuitively want to set to 1 only those bits in x that do not occur either in a[i] or a[i — 1] (and bits that do occur both in a[i] and a[i-1] are the result of the logical or of a[i] and a[i-1]).
•  » » » » 11 months ago, # ^ |   0 ok got it, thx
•  » » 11 months ago, # ^ |   0 i figured this out but it is giving tle. what are avx instructions that tutorial mentions??
•  » » » 8 months ago, # ^ |   0 have you got the answer bro?
•  » » 9 months ago, # ^ |   0 bro, you are the best, thanks!!
 » 11 months ago, # |   +16 G can be solved online using persistent DSU, which also allows us to solve the problem of finding the minimum energy needed to go from $a$ to $b$ in $O(\log n)$.Submission
•  » » 11 months ago, # ^ |   +8 Can you explain the algorithm you used in short please? I read about persistent DSU but I didn't see standard unite and find_set functions in your dsu and also what does the upper_bound at the end signify?
•  » » » 11 months ago, # ^ |   +8 Here is my implementation of persistent DSU:For each set, store its elements. Instead of storing the parent of each element store a vector of pairs $(time, parent)$. Since we want the length of the path from each element to its root to always be 1, when merging two sets update the parent of each element in the smaller set. This results in an amortized space and time complexity of $O(n \log n)$. Proof of complexityLet's count the contribution to the sum of each element. When a parent of an element is changed, the size of its set is at least doubled. Since the size of each set is bounded by $n$, an element can only change $\log n$ times, resulting in $O(n \log n)$ complexity.To get the root of an element at time $t$ we can simply binary search to find the last $(time, parent)$ entry with $time \le t$.
•  » » » » 11 months ago, # ^ |   0 Thanks a lot!
 » 11 months ago, # |   0 In my solution of D I have used slightly different checks to determine if the original array was a permutation.Let's assume we already know that the deleted element isn't the last one — it's either the first one or some element in the middle (same as in the editorial). For the original array to be a permutation there can't be two differences of 1 in the prefix array (we can't reduce a 1 to any other difference by inserting an element back to the prefix array). Given array of prefix sums can either have at most one difference bigger than n, or at most 2 equal differences (we need to insert back at least one element to fix either one of those, so if we have to insert more than one element, the original array isn't a permutation). I don't know how to prove that these 2 conditions are sufficient for the original array be a permutation, but apparently this is enough, my submission: 215583206
 » 11 months ago, # |   0 Someone please explain the secret solution of div2F mentioned in editorial.. what are AVX instructions ? :)
 » 11 months ago, # |   +1 unordered_map gives TLE while map worksCan someone help me understand why using unordered_map fails(TLE on case 13) but using map works in problem D. The logic in my code is to find answer based upon the occurance of a difference between adjacent elements > n and any repeated differences. I know that average time complexity for unordered_map is O(1) while in worst case it is O(n) and average for map is O(log n), however I always thought that using unordered_map is faster due to better average. What am I missing? My submission using unordered_map — 215744161 My submission using map — 215746951 Both the codes are same except the usage of map instead of unordered_map
•  » » 11 months ago, # ^ |   +8 Please refer to the classic blowing up an unordered_map
•  » » » 11 months ago, # ^ |   0 Thank you!
 » 11 months ago, # |   0 I thought editorial of D would have a better solution.
•  » » 11 months ago, # ^ |   0 You can check my submission for easier to understand and implement solution. Main idea is very similiar to the one in editorial. If last element was removed from prefix array, then we end up with permutation without 1 element, and if remove other element, then we will end up with permutation without 2 elements and a special element which should be the sum of missing elements of mentioned permutation.
 » 11 months ago, # |   0 can we simply solve e with simple array based approach. just trying to update values to 0 for potions which r unlimited and then calculate cost, and simply give the output by comparing.
•  » » 11 months ago, # ^ | ← Rev. 2 →   0 I think here, when we get some potion is unlimited by mixing other unlimited potions, then this will affect other similar potions so this would be a tree. Also, when we buy a potion, we get unlimited quantities of that potion too (I had misunderstood this during contest)
 » 11 months ago, # |   0 Is the solution 2 for F correct? Consider the case: 1 (0b000001), 31 (0b011111), and 32 (0b100000), with k = 6: Here the answer is ai=1, aj=32. If we look at a(i+1), it has a very high (non-minimal) xor with both aj and ai and the answer is not the adjacent numbers after sorting. Spoiler (my understanding)We have to minimize ai^aj, right? Or, maximize ai (bitwise equality) aj?
•  » » 11 months ago, # ^ |   +11 You're right about minimizing $a_i$ ^ $a_j$ part, but your example is wrong: 1^31 < 1^32
•  » » » 11 months ago, # ^ |   +8 Thanks, I made a very silly mistake there, blinded by maximizing the matching bits and ignoring their values
 » 11 months ago, # |   +8 good div 3!!but...i think problem A need to be easier to understand TvT
 » 11 months ago, # |   0 For problem F, I don't think solution 3 passes, as you typed O(n^2/16), or is it just a typo of the time complexity? Thanks.
 » 11 months ago, # |   +3 How can we use AVX instructions to solve 1851F?
 » 11 months ago, # | ← Rev. 2 →   -8 I feel that I have considered the d question comprehensively, but it's just not right. I'm so sad :（
 » 11 months ago, # |   +3 For problem E, since there can't be any cycles, the problem can also be solved by first doing a topological sort. And then a DP in that order, because for all $i$ we have already computed the answer for the potions it can be mixed by (namely some $j$ where $j < i$).(And great competition, thank you Vladosiya and team!)
•  » » 11 months ago, # ^ |   +8 Meanwhile you can perform a topological sort on a DAG, in case of a problem like this you don't need to. You can just traverse all the unvisited nodes with a simple dfs in a "topological sort"-like manner, compute the answer for each unvisited node based on prices of its children, memorize the answer (one of the main points of dp) and move onto the rest of the nodes.Because that's exactly what I did during the contest — wrote a topological sort on autopilot and didn't even use it afterwards (how silly of me).
•  » » » 11 months ago, # ^ |   0 can we simply solve e with simple array based approach. just trying to update values to 0 for potions which r unlimited and then calculate cost, and simply give the output by comparing.
•  » » » » 11 months ago, # ^ |   0 If by an array based approach you mean this solution: 215678317, then I don't think your approach is correct, namely, there are a couple of inaccuracies that stick out to me right away: You have multiple nested loops in your solution, which leads to at least $O(n^3 + k \cdot n^2)$ time complexity You seem to process all queries "online", i.e. you try to reduce prices for all potions after reading each new recipe I don't think you are reading the recipes in the right format Try to draw a couple of test examples on paper, look at what kind of graphs do you get and try to proceed from there.
 » 11 months ago, # | ← Rev. 4 →   -8 In F, we can actually prove that we need minimum XOR pair, let Ai = A and Aj =B and x = X (A ^ X).(B^X) = (AX' + A'X)(BX' + B'X) = ABX' + A'B'X From this, we can deduce that for whatever A and B, there exists X that can give a maximum value so, we only need to calculate the maximum value of AB + A'B' which is basically XNOR. Hence we need to calculate the minimum of XOR of all pair.
 » 11 months ago, # |   0 Solution of problem E using Trie. algorithm -> Before inserting a element into the TRIE we check the minimum xor we can form using the current element and the previous element inserted int the TRIE. We can also construct the other element and the number X(the element which satisfies the inqquality (ai⊕x)&(aj⊕x) ) while traversing the TRIE.submission Link : link
 » 11 months ago, # |   0 I need to code all 3 solutions for problem F
 » 11 months ago, # |   +4 I think F was Google-able, but I didn't have the insight to try it in the contest. Great contest! (if not a little frustrating)
 » 11 months ago, # |   0 can anyone help me in problem D?https://mirror.codeforces.com/contest/1851/submission/215772754wa in test case 3
 » 11 months ago, # | ← Rev. 3 →   +25 I tried coding this $O(n^2/16)$ solution to problem F using AVX instructions. However its taking 2.14s for 1.5e5 elements, but barely blowing the 3s TL for 2e5. Can anyone think of any optimizations?https://mirror.codeforces.com/contest/1851/submission/215775069Update: After only saving the block indices that gave the minimum (not the element indicies) and then reiterating separately though these two blocks, I managed to save just enough operations.$O(n^2/16)$ AC: https://mirror.codeforces.com/contest/1851/submission/215791929
•  » » 11 months ago, # ^ |   +5 curious to know how did you learn this AVX instructions thing and why?
•  » » » 11 months ago, # ^ | ← Rev. 2 →   +11 I dont really know AVX instructions (discovered it today).When I saw the editorial, I searched and found this blog https://mirror.codeforces.com/blog/entry/98594. After reading it, I implemented a solution using __m128 (which is $O(n^2/8)$ here). Later, googling a bit about 256 bit AVX, found more instructions and finally managed to optimize enough to pass.Why? No specific reason, guess I like playing around with these things on my spare time. It probably doesn't help me in CP though.
•  » » » » 11 months ago, # ^ |   0 Thank you for the link! You saved my day, I googled "AVL Instructions" only to find some pragma stuffs.
 » 11 months ago, # |   0 How unlucky you are-> My honest reaction: my solution= 215621385 and my submission just after the contest after just adding two brackets in if(i!=H) condition 215624902
 » 11 months ago, # |   0 My thoughts and solution for F:We can check by the truth table that we will get a plus in our final value if the $i$'th bit in $a$ and $b$ are equal (0 and 0) or (1 and 1). So, the final value is for each two values inside the array our value is: $\sum_{i=0}^{k-1} ((\text{ith bit of } a \text{ and } b \text{ equals to each other}) \ll i)$.So, the final output will be the indices of $a$ and $b$ that will give us the maximum value. We can notice that our sum is equivalent to: $2^{k} - 1 - \sum_{i=0}^{k-1} ((\text{ith bit of } a \neq \text{ith bit of } b) \ll i)$.Additionally, we can observe that $\sum_{i=0}^{k-1} ((\text{ith bit of } a \neq \text{ith bit of } b) \ll i)$ is the definition of $a \oplus b$ (the bitwise XOR operation between $a$ and $b$).So, how do we maximize our sum? We need to minimize the XOR, right? To do so, I know (from previous problems) that the minimum XOR occurs from a pair of values in the given array (but sorted).so we just do that :).sort the given array, get the minimum xor and build the X accordingly. my submission: 215782665
 » 11 months ago, # |   0 Has anyone solved F with a binary search(n log(n))?
•  » » 11 months ago, # ^ |   0 me
•  » » » 10 months ago, # ^ |   0 hi
•  » » » » 10 months ago, # ^ | ← Rev. 2 →   0 what?maybe you didn't find me in dm Because I added restriction that atleast LGMs can text me , now changed that to 'allow all'
 » 11 months ago, # |   0 can someone please explain me question D.
 » 11 months ago, # |   0 Someone please help me. This 215826166 really drives me crazy. I got tens of Runtime Error with GUN C++ 20 because of exit code: -1073741819 (STATUS_ACCESS_VIOLATION) but the same code could be Accepted with MS C++ 2017. I can't figure out why.
•  » » 11 months ago, # ^ | ← Rev. 2 →   0 I got it. It's actually a common problem in C++ when sorting if two elements are equal the result should be false!
 » 11 months ago, # |   0 Can someone explain why my soln to E 215608832 throws a runtime error on test 12
•  » » 11 months ago, # ^ | ← Rev. 2 →   0 mine too.[python]
 » 11 months ago, # |   0 I have an approach similar to solution-$2$ for Problem F but we don't need to sort. SolutionIt is obviously always betters to set higher bit even if we are unable to set any bit after that. Suppose that, the current bit we are handling is $k$ then we need only those elements in which either $k^{th}$ bit is set or unset because if $a_i$ has $k^{th}$ bit set and $a_j$ has $k^{th}$ bit unset then no matter what we do $(a_i \oplus x) \& (a_j \oplus x)$ will have that bit as zero.So, we partition our array into two arrays and solve for them independently, one having all elements in which $k^{th}$ bit is set and other having $k^{th}$ bit unset. If there are more than two elements in the "set" or "unset" array then you can set current bit in $x$ and check which path gets you the maximum answer. This can be done recursively, we need four things.Here is my solution.
 » 11 months ago, # |   0 Greedy solution for F:We need to find value = (a[i] ^ x) & (a[j] ^ x). Note that a[i] & value removes bits that do not affect the answer. Let's find value greedily starting from the highest bits. All we need to do is make sure that after adding next 1 bit, there is still at least two different i that produce a[i] & (value | (1 << u)). int value = 0; for (int u = k - 1; u >= 0; --u) { map q; for (int i = 0; i < n; ++i) { if (++q[a[i] & (value | (1 << u))] > 1) { value |= (1 << u); } } } map q; for (int i = 0; i < n; ++i) { int y = a[i] & value; if (q.count(y)) { cout << i + 1 << " " << q[y] + 1 << " " << (value ^ y) << "\n"; return; } q[y] = i; } 
•  » » 3 days ago, # ^ |   0 .ComplaintFrame { display: inline-block; position: absolute; top: 0; right: -1.4em; } .ComplaintFrame a { text-decoration: none; color: #ff8c00; opacity: 0.5; } .ComplaintFrame a:hover { opacity: 1; } ._ComplaintFrame_popup p, ._ComplaintFrame_popup button { margin-top: 1rem; } ._ComplaintFrame_popup input[type=submit] { padding: 0.25rem 2rem; } ._ComplaintFrame_popup ul { margin-top: 1em !important; margin-bottom: 1em !important; } Didn't understand your approach.. what is the variable value holding in the code snippet you shared, the logical meaning? value = (a[i] ^ x) & (a[j] ^ x) x is here is unknown, so don't think value is equal to that
 » 11 months ago, # |   0 Another opportunity to feel ashamed
 » 11 months ago, # |   0 Thanks for memorable contest first time touch pupil
 » 11 months ago, # |   +1 Anyone whose E shows runtime error on test case 12[python]
 » 11 months ago, # | ← Rev. 2 →   0 I have another solution for problem F. It relies on the fact that for any bit, if it is on or off in both the numbers we choose, than that bit will be on in the number we try to maximise. With this in mind we can define a recursive function that for some maximal suffix of bits of the number we try to maximse, keeps track of all the numbers with that suffix and which of those numbers have the next bit set to 0 and which bit set to 1 tries to turn on the next bit in the solution or just skip it if it is not possible (this happens when there are 0 or 1 numbers with the next bit being 0 and 0 or 1 numbers with the next bit being 1). This results in a solution of complexity O(nk). Here is my code: https://mirror.codeforces.com/contest/1851/submission/216012524
 » 11 months ago, # |   0 For task E, you can simply write a recursion
 » 11 months ago, # |   0 Very good contest! I learned a lot of new knowledge.
 » 11 months ago, # |   0 can anyone plz tell in problem g why cant we just simply perform dfs starting from a and check whether we are able to ever reach b or not with the necessary checks of energy .
•  » » 11 months ago, # ^ |   0 Because that would be too slow. In that case we will maybe visit all mountains in every query.
•  » » » 11 months ago, # ^ |   +1 thank you got it
 » 10 months ago, # |   0 Is there some tutorial for bitwise trie data structure from problem F? It's hard for me to understand the official solution.
 » 10 months ago, # |   0 in ques A why do we need to check if the vlag height(H) and escsalator person's height[i] is equal or not
 » 10 months ago, # |   0 in question A, I didn't understand the statement: "the difference between the extreme steps should not exceed the difference in height." why is this condition considered?
 » 10 months ago, # | ← Rev. 2 →   0 THese condition in question D is useless vector cntGt1; for (auto p: cnt) { if (p.second > 1) { cntGt1.push_back(p.first); } } if (cntGt1.size() > 1) { cout << "NO\n"; return; } if (cntGt1.size() == 1) { int x1 = cntGt1[0]; if (cnt[x1] > 2) { cout << "NO\n"; return; }  }
»
10 months ago, # |
0

Way too many edge cases for D.

My solution:-

# include

using namespace std; typedef long long ll; typedef pair<int, int> pii; typedef pair<ll, ll> pl; typedef vector vi; typedef vector vll; typedef vector vpii; typedef vector vpl; typedef vector vvi; typedef vector vvll; typedef vector vs; typedef vector<vector> vvll; typedef map<ll,bool> mllb; typedef unordered_set ucst; typedef unordered_map<ll,bool> umllb;

# define mp make_pair

const ll mod=10e9+7;

bool solve() { ll n; cin>>n; int m=n-1; vll arr(n-1); ll permSum=n*(n+1)/2; for(ll& x: arr) { cin>>x; }

for(auto x: arr) {
if(x>permSum) {
return false;
}
}

if(arr[m-1]>permSum) {
//Not a permutation
return false;
} else if(arr[m-1]<permSum) {
//Last element missing in prefix sum
arr.push_back(permSum);
unordered_map<ll,ll> mp;
for(int i=n-1;i>0;i--) {
mp[arr[i]-arr[i-1]]++;
}
mp[arr[0]]++;

for(ll i=1;i<=n;i++) {
if(mp[i]==0) {
return false;
} else if(mp[i]>1) {
return false;
}
}
return true;
} else {
//First or middle element deleted
arr.insert(arr.begin(),0);
unordered_map<ll,ll> mp;
for(ll i=n-1;i>0;i--) {
mp[arr[i]-arr[i-1]]++;
}

vector<ll> missing;
for(ll i=1;i<=n;i++) {
if(mp.find(i)==mp.end()) {
missing.push_back(i);
}
}

if(missing.size()>2) {
return false;
}

if(missing[0]+missing[1]>n && mp[missing[0]+missing[1]]==1) {
return true;
} else if(missing[0]+missing[1]<=n && mp[missing[0]+missing[1]]==2) {
return true;
} else {
return false;
}
}

}

int main() { ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0); ll t; cin>>t; //t=1; while(t--) { if(solve()) cout<<"YES"<<endl; else cout<<"NO"<<endl; //solve(); //cout<<solve()<<endl; }

return 0;

}

 » 10 months ago, # |   0 If anyone was curious, yes 1851E can be solved using modified dijkstra https://mirror.codeforces.com/contest/1851/submission/220974558
 » 9 months ago, # |   0 Hi, for problem E, why does the below code gives TLE but the tutorial solution works?223600037
 » 8 months ago, # | ← Rev. 2 →   0 You just need to do like this for prob B. Thank you for the contest but I hope you will do better editorial next time.https://pastie.io/hccsrs.cpp
 » 6 months ago, # |   0 Problem E,It is guaranteed that no potion can be obtained from itself through one or more mixing processes. but in 1st example case 3,potion 1 can be obtained from 2,4,5 , and potion 4 can be obtained from 1,5, doesn't that make 1 and 4 a circle???
•  » » 6 months ago, # ^ | ← Rev. 2 →   +12 Are you talking about this test case? 5 1 5 4 1 3 4 2 2 4 5 3 3 5 4 2 1 4 1 5 0 Read the statement again:"This is followed by $n$ lines describing ways to obtain potions by mixing.Each line starts with the integer $m_i$ ($0\le m_i < n$) — the number of potions required to mix a potion of the type $i$ ($1\le i\le n$).Then line contains $m_i$ distinct integers $e_1,e_2,\dots,e_{m_i}$ ($1\le e_j\le n, e_j\ne i)$ — the indices of potions needed to mix a potion of the type $i$. If this list is empty, then a potion of the type $i$ can only be bought."The first integer on those lines is is the number of remaining integers on that line, it's not one of the potions. Potion $1$ can be made from potions $\{2, 5\}$, potion $4$ can be made from potion $\{5\}$.
•  » » » 6 months ago, # ^ |   0 oh I thought the first digit of each line is a potion,thanks for replying!!!
 » 3 months ago, # |   0 I am Chinese.This competition is very interesting.I like it very much.It is gooder than luogu.
 » 2 weeks ago, # |   0 For F, the Python trie-based solution is very hard to make pass under the constraints (I searched a bit in AC solutions, did not find one). I did not manage to make mine pass, and ended up submitting the same algorithm in C++ to get AC.