### vovuh's blog

By vovuh, history, 4 years ago,

I'm really sorry about issues with problems E and F. Can't say anything more because I don't want to justify my mistakes.

1433A - Boring Apartments

Idea: vovuh

Tutorial
Solution

1433B - Yet Another Bookshelf

Idea: vovuh

Tutorial
Solution

1433C - Dominant Piranha

Idea: vovuh

Tutorial
Solution

1433D - Districts Connection

Idea: MikeMirzayanov

Tutorial
Solution

1433E - Two Round Dances

Idea: MikeMirzayanov

Tutorial
Solution

1433F - Zero Remainder Sum

Idea: MikeMirzayanov

Tutorial
Solution

1433G - Reducing Delivery Cost

Idea: MikeMirzayanov

Tutorial
Solution
• +115

| Write comment?
 » 4 years ago, # |   +5 Thank you Vovuh for all of your time and effort you put into these Div 3s, please don't stop :( Also, E is a good problem if OEIS didn't exist, I found it pretty constructive. F too hard for a brik like me
•  » » 4 years ago, # ^ |   0 How to solve E using OEIS? Never used OEIS before.
•  » » » 4 years ago, # ^ |   0 OEIS is an encyclopedia for integer sequences. You can search by entering a sequence and you will find all others with formulas also. As, for problem E there can be only 10 inputs, and 4 of them are given, so simply searching with the sequence can be find from OEIS
•  » » » » 4 years ago, # ^ |   +3 It could be found even by this number 12164510040883200
•  » » » 4 years ago, # ^ | ← Rev. 2 →   0 first, write a brute-force to generate the first few elements of the sequence or you can generate by hand in pen and paper, with those elements search in the OEIS. you will get all the sequences that start with these elements.
•  » » 4 years ago, # ^ |   +11 Would have used some MOD, then it's not straight forward!
•  » » 4 years ago, # ^ | ← Rev. 2 →   +6 without OEIS we can solve , by using simple math (n-1)!/(n/2) it's simple apply mathematics formula like combination of sitting in circular table (n-1)!/2
 » 4 years ago, # |   +3 Nice round vovuh really enjoyed the round !! Thanks for the round , hoping to have more such rounds with the interesting problems again.
 » 4 years ago, # | ← Rev. 8 →   0 In problem E how can we find sequence on OEIS ?? since we do not know the ans for i/p :6??
•  » » 4 years ago, # ^ | ← Rev. 3 →   +48 Underscore works as a wildcard:1, 3, _, 1260, _, _, _, _, _, 12164510040883200
•  » » » 4 years ago, # ^ |   0 Can you tell what is OEIS and how we can get anser from that??
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   0 oeis.org Web result with site links The On-Line Encyclopedia of Integer ... It contains information about integer sequences. For example the search https://oeis.org/search?q=1%2C+3%2C+_%2C+1260%2C+_%2C+_%2C+_%2C+_%2C+_%2C+12164510040883200&language=english&go=Search shows that such a sequence is known and displays formulas to get the nth term and more.
•  » » » » » 4 years ago, # ^ |   0 but how do we find that sequence what you searched in google specicifically??
•  » » » » » » 4 years ago, # ^ |   +1 Go to the website and enter the sequence or google the sequence and write oeis with it
•  » » » 4 years ago, # ^ |   +3 This was more useful than any above answers :)
 » 4 years ago, # |   +6 "Let dp[x][y][cnt][rem] be the maximum possible sum we can obtain if we are at the element ax,y right now, we took cnt elements in the row x and our current remainder is cnt."Shouldn't the current remainder be rem instead of cnt? I didn't AC this problem, and I want to know the solution. This round was fun, thank you. :)
•  » » 4 years ago, # ^ |   +3 Thanks, that was a typo, I fixed it.
•  » » 4 years ago, # ^ |   +3 Can someone suggest some problems similar to F one , i.e, combining the concept of dp and remainder ?
•  » » » 4 years ago, # ^ |   +3
•  » » » » 4 years ago, # ^ |   0 I think this is way simpler and a little different.
 » 4 years ago, # |   0 Very nice problemset, these round really help us (beginners) to improve and let us see that you really take care of this platform :)
•  » » 4 years ago, # ^ |   0 these were a bit too easy for mech keyboard merchants to type and get the precious rating.
 » 4 years ago, # |   0 Amazing Editorial!
 » 4 years ago, # |   0 E is a good problem. I really like it.
 » 4 years ago, # |   0 typo in D:"Let's find any district $i$ that $a_i \ne a_i$"It should be $a_i \ne a_1$
•  » » 4 years ago, # ^ |   +3 Thanks, fixed.
 » 4 years ago, # |   0 Test cases for problem F were weak, like I just calculate the maximum sum I can get in matrix choosing m/2 element in each row lets call it sum then I print (sum/k)*k , its like greatest multiple of k just smaller then sum, and boom I got AC :). PS:: Later it got hacked (:.
 » 4 years ago, # |   0 In the editorial of F it's written our current remainder is cnt. whereas it should be our current remainder is rem.
 » 4 years ago, # |   0 I didn't understand one thing. In the first problem, why did he write 'dig=x[0]-'0'',basically,subtract the zero.
•  » » 4 years ago, # ^ |   0 x[0] is a character....to get the correct integer.. char of 0 is removed. for eg: let x[0] = '3' so x[0] — '0' = 51 — 48 = 3
 » 4 years ago, # |   0 in C answer will be index of maximum element why is it showing wrong answer
•  » » 4 years ago, # ^ | ← Rev. 3 →   +3 TC15 5 5 5 2 TC22 5 5 5 5
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 The answer will be the index of maximum element adjacent to a non maximal element. If suppose, one prints only index of maximum then it could be right in between 2 maximal elements, such as index 1 in 4 4 4 3. Index 1 is maximum in the array but it is next to another maximum and cannot eat the number next to it.
•  » » » 4 years ago, # ^ | ← Rev. 2 →   0 There are multiple answers. 4 4 4 3 = 4 4 5 so index 3 is the answer
 » 4 years ago, # | ← Rev. 5 →   +1 The approach to sentence E is simpler 96159630: there are n people, calculate the number of ways to make 2 circles.The first person has 1 choice (because he is the first person in the first circle), the second person has n-1 choice, the third person has n-2 choices, ..., the n / 2 person has n / 2 + 1 choice, n / 2 + 1 person has 1 choice (since this is the first of the second circle), the n / 2 + 2 person has n / 2-1 choice .. .For example:With n = 6, the number of ways to choose will be calculated as 1x5x4x1x2x1.With n = 8, the number of ways to choose will be calculated as 1x7x6x5x1x3x2x1.=> The general formula is (n-1)! / (n/2).Sorry for my English not good.
 » 4 years ago, # |   0 Can someone please explain in problem E how we choose permutations as n!/(n/2)? (Sorry for the silly question)
•  » » 4 years ago, # ^ |   0 You can see more here
 » 4 years ago, # |   0 In Problem E why we have to divide our answer by 2 ?
•  » » 4 years ago, # ^ |   -8 You can see more here
•  » » » 4 years ago, # ^ |   +1 I got that logic. But I was trying to find out the tutorial
•  » » » » 4 years ago, # ^ |   +1 The Dancing pairs [1, 2 ,3 ,4] [5, 6, 7 ,8] is same as [5, 6, 7, 8] [1, 2, 3, 4] so, when we applied the formula, we are counting every possible pair twice. So, we need to divide the answer by 2.
 » 4 years ago, # |   0 There is a simpler formula in Problem E. Since we need exactly different round dances, we can notice that all of them can be represented as permutations with a fixed first element (for example, with a fixed number 1 at the beginning). Then the number of such sequences will be (n-1)!, now we just need to divide this number by (n / 2) (I don't know exactly how to prove why this works, but it works XD).
•  » » 4 years ago, # ^ |   0 Oh, I didn't notice that this solution was already shown above.
 » 4 years ago, # |   +6 Don't know why are you encouraging to find sequence on OEIS for E?It is bit surprising to see that
•  » » 4 years ago, # ^ |   0 Would've been a bit difficult if one didn't know the math. I think that should've been done though. It turned out to be too easy right now.
•  » » 4 years ago, # ^ |   0 You are very pretty <3
 » 4 years ago, # |   0 When will the changed rating be reflected in the account?
•  » » 4 years ago, # ^ |   0 About an hour after finishing system testing..
•  » » » 4 years ago, # ^ |   0 means in 20-25 min from now right??;-) , I am excited bcz it's my best rank yet.
 » 4 years ago, # |   0 Can anyone help me to understand how to approach for problem F?
•  » » 4 years ago, # ^ |   +1 In this dp problem, we are simply tracking the results on variation of every parameter(since constraints are low, seeing every possibility like Dr. Strange ;D).Firstly, we are traversing over every element in the matrix. Secondly, we are also traversing over all possible number of elements picked in a row. And also varying remainder of total sum. If a configuration with such properties (number of elements picked in that row and the total sum remainder) exists, $dp[i][j][cnt][rem]$ will have the max sum. Otherwise value will remain $-inf$.If you have solved dp problems before, you'll have an idea now. Otherwise, I urge you to practice basic dp problems, after which you'll solve this one.
 » 4 years ago, # |   0 In F , why the answer is stored at dp(n,0,0,0)?? Thank U
•  » » 4 years ago, # ^ |   0 dp[n][0][0][0] represents max sum, when we are at (n,0) [out of the matrix], i.e we have traversed the matrix and came out of it, with 0 elements picked in this row(obviously, because this row is not a part of the matrix), with the sum havin 0 remainder.
 » 4 years ago, # |   0 Это довольно стандартная задача на динамическое программирование. Пусть dp[x][y][cnt][rem] равно максимальной сумме, которую мы можем получить, если сейчас мы находимся на элементе ax,y, взяли cnt элементов в строке x и наш текущий остаток равен cnt (скорее всего хотели сказать rem).
 » 4 years ago, # |   0 Solution of problem G is giving compilation error given by you. Can you please recheck
 » 4 years ago, # |   0 I just spent whole 2 hours during the contest thinking on $F$ that $O(N^4)$ space solution won't work in $1$ second and wrote a $O(N^3)$ space + $O(N^4)$ time solution & kept debugging it till end and got AC when 7 minutes left. Yesterday i realised how bad i am at handling base cases in iterative dp.Can anyone share how do they identify whether their solution will work or not when no. of operations & size of the array used is of the order more than ~10 million.
•  » » 4 years ago, # ^ |   0 in one second computer can do O(10^9) operations. since most of the times time limit is 1 second or 2 second you can do as below most of the cases. If an array is given the size of 10^5 . so you can apply o(N)(10^5) or O(NlogN) (10^5*log(10^5)) kinda algorithm in that cases. If array size is like 5000 somthing you can apply O(N^2) or maybe sometimes O(N^3) algorithm. if size~100 you can apply O(N^4) algorithm. So main thing is in one second you can do 10^9 operations keeping in mind , on seeing constraints calculate upper bound and think accordingly. And one thing is no of test cases. It should be also consider in taking time complexity. Hope this helps.
•  » » » 4 years ago, # ^ |   0 got it. thanks :)
 » 4 years ago, # |   0 In G, if you use floyd with int values it will pass. But using long long is causing TLE.
 » 4 years ago, # |   +1 There's a typo in the editorial of F. The current remainder is denoted by cnt instead of rem
 » 4 years ago, # | ← Rev. 4 →   -23 [DELETED]
•  » » 4 years ago, # ^ |   0 Asked for help only to get downvoted. What's wrong in asking for help politely :(
•  » » » 4 years ago, # ^ |   0 Use "spoler" or just link the submission. It's not nice to scroll through codes in comment section
•  » » » » 4 years ago, # ^ |   0 Thanks for the tip! Updated.
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 Keep applying mod operation with function calls.I think this will be one of the easiest solutions of problem F to learn, I haven't use comments, but the code is clean though.
 » 4 years ago, # |   +5 Has anyone seen more problems like F? Editorial says it's a standard DP.Also, What are some other standard DP problems not commonly available?
•  » » 4 years ago, # ^ |   0 It's basically 0/1 knapsack DP in a 2d grid. For some other standard DP problem you can check CSES problemset DP section.
•  » » » 4 years ago, # ^ |   0 Actually, it's the variation in standard problems that's tough.It would be nice if you can share some problems which are a variation of these standard problems.
•  » » » 4 years ago, # ^ |   0 https://mirror.codeforces.com/contest/1433/submission/96212192 Can you tell me where i am doing wrong in my code for problem F??
•  » » » » 4 years ago, # ^ |   0 I think you must take sum in your dp with MOD; int val= solution(i,j+1,count-1,(sum+arr[i][j] % k)); (sum + arr[i][j]) % k try this
•  » » » » » 4 years ago, # ^ |   0 where to put this line?? my dp state storing sum of numbers why to store with mod??
•  » » » » » » 4 years ago, # ^ |   0 https://mirror.codeforces.com/contest/1433/submission/96227486I copy your code and set parentheses in this line
 » 4 years ago, # | ← Rev. 3 →   0 In the problem F,What happens when we initialize dp array with $0$ instead of INT_MIN or LLONG_MIN? I tried this and got wrong answer. But overall, the value of every state is going to become zero before we arrive that state during dp transition. So how does initialising values with zero go wrong?
•  » » 4 years ago, # ^ | ← Rev. 3 →   0 Never mind.
•  » » 4 years ago, # ^ |   0 int_min denotes that this state is impossible. If you initialise it as 0, you can distinguish between impossible states and states with sum 0.
•  » » » 4 years ago, # ^ |   0 But, if the state is impossible, it will have 0 sum, as we won't be able to create an impossible state. Also, the minimum sum that can be created is 0, not $-inf$.You can check the Accepted submission, and the one with Wrong Answer. You can see, the code with zero initialised dp array creates a difference of 1 in the first testcase.
•  » » » » 4 years ago, # ^ |   0 Try inputting some larger cases, deviation will be much greater than one.
 » 4 years ago, # |   0 In problem E, why are we dividing by 2 in last step? Can Anyone explain in detail?
•  » » 4 years ago, # ^ |   +1 Suppose we select n/2 members in first dance group and then the rest n/2 will automatically form the next dance group. But the formula here considers that the first n/2 and rest n/2 are 2 different cases. Hence we need to divide the value by 2 to make them identical.
•  » » 4 years ago, # ^ |   +1 Every selection is accompanied by a rejection.For eg — S = {1, 2, 3, 4}There (6C2) ways to choose 2 elements. But suppose you select {1, 2} then {3, 4} is automatically selected. Now you don't want to select {3, 4} again and {1, 2}, because this will be counted twice. So to avoid counting again, we divide by 2.
•  » » 4 years ago, # ^ | ← Rev. 3 →   +1 When we are taking C(n,n/2) we are counting how many ways we can take n/2 different elements from n elements. Let's say n = 1, 2, 3, 4, 5, 6, 7, 8we can take one round as (1, 2, 3, 4) and the other round will be (5, 6, 7, 8)we can take one round as (5, 6, 7, 8) and the other round will be (1 ,2, 3, 4)But both are the same configuration. So we are counting it twice, hence we need to divide the C(n, n/2) by two.
 » 4 years ago, # |   0 How much more time would be taken in reflecting the rating changes from this round 677? Or has the contest become unrated!?
 » 4 years ago, # | ← Rev. 2 →   0 Hello Everyone.... Can someone please explain why we multiplied with (n/2-1)! in E (two round dances) problem. UPD : Solved
•  » » 4 years ago, # ^ |   0 There are two round dances each time and each one can have (n/2-1)! permutations everytime.
•  » » 4 years ago, # ^ |   0 For n people in a circle, there are (n-1)! permutations in which they can be arranged
 » 4 years ago, # |   0 why does rating change for Div3 and educational rounds take sooooo long???
•  » » 4 years ago, # ^ |   0 because there is hacking phase of around 12 hours in these rounds.So after that system testing happens and ratings get updated.:)
•  » » » 4 years ago, # ^ |   0 It has been 5 hrs since the hacking phase ended.
 » 4 years ago, # |   0 Can F be solved with a greedy solution???
•  » » 4 years ago, # ^ |   0 I don't think it can. I tried sorting every row, taking last m/2 elements and if the sum is not divisible by k than removing the element which difference from the previous in its row is the smallest, but this doesn't work. It also doesn't work if you simply try to remove the smallest element you find. So I don't think you can solve it with some kind of greedy.
•  » » » 4 years ago, # ^ |   -8 We could do it this way, sort all the elements then for each max val "a", get a max val "b" from any row such the (a+b)%k == 0 and number of numbers taken from rows of "a" and "b" is not more than m/2.
•  » » » » 4 years ago, # ^ |   +1 It won't work, since it's possible for some "a", there does not exist "b" such that (a+b)%k==0 i.e, Cases where (a+b+c)%k==0, or sum of more number of elements will be div by k.
•  » » » » » 4 years ago, # ^ |   0 Ya that would surely fail my solution. Thanks for the replies
•  » » » » 4 years ago, # ^ |   +1 I don't see how that would work. What if the number of elements you can take is not even. What if for example you can take only 4 numbers (2 from 2 rows) and you can't find pairs such that (a + b) % k == 0 but sum of all 4 is divisible by k 2 4 10 1 1 5 9 1 1 2 4 Here the answer would be 20 but your algorithm would not even find that. Also you would have to pay attention to how many numbers have you taken from which row. So I don't think that works
•  » » » » » 4 years ago, # ^ |   0 Ya that would surely fail my solution. Thanks for the replies
 » 4 years ago, # |   0 @vovuh start system testing so we can get our rating :P..
 » 4 years ago, # |   +1 This contest is pretty easy. A speedrun is required like this contest.
 » 4 years ago, # |   0 for problem G, what is the disadvantage in this apporach: -> find shortest paths (not just lengths) for all delivery routes -> find the intersection paths among all and keep track of the largest weight -> set it to 0 then from the total cost , subtract , largest_weight*number_of_delivery_routes
•  » » 4 years ago, # ^ |   0 Your idea will fail in the 2nd sample. The largest weight will be 5 on the intersection paths but that's clearly not the edge we want to reduce to 0.
 » 4 years ago, # | ← Rev. 2 →   0 There is another way of thinking the partitioning of the given set in Problem E. Instead of dividing $\binom{n}{n / 2}$ by $2$, we can fix some arbitrary element, let's say $1$, and count in how many ways we can choose the other elements in $1$'s subset. That is $\binom{n - 1}{n / 2 - 1}$. This way, we don't have to worry about counting some configurations twice or more. And this idea can be applied in more complex recurrences involving set partitioning, like Bell Numbers, where the divide by $2$ thing won't work anymore.
 » 4 years ago, # |   0 vovuh there is a typo in F's Editorial, the first paragraph:...., we took cnt elements in the row x and our current remainder is cnt.(this cnt should be rem)! Thank you!
 » 4 years ago, # |   0 int idx = -1; for (int i = 0; i < n; ++i) { if (a[i] != mx) continue; if (i > 0 && a[i — 1] != mx) idx = i + 1; if (i < n — 1 && a[i + 1] != mx) idx = i + 1; }In C ,I can't understand this iteration , can anybody please kindly explain this part ? or why does it work?
•  » » 4 years ago, # ^ |   0 This loop only looks at the largest elements because they can surely become dominant. For a pirana to be dominant it needs to have at least one pirana(left or right) that has smaller value. First if cheks for piranas on left. Second if cheks for piranas on right.
 » 4 years ago, # |   0 Wyh Dijstra work well in problem G?Is't it O(n^2 log m)?I think SPFA is faster.
•  » » 4 years ago, # ^ |   0 Dijkstra on sparse graphs: O(ElogV).Running V times is O(VElogV)
 » 4 years ago, # |   0 When will ratings update?
•  » » 4 years ago, # ^ |   0 Same Question
•  » » » 4 years ago, # ^ |   0 It's done already
 » 4 years ago, # | ← Rev. 2 →   0 dp[x][y+1][cnt+1][(rem+ax,y)%k]=max(dp[x][y+1][cnt+1][(rem+ax,y)%k],dp[x][y][cnt][rem]+axy)I am not able to understand this transition. What does dp[x][y][cnt][rem]+axy exactly mean?UPD : Understood. We are looking whether or not it is beneficial to take the current element or not.
 » 4 years ago, # |   0 vovuh What would be the Worst Case Time Complexity for the problem F using this DP approach.
 » 4 years ago, # |   0 Hi, for problem F, I submitted 2 codes. One has dp array initialized to -1, other has it initialized to -2.-1 works and gives Accepted but -2 doesnt. Why so?-1 => 96209145-2=> 96263305
•  » » 4 years ago, # ^ |   +4 memset(dp, -2, sizeof(dp)); does not do what you think it does.memset sets the value of every byte, not the actual element on the array. 0 and -1 works as intended because their binary representations have all zeroes and ones respectively.
•  » » » 4 years ago, # ^ |   0 Ohhhhh..... thanks a lot. I understood now. so memset(dp, 5, sizeof(dp)) is also wrong?
•  » » » » 4 years ago, # ^ |   +1 It's not "wrong", it will do what it's supposed to do- set all bytes to 1, 0, 1 continuously.It won't replace every elements with 5.
•  » » » » » 4 years ago, # ^ |   0 oh so for array of bits of size 5 called dp, memset(dp,5,sizeof(dp)) will do-> 1,0,1,1,0 ??
•  » » » » » » 4 years ago, # ^ |   +1 Bytes, not bits. So, every 8 bits of the array will be converted to 10100000.I'm not sure if that's reliable enough though.
•  » » 4 years ago, # ^ |   +1 memset(dp, -2, sizeof(dp)); -2 is illegal， only 0, -1 or some Hexadecimal number like 0x3f3f3f3f
 » 4 years ago, # |   0 MyCode Can anyone help me? My code almost same as the standard solution, but get wrong answer on test 3
 » 4 years ago, # |   0 the tutorial for problem C is wrong for the example test case:n=5; arr=[5,3,4,4,5];resultant output=5 expected output=3any helps regarding this?
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 Both are correct!If the output is 5, 5 eats 4, 3, 2, 1.[5,3,4,4,5] -> [5,3,4,6] -> [5,3,7] -> [5,8] -> [9]If the output is 3, 3 eats 2, 4, 5, 1.[5,3,4,4,5] -> [5,5,4,5] -> [5,6,5] -> [5,7] -> [8]
 » 4 years ago, # |   0 96297177 code for F with comments
•  » » 4 years ago, # ^ |   0 Nice!
 » 4 years ago, # | ← Rev. 3 →   +1 The problems were fine, but I don't think E is an appropriate problem, 2 reasons.1)99% Math (Which is still fine, as most algorithm problems can be restated as Math problem, though this was a bit direct)2) OEIS (I hate using OEIS, and I hate problems that allows to use OEIS, because there is literally no fun in it, and we don't learn anything new and you are caught when asked to do the same problem in onsite, but Div3 is a sprint race, when most of them use oeis and you don't, your rating suffers)P.S. You don't even have to think of a few cases, you can copy paste 12164510040883200 in the OEIS search
•  » » 4 years ago, # ^ | ← Rev. 2 →   +1 I have no problem with a little of combinatorics, but OEIS feels like cheating. And our friend 1000000007 would have easily done away with that.
 » 4 years ago, # | ← Rev. 2 →   0 If we are (i,j) then why are we taking the state dp[i][j+1]? Shouldn't we take dp[i][j]?
 » 4 years ago, # |   0 This is the second time I joined in the Codeforces contests. Such a nice one!
 » 4 years ago, # | ← Rev. 2 →   0 I don't understand why the answer is dp[n][0][0][0] in problem F
 » 4 years ago, # | ← Rev. 2 →   0 In problem F: The transitions from the last element of the row are almost the same, but the next element is a[x+1][0] and the new value of cnt is always zero.Why cnt is equal to 0, even if we take the current element?
 » 3 years ago, # |   0 my solution for problem G is somewhat same as that given in the tutorial, however, I am getting TLE. any suggestion? link to solution — https://mirror.codeforces.com/contest/1433/submission/104609982