### Stepavly's blog

By Stepavly, history, 3 months ago, translation, ,

1282A - Temporarily unavailable

Idea: MikeMirzayanov Preparation: MikeMirzayanov

Editorial
Solution (MikeMirzayanov)

Idea: MikeMirzayanov, unreal.eugene Preparation: Supermagzzz

Editorial
Solution (Supermagzzz)

1282C - Petya and Exam

Editorial
Solution (Supermagzzz)

1282D - Enchanted Artifact

Idea: unreal.eugene Preparation: unreal.eugene

Editorial
Solution (Darui99)

1282E - The Cake Is a Lie

Editorial

• +47

 » 3 months ago, # | ← Rev. 3 →   +4 Could anyone explain why the problem D answer for the below case is 1?6 17 2 60 0 1 0 0 17 6 3 7 10 12When can the student leave to get 1? If the student leave T=6, he can solve 3th problem, but the 2th problem makes his score 0. Am I missing something?
•  » » 3 months ago, # ^ |   +4 Petya could solve one of easy problems and leave exam at time 2.
•  » » 3 months ago, # ^ |   0 In this case, you should prioritize easy problem over hard problem, which takes exactly 2 minutes.He can leave immediately when he solve a easy one at t=2.
•  » » 3 months ago, # ^ |   +10 Actually, it's problem C, not D.
 » 3 months ago, # |   +5 Solution to problem B is not very clear yet. Can someone please explain it in little more detail?
•  » » 3 months ago, # ^ | ← Rev. 3 →   +10 Sort the items by price. Let us say we wanted to buy only all of the first i items by paying the minimum price possible, call this minimum price needed dp[i]. If i >= k-1 (0 based index).We will definitely have to buy the most expensive item but on purchasing that we can get items from i-k+1 to i-1 for free.So dp[i] = cost of item i + dp[i-k]. If dp[i] <= budget, our answer is atleast i+1.If i < k-1 the case is simpler, check the code for this case.Also if there are n items on sale, there is no point of purchasing an expensive item but leaving out a cheaper one. In essence if you buy some ith item you will definitely buy first i-1 items also. Let me know if you'd like a proof for this too.https://codeforces.com/contest/1282/submission/67533829
•  » » » 3 months ago, # ^ |   0 Thanks , for the explanation. Can you also explain what is said in the editorial,I think it's a little different from this(dp) approach?
•  » » » » 3 months ago, # ^ |   +1 I feel it's almost the same. Not sure though.
•  » » » » 3 months ago, # ^ |   +4 dp is kinda parallel counting for all possible prefixes, while solution from editorial is serial: try first prefix, then run with step k; try second prefix, then run with step k, and so on.
•  » » » » » 3 months ago, # ^ |   0 Thanks, now i got it.
•  » » » 3 months ago, # ^ |   0 Wanted to ask that for i=1 to k-1 (in ur case: 0 to k-2), we can't apply the offer bcz. we don't have exactly K items to buy. So, in that case shouldn't it be 1 item(not applying the offer). Your code looks like that we can buy less than K items. Can you clarify if we can buy less than K items or not?
•  » » » » 3 months ago, # ^ | ← Rev. 2 →   +3 We can definitely buy a single item without using the offer. If you buy say g items, g < k, then you have to pay sum of individual cost of those items equivalent to buying each of them one at a time without using the offer.In my code dp[i] where i < k-1 is simply the prefix sum.Let me know if this clears your doubt. You have both the option of applying the offer or not applying it.
•  » » » » » 3 months ago, # ^ |   0 can you explain more how offer is used and not used in both the conditions (i >= k-1 and i < k-1 )
•  » » » » » » 3 months ago, # ^ |   0 You may only use the offer if you buy exactly k items. Say K=5, you want to buy 7 items, you can buy 2 items individually without using an offer and 5 items on offer. Not sure what exactly you are trying to ask but still hope it helps.
•  » » » 3 months ago, # ^ | ← Rev. 2 →   0 t=1 n=5 p=6 k=2 A=1 2 4 5 7for this case if we buy price of 5 we have {5,4,1} as purchased items and the optimal buy would be {4,2,1} ,here both types of purchase are optimal.So how can we be assured of the fact that higher purchase will not lead to best result
•  » » » 3 months ago, # ^ |   0 if dp[i]=min cost to buy first I items,how is it that that is the optimal way to buy max no. of items? I understood your dp, but I would like to have a proof. My dp is quite poor.
 » 3 months ago, # |   +16 My solution for problem E is similar but definitely much shorter, the main idea is to take an arbitrary triangle to remove it and recursively solve on the three sides, I have a list with the order of vertices and other for the order of triangles, in my recursive function I have a fixed side(edge) then I find one arbitrary triangle with that edge and solve similarly for both sides
•  » » 3 months ago, # ^ |   0 Crispy and tasty code. With inline functions too. Nice.
•  » » 3 months ago, # ^ |   0 similar with my Solution XD
 » 3 months ago, # |   +35 I have another approach for $D$.Assume length of $s$ is $l$. First, query with $t$ = "a" and let the response be $r$. There are 2 cases:1- string $s$ consisted only of "b" letters, then $l = r$.2- string $s$ had at least one "a" in it, then $l = r + 1$. Now query with $t$ of length $r$ consisting only of "b" letters and let the response be $e$. If $e = 0$ then terminate, otherwise, You have 2 pieces of information:1- $l = r + 1$.2- The distance between string $s$ and string full of "b" letters of length $l - 1$ is $e$.3- $s$ has at least one "a" in it.Now, let $t$ be a string of length $l$ of only "b" letters. The distance between $t$ and $s$ is also gonna be $e$ (I think the only case where this is not true if string $s$ consists only of "b" letters, which we know it already doesn't. Would love a proof for this tho). Now just like the rest of the solution in the editorial, iterate over $t$ from $1$ to $l$ and try to turn the character to "a" and query. If the distance in the response is less than $e$, then update $t$ and $e$. keep doing this until the response is $0$
 » 3 months ago, # |   0 For the question, Price of one easy version i tried to solve it using modification of knapsack but it gave tle on pretest 3. Did i miss something, code is as below:- private static int getMem(int[] a, int n, long w, Map map) { StringBuilder sb = new StringBuilder(); sb.append(n).append(":").append(w); String key = sb.toString(); if (map.containsKey(key)) return map.get(key); if (n <= 0 || w <= 0) { map.put(key, 0); return 0; } if (a[n-1] > w) { int value = getMem(a,n-1,w, map); map.put(key, value); return value; } else { int d = Integer.MIN_VALUE; if (n-2 >= 0 && a[n-1] <= w) { d = 2; } int val = max(1+getMem(a,n-1,w-a[n-1], map), getMem(a,n-1,w, map), d+getMem(a,n-2,w-a[n-1], map)); map.put(key, val); return val; } } 
•  » » 3 months ago, # ^ |   0 Note i sorted array before calling this method.
•  » » 3 months ago, # ^ | ← Rev. 3 →   0 You can optimize this to linear time, since the values of all of the items are equal. For example, since an item with cost 10 and another with cost 5 are always equal value, you always want to take the one with the smaller cost.
 » 3 months ago, # |   +13 In solution to problem B, the sort function is used which takes o(n*log(n)) time complexity. So how come the solution has linear time complexity? Please explain!
•  » » 3 months ago, # ^ | ← Rev. 3 →   0 It is not, but since the $a_i$s are only from 1 to 10^4, we can use radix sort or counting sort to sort the numbers in $O(10^4)$ timeUpdate : As the number of testcases is large($10^4$), so it won't probably pass, so nevermind, just use quicksort.
•  » » » 3 months ago, # ^ |   0 but the editorialist solution uses builtin quicksort function brother
•  » » » » 3 months ago, # ^ |   0 We can do, the solution is not linear though, it indeed is $O(nlog n)$
•  » » » 3 months ago, # ^ |   0 It will take 10^4 iteration per test case,which will give worse time complexity than nlogn.
•  » » » » 3 months ago, # ^ |   0 Not at all, it would give us $O(10^4+n)$ time which is linear in n
•  » » » » » 3 months ago, # ^ |   0 It will surely take O(10^4) to sort the array per test case, and total no test case can be max 10^4. so here worse time complexity is O(10^8) which is worse than nlogn.
•  » » » » » » 3 months ago, # ^ |   0 Ahh I see, then it sure is not linear, thanks for correcting :)
•  » » » 3 months ago, # ^ |   0 radix sort is $O(n \log(A))$ where $A$ is logarithm of number of possible values. counting sort is $O(n)$.You can do $O(n)$ here, but I came up only with offline solution. It's useless.Here is comparison: Python 3.7.2 (during round) 67536168 343 ms $O(n)$ 67625343 499 ms Microsoft Visual C++ 2010 67625686 61 ms $O(n)$ 67625691 41 ms
 » 3 months ago, # | ← Rev. 2 →   0 Someone Please Help me Correct my Solution For Problem B. I am unable to find which case I am missing . It would be of great help. My Solution :- (https://codeforces.com/contest/1282/submission/67578229)
•  » » 3 months ago, # ^ |   0 PROBLEM BCan someone Please Explain this Test Case?4 9 22 3 4 5How the answer is 4?
•  » » » 3 months ago, # ^ |   0 You can buy goods worth 4 & 5 for 5 coins and goods worth 2 & 3 for 3 coins.Hence answer is 4
•  » » » » 3 months ago, # ^ |   0 Is it allowed to do this operation multiple times?
•  » » » » » 3 months ago, # ^ |   0 Yes as long as you have ample amount of money left . You can buy as many items as you want
•  » » » » » » 3 months ago, # ^ |   0 Thanks.
•  » » » » » 3 months ago, # ^ |   0 it is written in question**Vasya can perform one of the following operations as many times as necessary**
•  » » » » » » 3 months ago, # ^ |   0 Yes exactly
•  » » » » 3 months ago, # ^ |   0 Can you please explain again in detail? I didn't get what you said.
•  » » » 3 months ago, # ^ |   0 As we can take 2,3 as one pair and 4,5 as another pair so re money is max(2,3)+max(4,5) i.e. is 8 which is less than 9 so we have one unit of money left.. Since we have no other toy this is our final ans;
•  » » » 3 months ago, # ^ |   0 First buy 5 and then you get 4 for free and then buy 3 you get 2 for free so total cost = 8 but you have 9 and so you buy 2 goods and get 2 for free
 » 3 months ago, # |   0 Could anyone explain problem B in more detail
•  » » 3 months ago, # ^ |   0 first sort the prices.let dp[i]= min cost to buy i items.for i
•  » » 3 months ago, # ^ |   0 Analyse this submission. It should be added in editorial. https://codeforces.com/contest/1282/submission/67531855
 » 3 months ago, # | ← Rev. 2 →   0 My submission for C giving WA. Someone please help!! Edit:: I think I got my mistake..
 » 3 months ago, # | ← Rev. 2 →   0 In problem C, why is the answer of this testcase $1$ ? 1 6 17 2 6 0 0 1 0 0 1 7 6 3 7 10 12 
•  » » 3 months ago, # ^ | ← Rev. 3 →   0 The same question has already been answered above: https://codeforces.com/blog/entry/72461?#comment-567159
•  » » » 3 months ago, # ^ |   0 Didn't notice it earlier! Thanks!
•  » » 3 months ago, # ^ |   0 Start solving an easy problem at t = 0. It'll be finished at t = 2. Then you can leave.
•  » » » 3 months ago, # ^ |   0 thanks!
•  » » » 2 months ago, # ^ |   0 what if i take another easy problem at t=2, and finish it by t=4 then if i leave i'll be having 2 questions done by me .
 » 3 months ago, # |   0 For the first problem(temporarily unavailable) in the fifth test case a=-10 b=20 c=-17 r=2 , the coverage will be from [-15,-19].so a to b must added, the answer must be 31, if I'm wrong case someone explain why I'm wrong.
•  » » 3 months ago, # ^ |   0 The time taken to travel from a to b is 30 units right? Cause you travel from 0 to 1 in 1 unit time
 » 3 months ago, # |   0 Please find the error in my code i don't know what's going wrong here My Submission for Problem D
 » 3 months ago, # |   0 For Problem B. first sort all the goods by its price.and take this array as p[]; then let f(x) means the least money he should pay to buy first x goods. so if x <= k , f(x) = f(x-1) + p[x] if x > k , f(x) = min(f(x-1) + p[x],f(x-k) + p[x]) // use offer or not then iterater all the f(x),find the most x satifies f(x) <= the money I have. does I make this problem more complex?
 » 3 months ago, # | ← Rev. 2 →   0 problem b order is $O(N^2)$ ?
•  » » 3 months ago, # ^ |   0 I was also confused initiallyBut notice you visit every index just once so the order is O(n).
•  » » » 3 months ago, # ^ |   +1 yes, order is (k) * (n / k) = n
 » 3 months ago, # |   +1 can someone explain C approach briefly?
 » 3 months ago, # |   +1 In tutorial of problem 1282D - Enchanted Artifact:Consider an arbitrary string $t$ of length $l$ and let the answer to its query be $q$. Then if we replace the letter $t_i$ with the opposite one (from a to b or from b to a), then we may have one of two situations: $q$ decreased by $1$, then the letter $t_i$ after the change coincides with the letter $s_i$. otherwise the letter $t_i$ before the change matches the letter $s_i$. I think that this is not truth. For example:$s=$ abababababababababab$t=$ babababababababababaEdit distance is $q=2$ (delete the first character and insert the last one).Let's replace some a in the middle with b (the letter $t_i$ after the change coincides with the letter $s_i$):$t'=$ bababababbbababababaEdit distance is $q'=3$ (replace this letter, delete the first character, and insert the last one).The letter $t_i$ after the change coincides with the letter $s_i$, but $q$ didn't decreased by $1$.Am I right?
•  » » 3 months ago, # ^ | ← Rev. 2 →   +11 does the author means the t is all letter 'a' at first.so it can't be babababababababababababa it can only be babababababaaaaaaaaaaaaa so convert from a to b will make difference?
•  » » » 3 months ago, # ^ |   0 Maybe, but he didn't write it.
•  » » » » 3 months ago, # ^ |   +8 but in his codes, he inits the t in this method. so maybe this is what he actually means.
•  » » » 3 months ago, # ^ |   0 BTW, how can this be proved?
•  » » » » 3 months ago, # ^ |   +8 I think because the s and t has same length. So the quickest way to transform from t to s is to replace each a to b where the char in them are different. if a change from 'a' to 'b' make the distance greater(we need more step to make make t equal to s cause our change), it means this position should be 'a', otherwise, it should be 'b'.
•  » » 3 months ago, # ^ | ← Rev. 2 →   +13 You are right. Actually, these statements are not true in general case, but they are valid if your initial string $t$ is monochrome (consists of letters of a single type) and has the same length so as $s$. Moreover, it is valid when string $t$ has a common prefix with $s$ and the remaining part is still monochrome. It is also can be proven (by induction I guess) that during these actions deletions and insertions are not necessary operations, because what can be done with these operations can also be done with changing letters in no more operations. I'll make the editorial more clean to understand soon.Additionally, it looks like when we are increasing a valid prefix in string $t$ on an initial monochrome string, the edit distance is actually equal to the Hamming distance.
•  » » » 3 months ago, # ^ | ← Rev. 2 →   0 Can you prove these facts or post a link to the proofs for the same, as proving things is as much necessary in a contest(especially for greedy problems) as is intuition?
•  » » » » 3 months ago, # ^ |   0 I would like very much a proof too :)
•  » » 3 months ago, # ^ |   0 Yeah I think you are right, take this case also: Edit distance between (aaaa , baba) is 2. The edit distance between (abaa , baba) is also two.Even though only one character has been changed from a to b.
 » 3 months ago, # | ← Rev. 4 →   0 Could anyone explain why the solution of problem C answer for the below case is 2?3 5 2 100 0 0 15 5 5 If student leave at 5, he must solve all the problem. Am I misunderstanding？
•  » » 3 months ago, # ^ |   +1 he can leave at 4, solve problem 1 and problem 2, which costs 4 minutes.he doesn't need to leave at 5
•  » » » 3 months ago, # ^ |   0 Gotcha! Thank you.
•  » » 3 months ago, # ^ |   0 If he leaves at 0, then he will receive 0 points. If he leaves at 1, then he will receive 0 points. If he leaves at 2, then he will receive 1 point for solving an easy question. If he leaves at 3, then he will receive 1 point. If he leaves at 4, then he will receive 2 points for solving both the easy questions. If he leaves at 5, then he will receive 0 points because all the problem becomes mandatory and solving a hard problem takes 100 minutes and You don't have that much time.
 » 3 months ago, # |   0 In problem B2: t = int(input()) for _ in range(t): n,p,k = map(int , input().split()) a = list(map(int , input().split())) a.sort() ans= 0 pr = 0 for i in range(k+1): s = pr if s>p:break cnt = i for j in range(i+k-1,n,k): if s+a[j]<=p: cnt+=k s+=a[j] else:break pr += a[i] ans = max(ans , cnt) print(ans) This code is giving runtime error in Test case 2 I know that error is in line 9 for i in range(k+1) but this is similar to the solution provided. What is the reason that this code gives AC in c++14 but RE in pypy3.6 ?
 » 3 months ago, # |   0 In Problem C, can anyone please explain the code after l is declared. I know we need to increase the count as the question becomes mandatory but I am unable to figure out the code. Kindly help.
 » 3 months ago, # |   0 explain test case no 8 of 1282C — Petya and Exam how it is 1 it ans should be zero
•  » » 3 months ago, # ^ |   0 I think it should be 0 too.
 » 3 months ago, # |   0 I am getting WA on test 2 on C.https://codeforces.com/contest/1282/submission/67600070Can Someone Please Help.Thanks
•  » » 3 months ago, # ^ |   0 instead of starting from the beginning, I have for once solved all the questions and then am deciding from the end, If the solution is permissible or not(depending on time limit, and if the next question is mandatory within that limit)
 » 3 months ago, # | ← Rev. 2 →   +10 Based on your editorial of problem E, I think one can do simpler things to solve the problem: To find $q$, consider the fact that whenever we cut off a triangle, we lose a vertex, i.e., it no longer appears in any further triangles. So, simply maintain the frequency of all the vertices (the no. of triangles it appears in) and also the triangles that contain the $i$-th vertex. At every step, choose a vertex that has a frequency of $1$, and cut its corresponding triangle. To find p, consider every triangle. It has $3$ edges. Since only a single edge corresponds to the vertex of the polygon, and it appears only once in the input, we construct a graph on n vertices. We compute the count/frequency of all the edges in the input and for all edges $(u - v)$ that have a frequency of $1$, we add an edge $(u - v)$ in the graph. Now, the graph will be in the form of a simple cycle corresponding to the order of the vertices in the polygon. So, simply perform a dfs to find $p$. Here, we have found $p$ and $q$ independently.Code: 67604806
•  » » 3 months ago, # ^ | ← Rev. 2 →   0 can you explain the steps to implement first part(to find q) of your exaplaination .I got what you are saying but i am not able to understand how to implement it efficiently.
 » 3 months ago, # | ← Rev. 3 →   0 Can anyone explain the problem A please? I am really confused on that.
 » 3 months ago, # | ← Rev. 3 →   0 I dont get why the answer to test case 8 of problem c is 1 and not 0. In that test case the problem with lowest mandatory time is the third one, so if we want to solve one or more problems we must solve this one, but because the time it takes to solve this problem is 6 seconds, and there is a problem which its mandatory time is 6, we can not solve the third problem without solving this one or getting a zero score.So why is the answer to this test case 1? I am probably missing something here but i cant figure it out. Can someone please help me? edit* I found out how it can be 1.
•  » » 3 months ago, # ^ |   0 did u find the answer?
•  » » » 3 months ago, # ^ |   0 Solve an easy question and leave the exam
•  » » » » 3 months ago, # ^ |   0 But after solving s become 6 which is equal to next ti,so it became mandatory,how can he leave the xm
•  » » » » » 3 months ago, # ^ |   0 He will solve an easy problem from t=0 to t=2(0-1-2) and leave at end of 2nd minute
 » 3 months ago, # |   0 queryforces
 » 3 months ago, # |   0 Please someone could explain to me the intuition behind the problem D.I am not able to understand how the query is coming and how we to find the spell(string) using t(string) and query of edit distance.In fact, I am not clear with input.
 » 3 months ago, # | ← Rev. 4 →   0 My soln. is similar to one which is mentioned in editorial and it is passing for problem B1 but failing for problem B2. I am not able to find out the mistake. Please help.
•  » » 3 months ago, # ^ |   0 I think sum += ar[i] should be replaced with sum = ar[i].
 » 3 months ago, # |   0 please, can anyone tell me that why in b2 editorial solution first loop run upto only k ? why ? please, anyone reply guys.
•  » » 3 months ago, # ^ |   0 Because if you buy greater than $k$ items without using the offer, you would spend more money, rather buy $k$ items using the offer and buy others individually.
•  » » » 3 months ago, # ^ |   0 thanks bro
 » 3 months ago, # |   0 For me personally, finding permutation of vertices was far easier than finding an order of cuts.
 » 3 months ago, # | ← Rev. 2 →   0 Do I have correct idea for problem E?https://codeforces.com/contest/1282/submission/67669011Idea:Store vertices like graph. 1 piece = 6 new edges. Sort vertices by number of neighbors. On each step poll vertex V with smallest number of neighbors. V must have only two neighbors. Delete V from its neighbors. Add these two neighbors back to priority queue. Repeat while there wont be only two vertices.
 » 3 months ago, # |   0 Why does the link to the editorial of this round appear twice? https://codeforces.com/contest/1282
 » 3 months ago, # |   0 Can someone figure out what's wrong with this solution for D? 67670033It has a similar idea as given in the tutorial.unreal.eugene
 » 3 months ago, # |   0 Nice testcase 67 in D.
 » 3 months ago, # | ← Rev. 3 →   0 why is my answer for problem E got wrong answer Input 3 6 3 6 5 5 2 4 5 4 6 6 3 1 6 2 5 6 2 5 1 4 1 2 1 3 5 3 1 2 3 Participant's output 1 3 5 2 4 6 3 1 4 2 1 4 2 6 5 3 1 4 2 3 1 2 3 1 Jury's answer 1 6 4 2 5 3 4 2 3 1 1 4 2 6 5 3 3 4 2 1 1 3 2 1 I think it said clockwise or counter clockwise
 » 3 months ago, # | ← Rev. 2 →   +3 -3 1 2 0 how is the answer to this query for problem A 4?shouldn't it be 5? while travelling : -3 -2 -1 0 1 ,nowhere is tower signal available.
•  » » 3 months ago, # ^ |   0 same Confusion!! Did you find the answer?
•  » » » 3 months ago, # ^ |   0 I got the idea. We need to count distance and not number of points. So, -3 to -2 -> 1 -2 to -1 -> 1 -1 to 0 -> 1 0 to 1 -> 1. So, answer is 4.
•  » » » » 3 months ago, # ^ | ← Rev. 3 →   0 Well if that so in this test case answer should have been 6 : 1 10 7 1 : as, 1->2,2->3,3->4,4->5,(available for 6,7 and 8)8->9,9->10.
•  » » » » » 3 months ago, # ^ |   0 Invalid distance are 6->7 and 7->8. valid pairs 1->2,2->3,3->4,4->5,5->6,8->9,9->10. hence,answer is 7.
•  » » » » » » 3 months ago, # ^ |   0 Yes you are right, thankyou
 » 3 months ago, # |   0 anyone please explain B2 in Recursion DP <3
 » 3 months ago, # |   0 Could anyone help me to see why I got RE? TIA! 68362298
 » 3 months ago, # |   0 I doubt that the checker of Problem D calculates incorrect Edit Distance when $|t|>|s|$ because 68363205 got WA on test 1. Can anyone help me about this? Thanks.
 » 3 months ago, # | ← Rev. 2 →   0 In problem b2 for test case : 5 2 310 1 3 9 2should the answer not be equal to 2 as by buying the element with cost 2 we could also buy the element with cost 1 i.e buying element of cost 2 and getting the element with cost 1 as free.
•  » » 2 months ago, # ^ |   0 Note that the announcement says: You must buy exactly k gifts to use the offer. So the answer is 1.