### vovuh's blog

By vovuh, history, 3 years ago,

1077A - Frog Jumping

Tutorial
Solution

1077B - Disturbed People

Tutorial
Solution

1077C - Good Array

Tutorial
Solution

1077D - Cutting Out

Tutorial
Solution

1077E - Thematic Contests

Tutorial
Solution

1077F1 - Pictures with Kittens (easy version)

Tutorial
Solution

1077F2 - Pictures with Kittens (hard version)

Tutorial
Solution

• +27

 » 3 years ago, # | ← Rev. 2 →   +64 In spite of being (of course) an overkill for problem F1 and F2, it's really interesting to point out that many problems of that kind are solvable with a linear programming approach. For example see this 45824522.Of course, problem F2 has to be solved with a randomized-pivoting version of simplex algorithm (mine's is much simpler and shorter so it TLE's on Test 37).The thing is that when you have this kind of conditions, the problem actually becomes solvable, because the matrix you have is totally unimodular (I didn't try to prove it this time, but I relied heavily on Problem D "Delight of a cat" from 2016-2017 NEERC Problemset).Specially in online-contests on which you are allowed to copy-paste these long pieces of code, it's useful to be aware of this, because the actual writing part of the program is indeed straightforward.
•  » » 3 years ago, # ^ |   +11 Wow, that's quite cool and interesting!
•  » » 17 months ago, # ^ |   0 can anyone please explain this simplex algorithm? which type of problem can be solved using this algorithm?
 » 3 years ago, # |   0 Could someone explain to me the logic behind this if statement in problem C: if (sum % 2 == 0 && sum / 2 <= MAX && cnt[sum / 2] > 0)?.
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 Let i be the index of the number being removed from the array. Hence, the new sum will be equal to sum(original sum)-arr[i]. Now, let arr[j] be the number for with the property holds true. Therefore, arr[j] = sum(new sum) — arr[j] => arr[j] = sum/2. As all the numbers in the array are integers with their maximum value being 10^6, if such an arr[j] exists, sum should be divisible by 2 and sum/2 should be less than or equal to 10^6. cnt[sum/2] simply tells whether such a number exists or not.
•  » » » 3 years ago, # ^ |   0 why do we have check sum/2 is less than 10^6?
•  » » » » 2 years ago, # ^ |   0 we are looking for a number that equals to the rest of the array. so sum of all elements(including that number) sum = (our number plus rest of array). sum/2 = our number and sum/2 = sum of rest array. but our max value of number equals to 10^6 so that why we check sum/2 < 10^6. idk if it helps, but I tried
 » 3 years ago, # |   0 I am having trouble understanding tutorial logic,can anyone explain me problem 1077C — Good Array, Thanks
•  » » 3 years ago, # ^ | ← Rev. 8 →   0 So, think in this manner. For every i, can it be a nice index or not ? How can we think to solve it for every 'i' in reasonable time ? Well, let's try to think what all do we need to calculate to answer that if 'i' is nice or not? So, we need to remove 'i'th element and then check whether the remaining sum is exactly twice of a single element in the array(after excluding ith element).Finding the remaining sum is easy.(We can have a global sum variable and we will subtract a[i] from it).Now, How to check whether this is twice of single element or not ? (hash table ?) Sure, we can have a cnt[] array which will say, cnt[5] = 3, that means three 5's are in array. (think of it as occurence/ frequency array).Now, how this will help in finding the previous question ?Ans: We have sum(remaining sum after excluding ith element), we can check that if(cnt[sum/2] >1 and sum%2==0). i.e.(sum should be even and there should be an element in array equal to sum/2). But there is a catch. It could happen that ith element itself = sum/2. So. we need to tackle that too. Sure, we can have nasty if else.Tutorial presents an elegant way, which is:Find the sum(remaining sum)subtract this element from cnt array(by decrementing it)Now check for if(cnt[sum/2] >1 and sum%2==0)again add this elememt to cnt array.again add a[i] to sum.In case of some doubt, feel free to ask.
•  » » » 3 years ago, # ^ |   0 can you provide me solution in java?
•  » » » » 3 years ago, # ^ |   0
 » 3 years ago, # |   0 Can someone please explain why this code gave WA for problem F1 ?
 » 3 years ago, # |   +3 Another greedy approach for problem E:If we have k distinct topics, let's call number of problems for j-th topic k[j] such that 0<=j<=k-1.the following solution is built on an observation: if the optimal solution is to create X contests, then these X contests could be made from the X topics which have the largest number in array k.So we need to sort array k to have the largest number in the beginning.Then we need to know for every X (number of contests) what is the maximum number of problems we can use, let's call this number M.We already know the answer to make 1 contest M = k[0]. Then for making 2 contests: M = min(M/2, k[1]). Generally M = min(M/2, k[X-1]). And finally for every X compute the answer as follows: ans= max( ans , M * ( 2^(X+1) — 1 ) ).
•  » » 3 years ago, # ^ |   +1 Great solution! Easier to understand than the tutorial one, thanks mate !
•  » » 3 years ago, # ^ |   0 I solved the problem using exact same approach! However, I cannot understand the solution in tutorial :(
 » 3 years ago, # |   0 Can anyone please tell me why problem F2 cannot be solvable with Segment Tree(or any RMQ Data Structure) ?
•  » » 3 years ago, # ^ |   0 Author was not saying it was unsolvable using segment tree, just that their method was easier to code
•  » » 13 months ago, # ^ |   0 i got TLE with these data structures :)
 » 3 years ago, # |   0 Another solution for D without Binary search using priority_queue, 45895914.
•  » » 3 years ago, # ^ |   0 why are you doing f[i]/j? please explain.
•  » » » 3 years ago, # ^ |   0 To include every possible solution can get it from these Frequency, suppose the freq = 5, so i push 5, 2, 1, 1, 1 in priority queue if freq 1 is in the solution this guarantee that 2 and 5 in the solution so the best is to take this number 5 or k times in this case.
•  » » » » 3 years ago, # ^ |   0 Ok, I think I got it. It should work this way. when freq 5 is chosen it says that we can have five copies of that number at max but since if k is greater, 2 will be chosen which implies that we can have 2 copies of that [number, number].Same goes dynamically for every number. This way we'll have an array having numbers which will have maximum copies.Your solution is small, clean and much easier to understand. Thanks!
•  » » » » 17 months ago, # ^ |   0 Masalmah How does that maximize the answer?
•  » » 3 years ago, # ^ |   0 Can you explain me your idea , please i want to learn
•  » » » 3 years ago, # ^ |   0 if we can divide the freq of any letter in small group of freq(with same letter), then if the smallest one is in solution the greater also in solution.
•  » » 11 months ago, # ^ |   0 thanks this is easy and understandable.
 » 3 years ago, # |   0 Can anyone tell me why am I getting wrong for problem D? https://codeforces.com/contest/1077/submission/45898888
•  » » 3 years ago, # ^ |   0 Got it, thanks anyways
 » 3 years ago, # |   0 solved F2 using DP DnC
•  » » 3 years ago, # ^ |   0 what is DP DnC ?
•  » » » 3 years ago, # ^ | ← Rev. 2 →   +3 It's a dynamic programming optimization technique. The DnC stands for divide and conquer. You can learn more from here https://codeforces.com/blog/entry/8219
•  » » » » 3 years ago, # ^ |   0 Thanks Kelly! Your comment helped me to learn Divide & Conquer Optimization and solve this beautiful problem for Vova. ^_^
 » 3 years ago, # |   0 I am having trouble understanding the logic of 1077D — Cutting Out. Can someone please explain it and help me. Thanks!
•  » » 3 years ago, # ^ |   +3 NOTE: I was not able to solve it during the contest, but I was able to do it afterwards. This is one type of binary_search problem. You will encounter many such problems. General binary_search which you might have studied(or will study) in the university is done over an array of numbers. In this kind of problems, you need to perform the binary_search over the entire solution space. For e.g.: In this problem, you can consider the entire solution space as all numbers ranging from 1 to n (as you cannot have more than "n" copies). So, step 1 of this problem is to find the maximum number of copies which can be cut out from the the given array. Let me call this value as val. How we are going to find val, I will come back to that later. Let us move to Step 2Step 2Now, given that if we want to cut out val copies from the given array, can we do that ?In order to answer this question we need to first construct a frequency array (In this we will store the frequency of all given numbers). We need to construct an array of length k We iterate over all elements over the frequency array. First we check that whether the given key in frequency array has value more than val. If it has a value more than val. Then this can be used in the construction of the array. But there is an interesting catch here. Since, duplicates are allowed in the given answer, we might be able to use the same value again. So, we can use that value atmost floor(value/val) number of times. So this is basically the can function being talked in the above tutorial.Now coming back to Step 1.We need to figure out what values of val are valid. Minimum Val = 0 Maximum Val = n If we do a binary_search over this, then we will be able to get the MAXIMUM possible value of val.Another thing to note is binary_search can only be done when the array(solution space) is monotonically increasing. In this case, as mentioned in the tutorial, if you can create, val copies you can also create val - 1 copies. Here is my solution: http://codeforces.com/contest/1077/submission/45915120
•  » » » 3 years ago, # ^ |   0 Thanks!!!
•  » » » » 3 years ago, # ^ |   0 Hello:Please help me honeysingh123 What do you recommend me?: Lately I have had problems when it comes to solving problems, it's like I'm at a stop, and I can not go any further, but I really like competitive programming, and I'd like to keep improving my skills, so I ask you what do you recommend me ?, Thanks in advance
 » 3 years ago, # |   0 I do not understand F2"each segment of the news feed of at least k consecutive pictures has at least one picture reposted by Vova;",if n=4,k=2 then we should choose one of 1,2 and one of 3 4.or choose one of 1 2 and one of 2 3 and one of 3 4?
•  » » 3 years ago, # ^ |   0 I know it now.
•  » » » 3 years ago, # ^ |   0 Can u tell me ?? Unable to understand it properly
 » 3 years ago, # |   0 Can anyone please explain the problem statement of 1077B, I find it confusing especially that explanation of first test case and what does k pairwise district flat means? please illustrate with a test case. Thanks in advance!
•  » » 3 years ago, # ^ |   0 As I understood pairwise distinct simply means that none of them should be counted (turned-off) twice or more.Let's consider the first example101 1 0 1 1 0 1 0 1 0(Their index is 1-based)As the statement, No. 3 and No. 6, No. 8 are disturbed (because both people next to them turned on)To make sure there are nobody "disturbed", you must turn off some lights. One way : we can turn No. 2, 4, 5, 7, 9 off. This way, the testcase turns to101 0 0 0 0 0 0 0 0 0This was simple way to do the task : turn every light off which contributes making someone 'disturbed'However, we can do better. note that if we turn No.2 off, we don't have to turn No.4 off. No. 3 is already in peace.Note one more : if we turn No.7 off, we can make both No.6 and No.8 'not disturbed' by turning only one light off.If we turn 2 and 7 off,101 0 0 1 1 0 0 0 1 0Nobody is disturbed.If we turn 4 and 7 off101 1 0 0 1 0 0 0 1 0Nobody is disturbed.We can't do better (i.e. task is impossible by turning 1 light off), so the answer is 2.Hope this is helpful :)
 » 3 years ago, # | ← Rev. 2 →   0 In Q.1077-D I can't understand exactly how are we calculating value of 'val'.Can someone please explain it to me. I used a brute force approach for that and it's getting TLE.
•  » » 3 years ago, # ^ |   0 Figured it out...thanks anyway
 » 3 years ago, # |   0 I can't understand, why in problem D solution this line is written ??" if (!can(r)) can(l); " at the just end of binary search!
•  » » 3 years ago, # ^ |   0 cause in binary search, the solution used l = mid; if l = 2, r = 3, mid = 2 and can(mid)=true, this will be infinity loop.so you can see the loop ends where r-l>1 and after that r maybe an answer and l must be an answer.but at condition above, we didn't check whether r can be an answer, if it can, this will be the real answer because r>=l
 » 3 years ago, # |   0 I am trying to submit solution for problem c in C but I am getting runtime error(46049311). Can any body let me know why this may be coming. It runs fine in my laptop.
•  » » 3 years ago, # ^ |   0 Got the issues after compiling with clang instead of gcc.
 » 3 years ago, # | ← Rev. 2 →   0 I understand the solution of problem F1, but I don't know why my source code got 'Wrong Answer on test 45'(46061836).Can anyone please explain me the reason? I've been thinking about this for two days, but I have not found the answer :(
 » 3 years ago, # |   0 Sorry but please explain to me why the dp[i][j] can equal to dp[p][j+1] + a[i-1] // why (j+1) and a[i-1] // Thank you so much
 » 3 years ago, # |   0 vovuh please add this editorial to the dashboard . First I thought editorial is not published but when I searched it on google I found it .
 » 3 years ago, # | ← Rev. 2 →   0  comment was incorrect.
 » 3 years ago, # |   0 can anyone please explain why i am getting WA. my approach: iterate through X let it be x and iterate through n let it be i. Then find maximum elemment in dp[i-k-k-1: i-1] [x-1], add ara[i] with it and save in dp[i][x]. submission
 » 3 years ago, # |   0 C isnt just a max segment tree?
 » 3 years ago, # | ← Rev. 2 →   +3 Can anyone help, why i am getting wrong answer at test case 12 in ques E. Here is my solution-https://codeforces.com/contest/1077/submission/54516795
•  » » 3 years ago, # ^ |   0 I found the mistake.Here is accepted code-https://codeforces.com/contest/1077/submission/54522333
 » 2 years ago, # |   0 I can't understand solution of D please someone explain it
 » 2 years ago, # |   0 Getting TLE on using unordered_map, while AC on using map for problem E. Can anyone explain why??(I think unordered_map should be faste than map bcz its time complexity is O(1) for all operation.)Link for TLE SolutionLink for AC Solution
•  » » 2 years ago, # ^ |   0 unordered_map's time complexity is O(k) where k is the size of the bucket, usually it is small hence we consider it to be O(1). In your case, there must've been alot of collison hence getting tle. Ordered_map is LogN always.
»
20 months ago, # |
0

why did i get runtime error in this code

# include <bits/stdc++.h>

using namespace std;

int main() {

int n,i,c=0;
vector<int> v(n);
cin>>n;
for(i=1;i<=n;i++)
cin>>v[i];
for(i=1;i<=n-2;i++)
{
if(v[i]==1 && v[i+1]==0 &&v[i+2]==1)
{
c++;
v[i+2]=0;
}

}
cout<<c;
return 0;


}

•  » » 20 months ago, # ^ |   0 increase the size of vector by one in beginning i.e write v(n+1) instead of v(n)
•  » » 20 months ago, # ^ |   0 such problems should be first looked at by you only.. this will improve your debugging skills and refrain from asking such "SYNTAX ERRORS" doubts and start putting some more efforts.
 » 19 months ago, # |   0 import java.util.Scanner; public class test40 { public static void main(String[] args){ Scanner scan = new Scanner(System.in); int t = scan.nextInt(); double z[] = new double[t]; for (int i = 0 ; i < t ; i++) { double a = scan.nextDouble(); double b = scan.nextDouble(); double k = scan.nextDouble(); if (k % 2 == 0) z[i] = (a - b) * (k/2); else z[i] = (a - b) * (k - 1)/2 + a; } for (double x : z) System.out.println(String.format("%.0f", x)); } } Look at my code. what is wrong with it. it is showing wrong answer at 2
 » 11 months ago, # |   0 May anyone please tell me two criteria used 1. on what basis binary search was done 2 . what was second criteria that was used to move the left of the array or the right of the array after calculating the mid value
 » 10 months ago, # |   +3 Hey guys. I used max queue + DP to solve 1077F2 - Pictures with Kittens (hard version) in $\mathcal{O}(nx)$. Can someone please tell me why 106266241 TLEs and why 106266040 passes? Both codes are the same. The first submission is in C++14 (TLEs for TL = 2500 ms) and the second one is in C++17 (64) (which passes in just 982 ms). Why is the difference so huge? What can I do to avoid this in the future?
 » 7 months ago, # |   0 SOLVED THIS TODAY ( QUESTION C ) :FOR FUTURE PPL WHO MIGHT ATTEMPT THIS ( since the editorial has a weird soln) :In C , try to think that we can only converge the sum to -> Largest and second largest value .My soln. : 114659984