In this problem you should just code what was written in the problem. For every point you can check if it is supercentral. Consider every point consecutively and find neighbors from every side. The complexity is *O*(*N*^{2}).

This problem can be solved using binary search for the answer. obviously, if number *v* is an answer than every number *w* > *v* is also the answer, because the number of written lines of code could only become more. To check some number *v* you can use formula given in the problem, because it will have less than *O*(*logN*) positive elements. The complexity is *O*(*log*^{2}(*N*)).

165C - Another Problem on Strings

You was to find number of segments [*lf*;*rg*] of the strings on which the sum equals to *k* (we are working with array of integers 0 and 1). We will count array *sum* where the value *sum*[*i*] equals to sum on segment [0;*i*]. We will count the answer going from left to right. Let's say we are in position *pos*. Now we will add the number of segments on which the sum equals to *k* and ends in position *pos*. To do it we will count array *cnt* where *cnt*[*i*] — number of occurrences of *sum*[*i*]. Then in position *pos* we add *cnt*[*sum*[*pos*] - *k*] to the answer. The complexity is *O*(*N*).

Beard-graph is a tree. It consists of one root and several paths from this root. There is a single path between every pair of vertices. That's why you should check whether every edge on the path between two vertices is black. If some edge is white there is no path between two vertices now.

The distances could be found separately. For every vertex *v* precalc such information: index of path from the root where *v* is situated and *d*[*v*] distance between root and *v*. If you know such information you can find distance between any two vertices.

To check whether every edge on the path between two vertices is black we will use segment tree. Mark black edge with value 0 and white with 1. Than repainting some edge — update in some point. The query — sum on some segment (the path between two vertices). If the sum equals to 0 there is a single path. Else the answer is -1 now. Complexity is *O*(*NlogN*).

Consider some number *x* from the array. Inverse all bits in *x* and say it is number *y*. Consider an integer *a*[*i*] from array. It can be an answer to the number *x* if for every position of zero bit from *y* there is zero bit in *a*[*i*] in the same position. Other bits in *a*[*i*] we can change to ones.

Then we will use such dynamic programming *z*[*mask*] = {0, 1} which means if we can change some zero bits to ones in some integer from given array *a* and get mask *mask*. Initial states are all integers from array *a* (*z*[*a*[*i*]] = 1). To go from one state to another we consider every bit in *mask* and try to change it to one. The length of bit representation of all integers in *a* is less or equal than 22.

To answer the question YES or NO for some number *x* you need to get value [*z*(*y*)&(1«22) - 1] (inverse all bits in *x* and make the length of the number 22). If you want to know the exact answer what number *a*[*i*] you should choose you can save previous states for every state *z*[*mask*] or just save it in *z*[*mask*]. Complexity *O*(2^{K} * *K*), where *K* – length of bit representation of numbers (*K* < = 22).

I'm trying to solve problem D "Beard Graph". I understand the idea but i don't see how to use a segment/fenwick tree in this case. Probably my problem is on understanding what HolkinPV means by " index of path from the root where v is situated". Can anyone give me some help on this please?

Can anyone explain the idea behind this O(n) solution for problem C by Scott Wu: http://ideone.com/sysG1b Thanks.

Actually this solution is incorrect. It gives wrong answer even on test #1, anyway idea is quite simple. If for position

isum of 1's from position 0 toiis equalsands>=kthen you need to find number of positions with sum of 1's equals-k(let's denote it asC[s-k]) and add it to the result. We will get full result after processing all positions.Instead of processing each position individually we can count the number of positions with sum of 1's equal to 1, 2, ... etc. and iterate over them. Then if we are processing some sum equal

swe should add to the result valueC[s]*C[s-k]. This solution will not work fork= 0. In this case we should add value(C[s]-1)*C[s]/2. Why? Because if we've gotC[s]positions with sum equal tosthen one of this position contains 1, so we need to find how many different sub strings we can get fromC[s]-10's.Correct solution: 11487947

why i need to find number of positions with sum of 1's equal s-k??

Can you please again the idea behind adding C[s]*C[s-k] to the result. This part is not clear to me.

Why we are adding C[s-k] to the result ? What does it represent ?

I will try to explain my solution using the following example:

2

10100100010000

First, let's calculate prefix sums of 1's. We get the following values:

11222333344444

Our C table looks like this:

C[0] = 1

C[1] = 2

C[2] = 3

C[3] = 4

C[4] = 5

Now we iterate over C table starting from i = 2 (which is our k) to 4 (which is maximum sum of 1's). Value C[i] is the count of positions which can be right end of our substring. Now we need to find the number of positions which can be left end of the substring. Number of such positions is C[i — k]. For example for i = 3 we get the following possible left and right end positions:

1

0 10 01 0 0 01 0 0 0 0So for any i we've got C[i — k] (left ends) * C[i] (right ends) possible substrings with k 1's. Answer is the sum of such values for all i.

Special case is k = 0. If I have for example C[2] = 3 it means that I have one position with 1 and two positions with 0s. So what I need to do now is to count number of all substrings of these two 0's. If Let's say I have n 0s, then I will have:

n substrings of length 1 n — 1 substrings of length 2 n — 2 substrings of length 3 ... 1 substring of length n

so generally the number of substrings of string of length n is sum of numbers from 1 to n which is n*(n+1)/2.

Thank you for the explanation with the example

Please Explain why is C[0] = 1 in your solution

It is needed for substrings which begin from the first character of the original string. C[0] = 1 means that I have one left end with 0 1's = the empty string. Of course if the original string begins with some 0's the value of the C[0] will be bigger.

Please explain 165C — Another Problem on Strings. Unable to understand approach.

Can anyone explain Problem -C . Not getting the approach

Consider two different cases when k=0 and when k!=0; When k=0, it is simple to calculate the number of strings by storing the position of occurrence of '1' in a vector and then iterate over the string and calculate the answer.(try it yourself it is simple). When k!=0,let's assume dp[i] = number of strings ending at position i and containing k 1's. Make a vector which stores the position of occurrence of 1 and use to calculate the answer. try it yourself.

A lot of sources which got AC at problem D returns different answers on this input :

7 1 5 6 4 2 3 3 5 5 6 3 7 1 3 4 2

I think is a good idea to introduce this test-case as official test.

UPD :Three samples of codes which got AC : First Second Thirdcan someone explain more detail problem E?thanks.

Let's suppose you have a number X and we invert the binary digits of X, now we have a number Y that has some 1s and some 0s, we want to find another number Z that has a 0 in every position where Y has a 0. It is important to note that every 0 in Y must strictly be a 0 in Z, but each 1 can be either a 1 or a 0 in Z.

A naive solution would be for every number Y in the range [1-4*10^6] iterate over every number Z in the same range, and for every valid pair, check if that number exists in a, if it exists you found an answer for Y, otherwise it's -1. Then you can just iterate over a and look at the solutions you precalculated by inverting the bits of each a[i].

Of course, this wouldn't work because it's too slow, time complexity would be O(2^2*maxBits) if you iterate over all numbers in the range [1-4*10^6], or O(3^maxBits) if you use bitwise operations to ignore invalid Zs. Both too slow, so we need to optimize it.

We will take similar idea, but change it slightly, for every Y that has an answer we say that we can propagate the answer to another number W, if and only if Y&W=Y and W has exactly 1 more bit turned on compared to Y. This works because as we stated earlier, we want Z to have zeros where Y has zeros, but if Y has a 1, we don't care what we have in Z, and because W is the same as Y, except that one 0 is now a 1, we know that Z will still be valid. We code this idea with DP and it becomes O(maxBits*2^maxBits)

http://www.codeforces.com/contest/165/submission/18102743

thanks for great explanation.

can u tell me how dp works? what is dp[i]? why dp[input[i]] = input[i]? thank you.

The man who gave such great explanation didn't get any upvote(before I gave one), but the man who thanked him for his great explanation got 5 upvotes(now 6 including mine) !! Just wow !!!

I get the idea in problem D but i don't understand the segment tree part. Can anyone help me out with this part?

Worrst tutorial ever!No one couldn't understand any problem specially problem C.

Read code of top rating, maybe it helps (at least for me) =))