### vovuh's blog

By vovuh, history, 3 years ago,

1296A - Array with Odd Sum

Idea: vovuh

Tutorial
Solution

Idea: vovuh

Tutorial
Solution

1296C - Yet Another Walking Robot

Idea: MikeMirzayanov

Tutorial
Solution

1296D - Fight with Monsters

Inspiration: 300iq, idea: vovuh

Tutorial
Solution

1296E1 - String Coloring (easy version)

Idea: MikeMirzayanov

Tutorial
Solution (dp)
Solution (greedy)

1296E2 - String Coloring (hard version)

Idea: MikeMirzayanov

Tutorial
Solution

1296F - Berland Beauty

Idea: MikeMirzayanov

Tutorial
Solution
• +97

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 » 3 years ago, # |   +39 It was really nice round. Wish everyone's rating to increase.
 » 3 years ago, # |   0 I can solve F like this: for every edge E find the maximum number that appears in the pathes that contain E and set that number for e. After all check if there is a contradiction the answer is -1, otherwise output the edge numbers. Its complexity is O(n*m). Does anybody has any better solution?
•  » » 3 years ago, # ^ |   +2 I used a maximum matching algorithm between the queries (A set) and the edges of the original graph (B set). In the flow network there is an edge with capacity 1 between an element of the first set and the second set if the value of that query is greater or equal than the minimum value that an edge from the B set must have.After running the maximum matching algorithm, assign the values of the queries matched to the elements of the B set and check if the resulting graph doesn't have any contradiction.Using Hopcroft-Karp the complexity is around the same as you said.
•  » » » 3 years ago, # ^ |   +13 What wrong with the community? Although the comment's approach may be (overly) complicated, it offers a new perspective to the problem. We should point out what makes this approach bad or what can be done to improve. Even you cannot understand it, you can ask for clarification. What's the point for just down-voting without making any comments?Wanna ask why your approach's complexity is around $O(nm)$. I assume you are building a flow network with $n \cdot m$ edges, and the complexity of running Hopcroft-Karp in your network should be $O(nm*\sqrt{n+m})$. Also, did you do anything to keep your constants low?IMO the observation 'the minimum value that an edge can have is a valid config if and only if there exists a valid config' is not that intuitive. The proof is not that straight forward either. What the comment writer did is to use maximum matching to construct an answer that would be 'more probably' to be correct. Anyway, for me personally, I would not code a flow for that, at least in a div 3 contest.Sorry for you getting so many downvotes. :C
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   +5 Oh... I mistook the square root on the edges not the nodes so I thought it was around the same complexity. Anyway, this solution uses a lot of memory, I had to reduce my variables to short to pass.
•  » » » » » 3 years ago, # ^ |   +5 I also thought that a match would not occur in two possible situations: when the query on set A is the result of a combination of other queries paths or when it's really impossible to set the edge as a value and not contradict other queries. So at the end I would check if all paths do not contradict which would also check if the queries that were combination are consistent with the values.
•  » » 3 years ago, # ^ | ← Rev. 2 →   +12 Okay. At first set all edges to $10^6$ Let's iterate over paths in non-decreasing order of $g_i$. Now notice that we can simply set $g_i$ to all edges in the current path to make tree correct for the current path. After making this with all queries find the minimum on each path and compare it with the needed one. Now we have queries of 2 types: Set some value on the path Get minimum on path This can be done using Heavy-Light Decomposition and a segment tree in $O(mlog^2n)$ :D
•  » » » 3 years ago, # ^ |   0 Great solution, can you pls tell me why should we go only in non-decreasing order of gi ?
•  » » » » 3 years ago, # ^ |   +1 You notice that for every query, all of the edges on that query must be at least the number given. Therefore, when we go in non-decreasing order, we never violate any previously processed queries unless the whole path of the original query is covered. This allows us to process each query only once in O(log^2 n) with HLD + Segment Tree as mentioned above, giving a total complexity of O(mlog^2 n)
•  » » » » 3 years ago, # ^ |   +1 If the explanation above isn't enough, you can read my explanation here
•  » » 3 years ago, # ^ |   0 Could you please explain how to check if there is a contradiction and what's the time complexity of it?
•  » » » 3 years ago, # ^ |   0 Do you mean checking if the answer is $-1$? You have an information that the minimum on the path between $a_i$ and $b_i$ is $g_i$. Heavy-light decomposition can answer query $minimum$ $on$ $the$ $path$ in $O(nlog^2n)$ time using a segment tree. To check if our answer is correct just iterate over all given paths and check if the minimum on the $i$th path is equal to $g_i$. If not, the answer is $-1$
•  » » » » 3 years ago, # ^ |   0 but the cfmaster guy said the complexity is $O(mn)$, so idk how he did it, or maybe he just made a mistake?
•  » » 3 years ago, # ^ |   0 Now I get you
•  » » 3 years ago, # ^ |   0 i used this solution too, but i can't prove that true.
 » 3 years ago, # |   +9 For F, Can anyone tell me any optimization in my code!Time limit exceeded on test 268 :( https://codeforces.com/contest/1296/submission/70317535Thanks!!
•  » » 3 years ago, # ^ |   +6 probably your solution runs in $O(n^2logn)$ which is a bit high if your constant factors in time complexity are large. Using unordered_map + a good hash function will pass the tests.
•  » » » 3 years ago, # ^ |   0 actually my solutions has the complexity that you said and i didnt use unordered_map and it got AC !
•  » » » 3 years ago, # ^ |   +1 Yeah, I was using Maps for no reason while I could have used 2D array. I changed that and now its working.
•  » » » » 3 years ago, # ^ |   0 To store the edges you could just store the edge that goes from v to i that v is parent of i as f[i]
 » 3 years ago, # |   +1 Thanks for this round! Waiting for another Div.3 rounds :)
 » 3 years ago, # |   +33 We have a direct formula in B, that is (s-1)/9+s;.
•  » » 3 years ago, # ^ |   -7 snorkel can you give a proof for this claim?
•  » » 3 years ago, # ^ |   +1 can you explain the formula please?
 » 3 years ago, # | ← Rev. 2 →   +10 Well this is a bit off-topic ; but is anyone else also experiencing some un-expected indentations in their codes in codeforces submissions ?As for example : code 1 — https://codeforces.com/contest/1296/submission/70274974code 2 — https://ideone.com/126Qo1These are exact same codes ; just codeforces does some un-usual indentationsExample 2 : code 1 — https://codeforces.com/contest/1296/submission/70317304code 2 — https://ideone.com/DtpiMqAny take on this ? maybe MikeMirzayanov ?
•  » » 3 years ago, # ^ |   +17 The problem is that the width of tabs on codeforces are 8 while on ideone are translated into 4, and your editor uses a mixture of spaces and tabs.
•  » » » 3 years ago, # ^ |   +1 But then if this is the case ; then I think this should have happened with every single code every single time. But actually this problem never occurred before. I am just seeing it now.
•  » » » 3 years ago, # ^ |   0 Do you know how to fix the problem?
•  » » » » 3 years ago, # ^ |   +5 The key is consistency.You may either try to replace all occurences of $4$ consecutive spaces to tabs, which will look the same on ideone. On Codeforces it will look much wider (tabstop size = $8$).ORYou may convert all tabs to $4$ consecutive spaces. It will look the same both on ideone and Codeforces.:)
 » 3 years ago, # |   0 Do basically in D, we could kill monsters in any order?
•  » » 3 years ago, # ^ |   0 You must kill monsters in given order, but the final answer has nothing to do with the order. After the sort() you could know which monsters to kill(by using secret technique) to get the maximum answer, then you could kill them in the previous order and the answer will not change.
•  » » » 3 years ago, # ^ |   0 I did not understand why the order doesn't play a role?
•  » » » » 3 years ago, # ^ |   0 It is because we can let our opponent kill the target we don't want to kill and kill only the ones which our greedy strategy provided......that's the reason why order doesn't matter because we can manipulate our opponent as we wish making him kill the unwanted enemies...cruel XD
 » 3 years ago, # |   +1 I used Bubble Sort to solve E1 ->70281553 and I wonder if it's correct.By the way, What are A and L represent in E1's tutorial Thanks qwq
•  » » 3 years ago, # ^ |   +2 alphabet (AL) size, i.e. 26 (a, b, ...., z)
•  » » » 3 years ago, # ^ |   0 oh thanks
•  » » 3 years ago, # ^ |   +1 ya that will work, you can check my solution for reference
 » 3 years ago, # |   0 Hello.I got WA with my C's submission and i don't even know what is the problem. Is there any body can help me indicate where i should fix? Please. 70292288
•  » » 3 years ago, # ^ |   0 omg.nevermind.
 » 3 years ago, # |   0 DSU solution for E1: https://codeforces.com/contest/1296/submission/70272613If there exists i, j that $s_i > s_j$ and $i < j$, there must be $c_i \neq c_j$ , in other words, $c_{i, 0}$ always exists with $c_{j, 1}$ and $c_{i, 1}$ always exists with $c_{j, 0}$. $s_i$ is the i-th char. $c_i$ is the color of the i-th char. We use dsu to union $c_{i, 0}$ and $c_{j, 1}$, $c_{i, 1}$ and $c_{j, 0}$. Before union if we find $c_{i, 0}$ and $c_{j, 1}$ is already united, we print "NO".Then, we just color 0 or 1 for every char. We color i-th char into 0, and color any char united with i for every i-th. This can be implemented in O(n), although my submission is O(n^2).
 » 3 years ago, # |   0 E2 Hard Version I wonder if the solution it to calculate the longest decreasing subsequence? Not the least decreasing subsequence? Is it right?
•  » » 3 years ago, # ^ |   +1 I think so, too.
 » 3 years ago, # | ← Rev. 5 →   +8 There is an $O(1)$ per test case solution for B.If $9$ is a factor of $s$, the answer is $\lfloor \frac{s}{9} \rfloor \times 10 +s\bmod 9-1$.Otherwise, the answer is $\lfloor \frac{s}{9} \rfloor \times 10+s\bmod 9$.
•  » » 3 years ago, # ^ |   +7 In fact, answer is $s + (s - 1) / 9$
 » 3 years ago, # |   +6 I've found few Guys Rampantly Cheating during Contests. Can Anyone please guide me, how to take this forward and make sure that this is seen by MikeMirzayanov , and there accounts are banned!Manthan_144andrajan_ladaniHere are the Submissions they did in Codeforces Round 617 (Div. 3), and it clearly Shows that they are Big Cheats!The Submission Of E1/E2 is clearly copied, and several redundant while loops have been added in order to escape from the copy-checker mechanism!Apart From this they have also copied A,B,C,D Guys please dont be cheats, and respect the community!
 » 3 years ago, # | ← Rev. 2 →   +9 [deleted]
•  » » 3 years ago, # ^ |   0 70268224 Like this solution.
•  » » 3 years ago, # ^ | ← Rev. 3 →   +1 [deleted]
•  » » » 3 years ago, # ^ |   0 This solution does not work because persistant segment tree couldn't do it either... Sorry for all my nonsense here.
•  » » » » 3 years ago, # ^ |   +8 Just wondering, since all you need to do is set_max on a path, why not just use a straightforward LCT to do that.
•  » » » » » 3 years ago, # ^ |   0 Well, stupid me.
 » 3 years ago, # | ← Rev. 5 →   0 In Easy version tutorial, it said, "Let's go from left to right and carry two sequences s1 and s2. If the current character is not less than the last character of s1 then let's append it to s1, otherwise, if this character is not less than the last character of s2 then append it to s2, otherwise the answer is "NO".While in Hard version, I also see the solution like find the minimum position i which satisfies the current character is not less than the last character of s_i by binary search, and append it to s_i. If not found, create a new sequence s_cnt and append it to s_cnt. How to prove it with Dilworth theorem?
 » 3 years ago, # |   0 In problem D, the number of secret technique needed can be simply calculated by (h-1)%(a+b)/a
•  » » 3 years ago, # ^ |   0 Could you please explain how did you arrive at this. Thank you!
•  » » » 3 years ago, # ^ | ← Rev. 2 →   0 It's similar to the original editorial, the only difference is that I used (h-1). Here's the reason: if $1\leq h\leq a$, we don't need to use secret technique, but if $h=a$, $h\bmod a=1$, so if we use (h-1)%a, that issue gets fixed.
•  » » » » 3 years ago, # ^ |   0 Im having trouble understanding of editorials solution too :(. I wanted you to explain how does this formula work.
•  » » » » » 3 years ago, # ^ | ← Rev. 2 →   0 Alright, so the module works because the the part of hp that greater than $a+b$ can be removed by attacks and it doesn't affect the result. Now the hp is $(h-1)\bmod (a+b)$ and we need attack (h-1)%(a+b)/a+1 times to kill the monster so we need to use secret technique (h-1)%(a+b)/a time.
 » 3 years ago, # |   0 Hey, so for problem C, I have thought of an alternative solution where, I have made prefix arrays of l,r,u and d. Now, since the robot ends at the same location in a substring, I just need to see substrings of powers of 2(as only then he will be back at the same location) from 1 to 18(according to constraints) in the whole array. So if at any time, delta(l)==delta(r) and delta(u)==delta(d). Those indeces are my answer.However, I am getting WA on a test case whereas I cant find any testcase to contradict. All the non-truncated test cases are passing. Please help: https://codeforces.com/contest/1296/submission/70330963
•  » » 3 years ago, # ^ |   0 Its possible that the optimal answer's length is not a power of 2. Consider RRUULULDDD
 » 3 years ago, # | ← Rev. 3 →   +19 For E1, there is a O(N) dp solution 70333509
•  » » 3 years ago, # ^ |   0 please explain as well
•  » » » 3 years ago, # ^ | ← Rev. 2 →   +16 dp[i] = if I split a[0], a[1], ..., a[i] into two non-decreasing subarrays, what is the minimum possible value of last element of the subarray other than the one a[i] belongs to.pre[N] and pre1[N] are just recording how the dp transit for restoration.I hope that you can understand my poor English.
 » 3 years ago, # |   0 I still don't understand the parity stuff in A. Someone please help a noobie
•  » » 3 years ago, # ^ |   0 Let me tell you what i have done for AFor NO: Two cases possible: 1)all numbers are even 2)all numbers are odd and there are even numbers In both cases sum is even and we also can't change the parity(even to odd) as all numbers in both cases have same parity.For YES: So now we have at least one even and one odd number(the case with all numbers are odd and they are odd in number has sum always odd) and we can change every odd number to even except one in this way we can get odd sum always.(one odd and remaining even)
•  » » » 3 years ago, # ^ |   0 Well I am completely embarrassed, but i think that I did't understand the problem statement even now. I thought that the question was too simple like "YES" for odd sum and "NO" for even sum.Specially I didn't understand the this line of the problem "In one move, you can choose two indices 1≤i,j≤n such that i≠j and set ai:=aj. You can perform such moves any number of times (possibly, zero). You can choose different indices in different operations. The operation := is the operation of assignment (i.e. you choose i and j and replace ai with aj)."
•  » » » » 3 years ago, # ^ |   0 Well, the array has an odd sum of elements if the number of odd elements is odd. To know this we count them.If this count is not odd we would like to make it odd.To make it odd, we can "overwrite" an even element by an odd one (or vice versa), which is "one move". We can make such a move only if there exists an even element, and an odd one.
»
3 years ago, # |
-8

# include<stdio.h>

void main() { int t, i, n, j, a, t1=0; scanf("%d", &t); for (i = 0; i < t; i++) { t1 = 0; scanf("%d", &n); for (j = 0; j < n; j++) { scanf("%d", &a); if (a % 2 == 1) t1++; } if (t1 % 2 == 1) printf("YES\n"); else printf("NO\n");

}


} THANK YOU;

•  » » 3 years ago, # ^ |   +8 Perhaps it is caused by "void main()" which usually is "int main()".It seems strange that you get the runtime error after printing the last result.Please note that is is usually more usefull to post the link to the submission instead of the code itself, like 70333640
•  » » » 3 years ago, # ^ |   +1 Thanks.Your suggest is truely right.But this code is wrong.In my first edition , the answer is right.I feel sorry about to post my code.I haven't expert how to post link.sorry
•  » » » » 3 years ago, # ^ |   +8 On top of the edit area there are some symbols. Click them, they won't bite you ;)
•  » » » » » 3 years ago, # ^ |   0 hhh,thanks
 » 3 years ago, # | ← Rev. 3 →   +5 It's OK to solve F in $\mathcal{O}(n\log^2 n)$ with Segment Tree Beats! (introduced here: https://codeforces.com/blog/entry/57319) and HLD. But it's rather long...Maybe set or other DS can solve it, too.My code: 70340529
•  » » 3 years ago, # ^ |   0 You can iterate over paths offline in non-decreasing order of $g_i$, so the STB isn't needed ;)
•  » » » 3 years ago, # ^ |   0 Well... I can't manage to process the situation when an edge is re-coloring without STB, would you please explain it more clearly? Many thanks.
•  » » » » 3 years ago, # ^ | ← Rev. 3 →   0 For what do you need STB? For each path, you do such thing: set all edges on this path to at least $g_i$. To do this, you use HLD + STB, am I right? What if we sort all paths in non-decreasing order of $g_i$? Now to set all edges to at least $g_i$ we can assign $g_i$ to each edge. Why? Let's take a look at 2 paths with numbers $i$ and $j$, $j < i$ and $g_j > g_i$. They have a path $l_1 - l_2$ in the intersection. Consider an edge $e$ in this intersection. After proceeding the path $j$, $f_e \geq g_j$. To make $f_e$ correct for the path $i$, we have to make $f_e := max(f_e, g_i)$. But what if we proceed path $i$ before path $j$? After the path $i$, $f_e \geq g_i$. $g_i < g_j$ so $f_e := max(f_e, g_j)$. If we sort all paths in non-decreasing order, for each $t$ ($t < j$), $g_t \leq g_j$. Thus $f_e := max(f_e, g_j)$ is equal to $f_e := g_j$, where $:=$ is an assignment operator. This leads us to a solution: assign $g_i$ to each edge for each path, iterating through paths in non-decreasing order of $g_i$. This is a HLD + standart Segment tree.
•  » » » » » 3 years ago, # ^ |   0 As this example shows: the tree is a chain, let its length equals to 4. 1<->2<->3<->4 And the coloring order is: 2 3 2 1 4 3 Firstly, we color the path 2 to 3 with 2. When we color the path 1 to 4, we can't color the path 2 to 3 with 3, but in HLD we can't iterate each edge on path. However we change the order we iterate, it can't avoid the problem. So it should use STB I think.Anyway, STB works seems quite fast (as fast as standard segment tree, though it's looooooong). Thanks for your reply :D
•  » » » » » » 3 years ago, # ^ |   0 Are you sure that your example's answer is not $-1$? My code will paint edge 2-3 with 3 after the second query. After all it will see, that our answer is not correct (min on the path 2-3 is equal to 3) and print $-1$
•  » » » » » » » 3 years ago, # ^ |   0 Oops sorry. I puts a wrong index. 2 4 2 1 4 3 3, 2, 2 is a valid solution.When we color the path 1->4, we only color edge 1->2, but when doing HLD, we will color all the edge 1->4 with color 3.
•  » » » » » » » » 3 years ago, # ^ |   0 3, 2, 2 is invalid :D. Minimum on the path 1-4 is 2, but not 3
•  » » » » » » » » » 3 years ago, # ^ | ← Rev. 2 →   0 Ahhhhh I'm almost dizzy...I try to implement your idea, but met many problems and can't pass the sample 2... You can check my code here.This one can't pass sample 3...
•  » » » » » » » » » 3 years ago, # ^ |   0 Ethylene, you have to iterate over all queries once again to check if the ans is -1. Do not do this in the same for
•  » » » » » » » » » 3 years ago, # ^ |   +1 70606132 passed.Really thanks for your help :D
 » 3 years ago, # | ← Rev. 2 →   -7 For E1/E2, I simply used the strategy that when we consider elements from left to right, we need to "sink" each element into the correct position, which requires the current element to be a different color than any previous elements that are greater than it. We then set this color in both the answer and in another array that indicates the maximum color used for each letter. Any subsequent letters less than this letter must be a new color.This reduces the problem to range-maximum and point update for an array of size $AL$. If we had to sort integers instead of characters, a segment tree can be used instead, but due to the small size of the alphabet one can just do this naively on a plain array.
 » 3 years ago, # | ← Rev. 2 →   0 [Deleted.]
•  » » 3 years ago, # ^ | ← Rev. 2 →   +8 [Deleted.]
 » 3 years ago, # |   0 For problem E2, there is a greedy solution :https://codeforces.com/contest/1296/submission/70332678 while read in a new alphabet, check all colors that have been chosen, if all colors are greater than it, choose a new color, then record its number to print.
•  » » 3 years ago, # ^ |   0 Besides, in E2 tutorial, I think dp maybe mean the longest decrease but not the least.
 » 3 years ago, # |   0 There is also $O(n^2)$ solution for E1, basically find position of max element (if there's more than 1 pick rightmost), then we want to make suffix pattern from that position like this "100..000" so we can move it to the last element (if we can't then answer is NO) then bring that maximum element to the end and recurse to subproblem with string length n — 1.
 » 3 years ago, # |   0 In Problem D. when h%(a+b) is 0 , why we have to rollback. As the monster is killed by us and we got the point. Can you please explain.
•  » » 3 years ago, # ^ |   +1 it's killed by opponent, because if h % (a + b) = 0, there is time when h equal a + b, after that we attack, now h equal a + b — a = b, then opponent attack, then monster die.
 » 3 years ago, # |   +4 Wow there are a lot of different solution to E1. Here is another one:For all j < i, if s[j] > s[i], add an edge between i and j. The answer is the two coloring of the resulting graph, if it exists.
 » 3 years ago, # |   0 regarding the colouring strings question,If we simply just store maximum element in two colour after each i ,0=a. Initially assign them a=-1 and b=-1.Then if s[i] using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int n,a=-1,b=-1; string s,g=""; cin>>n;cin>>s; for(int i=0;ib)swap(a,b); if(s[i]=b){b=s[i];g+='0';} else {a=s[i];g+='1';} }cout<<"YES\n";cout<
 » 3 years ago, # |   0 Hi, what's wrong with my solution. It gets TLE, but works in O(n^2). https://codeforces.com/contest/1296/submission/70360668
•  » » 3 years ago, # ^ |   0 Did it resolve?
 » 3 years ago, # |   0 #include using namespace std; int main() { int t; cin>>t; while(t--) { int cnt=0; int num; cin>>num; int rem=num; while(num-10>=0) { cnt++; num=num-10+1; } cout<
•  » » 3 years ago, # ^ |   0 are u serious brooo..... u r just subracting 10 each time ... do u understand how many iterations would it take for a number of order 18....
 » 3 years ago, # |   0 why g pushback index i , i do not see connection
 » 3 years ago, # |   0 Could anyone explain what is the least decreasing subsequence in problem E2
•  » » 3 years ago, # ^ |   0 longest I believe
 » 3 years ago, # |   0 Another cool trick for Problem C avoiding maps with pairs as keys.Let's replace all the L elements with -1, the Rs with +1.Similarly, replace all the Us by -1,000,000 the D's by +1,000,000.The problem is to find the shortest subarray with 0 sum.
 » 3 years ago, # |   0 Can someone explain the solution for Problem C??
•  » » 3 years ago, # ^ | ← Rev. 3 →   0 We want to remove the minimum-distanced subsequence that does not affect the path. That is, a sequence of letters starting from a position and ending at the same position — included in the whole path, AKA a Cycle. We want to store the starting index of the Cycle starting at any position. so while iterating over the letters, keep updating the current position and its starting index depending on the current letter. Initially the position is (0,0) and the starting index is 0. If the first 2 letters are "UD", In the first iteration the position will change to (0,1) and the starting index of (0,1) is 1 because you will continue the path from position (0,1) depending on the letters starting at index 1. In the second iteration the position will change to (0,0) and the starting index should be 2. Now, this is a Cycle. so maintain a visited map to know if the current position is previously visited. If it's visited, then that is a Cycle, Check if the length of the Cycle is less than any previous one, so maintain a res int which stores the min length of a Cycle — initially it can be infinity — INT_MAX, and when a Cycle is found just check if the distance is less than res, If it is, then set res to that distance and set l to the starting index of the current position which is the starting index of the Cycle and set r to i which is the ending index of the Cycle, if it's not visited, mark it as visited. Be sure to update the starting index depending on the position in both cases to guarantee a min-length Cycle. For example if the path is "URDLUD", "URDL" forms a Cycle of length 4, "UD" also forms a Cycle of length 2, so the second is minimum. if the starting index of position (0,0) is not updated after finding the first cycle, the second Cycle's length would be 6 rather than 4, and that is wrong. Finally if the distance is infinity, that means that no Cycles were found, so the answer is -1, else the answer is l r.
•  » » » 3 years ago, # ^ |   0 Thanks a lot for the explanation with starting index update. Overlooked that case and didn't get AC during the contest.
•  » » » » 3 years ago, # ^ |   0 Anytime. Glad it helped you.
 » 3 years ago, # | ← Rev. 2 →   0 Thanks for the nice set of problems. In problem 1296D - Fight with Monsters, it was understood implicitly that each player hit decreases the health points of the monster by the hitting power of this player. Nonetheless, it might have been worthwhile to mention this rule explicitly in the problem statement.
 » 3 years ago, # |   0 Can someone explain me problem C(Walking Robot)? In simple words.
•  » » 3 years ago, # ^ |   +1 We want to remove the minimum-distanced subsequence that does not affect the path. That is, a sequence of letters starting from a position and ending at the same position — included in the whole path, AKA a Cycle. We want to store the starting index of the Cycle starting at any position. so while iterating over the letters, keep updating the current position and its starting index depending on the current letter. Initially the position is (0,0) and the starting index is 0. If the first 2 letters are "UD", In the first iteration the position will change to (0,1) and the starting index of (0,1) is 1 because you will continue the path from position (0,1) depending on the letters starting at index 1. In the second iteration the position will change to (0,0) and the starting index should be 2. Now, this is a Cycle. so maintain a visited map to know if the current position is previously visited. If it's visited, then that is a Cycle, Check if the length of the Cycle is less than any previous one, so maintain a res int which stores the min length of a Cycle — initially it can be infinity — INT_MAX, and when a Cycle is found just check if the distance is less than res, If it is, then set res to that distance and set l to the starting index of the current position which is the starting index of the Cycle and set r to i which is the ending index of the Cycle, if it's not visited, mark it as visited. Be sure to update the starting index depending on the position in both cases to guarantee a min-length Cycle. For example if the path is "URDLUD", "URDL" forms a Cycle of length 4, "UD" also forms a Cycle of length 2, so the second is minimum. if the starting index of position (0,0) is not updated after finding the first cycle, the second Cycle's length would be 6 rather than 4, and that is wrong. Finally if the distance is infinity, that means that no Cycles were found, so the answer is -1, else the answer is l r.
•  » » » 3 years ago, # ^ |   0 Thanks for explaining it to me. I will try it today. Glad I asked.
•  » » » » 3 years ago, # ^ |   0 Anytime. If you still have any confusion, feel free to ask.
 » 3 years ago, # |   +3 problem e, any two neighboring (!!!) neighboring, is the key word for me, I've missed that word
 » 3 years ago, # |   0 E1 can also be solved using bubble sort because it's based on swapping successive elements. All elements that are swapped should have different colors.
 » 3 years ago, # |   0 The n^2 solution to F TLE on test case 271 in Java. I was able to have it just about pass by lazily computing the parent arrays instead of precomputing all of them.
 » 3 years ago, # |   0 Could anyone explain me how the formula h[i] = ((h[i] + a — 1) / a) — 1 came in Question D?
•  » » 3 years ago, # ^ |   0 If you want to calculate ceil(X/Y) then you do this: (X+Y-1)/Y, this works because division is rounded down if you cast the result to int
 » 3 years ago, # |   0 Why problems aren't rated yet?
 » 3 years ago, # |   0 В авторском решении мы сортируем монстров по значению секретной техники. Но в условии говорится, что бой идет от первого к последнему поочередно. Или я что то не так понял. Объясните кто нибудь. Спасибо.
•  » » 3 years ago, # ^ |   0 We sort it to find out cheapest set (in terms of secret technique usage). We will kill monsters in original order but only on those which are in that set we will use secret technique.
 » 3 years ago, # | ← Rev. 2 →   0 Did anyone solve problem F in python? I'm getting exceed memory limit on test 148, mostly solution is same as in editorial. solution
•  » » 3 years ago, # ^ |   0 the f word time limit exceed on 238
 » 3 years ago, # |   0 There is a much more elegant solution to E2. Assume we are processing a certain character at an index say i. Also assume that the previous characters have been alloted best possible colors. Now in order to give this the best possible color we need to consider some possibilities such as the color given to it should not have been given to any character lexicographically greater than it. Now in order to achieve this we consider the mex or the minimum possible color that we could give it. To find the mex, we make sets consisting of : {z} , {z,y} , {z,y,x} and so on till {z,y,x,...,a}. We will put the minimum possible values that have not been given to any character in the specific set. Then if we have this info the color is easily found bu using the set {s[i] + 1 — 'a' , ... ,x,y,z} . Then comes update. It is pretty easy too. Increase all the sets before the current set if their value is lower than the current color. https://codeforces.com/contest/1296/submission/70695446 . If anyone has further queries they are more than welcome.
•  » » 2 years ago, # ^ |   0 Hi. I would really like to understand why such a greedy solution works. I can intuitively see, that the the colours assigned are made are locally optimal but I couldn't prove why it reaches the global optimum. I'm pretty uncomfortable with greedy approaches in general, so any advice regarding it is immensely welcome. Thank you for your time.
 » 3 years ago, # |   0 We can solve E2 by DP: SolutionLet calculate for each position color[i] — optimal color for position i.If at a position less than i there is a letter larger than s[i], this means that they must have different colors, because we will swap them in future. Let's arrange the colors in ascending order, then color[i] = max(1, color[j] + 1), for all j such that j < i and s[j] > s[i].That gives us O(n) solution.
 » 3 years ago, # |   0 why my code is accepted (E2) Should it give time limit exceeded error.. void solve() { long long n,i,j,k,l; string arr; cin>>n>>arr; std::vector v; set st; char last[n]={'a'}; for(i=0;i= last[j]) { v.push_back(j+1); last[j]=arr[i]; st.insert(j+1); break; } } } cout<
 » 3 years ago, # |   0 So excited! for E2 i think i get a really easy way to solve. the main idea which inspires me is MikeMirzayanov's tutorial in E1 Note that the actual problem is to divide the string into two subsequences that both of them are non-decreasing and the beautiful greedy solution. for E2 we just need to divide the string into more parts, so why not we repeat and continue the E1's greedy work. here is my submission the key part is the double loop by which we can find the first character no bigger than s[i] in array k ,once we find ,the work is just the same as E1. Also there is another important thing. Though it's a double loop, the time complexity actually is not n^2 thanks to the second loop variable j is within 26(so the size of k doesn't need 2*10^5). and this promise that it won't TLE.(you can rewrite it by a binary search if you like) though i can't exactly explain why greedy still works .Hoping for someone to add.
•  » » 3 years ago, # ^ |   0 well i find i'm not the only one to find this. never mind
 » 3 years ago, # |   0 For E1,you can just connect the inverse counts(jstr[i]) and see if this is bi-partitie graph71648393
•  » » 3 years ago, # ^ |   0 Could you please explain why did you use Bipartite Graph?
•  » » » 3 years ago, # ^ |   0 If there is say x at position i, and y at position j with x < y and i > j, so you should definitely swap those 2 so they should be in different colours, so you can find all such pairs, and connect them, no adjacent nodes in this graph should be in same colour so that you can swap, that is nothing but bi partite graph
 » 3 years ago, # |   0 In E1/E2 Can someone explain the proof that number of colors is equal to number of non decreaing subsequences?
 » 3 years ago, # |   0 Problem F can also be solved in $O(n \log^2(m))$ with smaller larger optimization: 74863530
 » 22 months ago, # | ← Rev. 2 →   0 in Problem E1 we can also solve using bubble sort and bipartite graph. My solution123216223
 » 11 months ago, # | ← Rev. 2 →   0 h[i] = ((h[i] + a — 1) / a) — 1; i'm not getting this line , is there anyone to explain pls ? (Editorial D )
 » 11 months ago, # |   0 Great round
 » 11 months ago, # |   0 nice round
 » 2 months ago, # | ← Rev. 4 →   0 Proof problem D with Dilworth's theorem ( the only way i can find out ) Might be wrong, it's my first time write such thing. 1. Lemma 1: swap only happened if the two numbers are consecutive initially: Consider if they are not consecutive there will be three types (in each type we need to remove everything beteewn and swap them ,finally got a 0 1,lets proof they are all bad) — (1) 1 0xxxx0 0(xxxx means some 1s and 0s maybe empty)(in this case 0xxxx0 can be only one 0 too) Once you remove every 0 between them ,and swap them ,it is as same as swap the first 0 with 1 and remove every 0 after them — (2) 1 1xxxx1 0 Same as (1) — (3) 1 0xxxx1 0 Once you remove 0xxxx1 it will cost more than 2 remove ,so we can remove the first 1 and last 0 and the xxxx instead 2. Lemma 2 : We can do the swap at first (this is pretty easy by using Lemma1 let's skip it) 3. Lemma 3 : After all the swap the answer will be "input.size()-length of Longest Increasing Subsequence" (easy to proof) 4. Lemma 4 : If the numbers of operation is higher , the cost will be higher. The cost of operation is 1e12(+1) if there is any case that has higher number of operation but has a lower cost , than there will be at least 1e12 swap there,which obviously won't happend 5. Lemma 5 : The swap won't decrease the numbers of operation : Let's say the maximum numbers of opertion is x and we can remove one of the 0/1 and get a greedy better condition so the maximum numbers of opertion is at most x. 6. Lemma 6: Finaly, lets proof there will be no more than 1 swap : — Lets consider the antichains of the array , there are only 3 kinds : "1" "0" "10". — Once we execute one swap we may increase 1 antichain (and decrease the length of Longest Increasing Subsequence) witch is ,split "10" to "0" and "1". — And it may not change the amount of antichain,according to lemma 3 and Lemma 5 this will increase the numbers of operation ,and according to Lemma 4 this is harmful — Lets consider when will the harmful swap happen, once we split the "10" and got "0" and "1" ,if there is a free "1" antichain in front of them ,the "0" will combining with it;And if there is a free"0" after it the "1" will cobining with it — If there are more than one swap ,and both are not harmful than the "1" of the front one will combining with the "0" of the back one ,and that means one of them are harmful So , there won't have no harmful swap if there are more than one swap ,and according to Lemma 4 ,we can't execute two swap.So we can finally solve the problem We are going to record two array :how much number 0 before i and how much number 1 after i — the least cost with out swap will be (1e12+1)*(input.size()-(biggest before[i]+after[i]+1)) — and lets find every consecutive number "10" in the initial array,the least cost is (input.size()-2+before[i-1]+after[i+2])*(1e12+1)+1e12(i is the position of "1") The answer will be the minimum cost of them. Just write this for fun , there should be some easier ways to proof it ,but i can't find out them
 » 2 months ago, # |   0 Problem C can be reduced to finding smallest subarray with sum 0 which is a standerd problem if we replace the the character 'L' with 1 and 'R' with -1 and 'U' with some bigger values like 10^7 and 'D' with the negative of 'U' ie -10^7 or vice versa submission link