### vovuh's blog

By vovuh, history, 5 years ago,

1095A - Repeating Cipher

Tutorial
Solution

1095B - Array Stabilization

Tutorial
Solution

1095C - Powers Of Two

Tutorial
Solution

1095D - Circular Dance

Tutorial
Solution

1095E - Almost Regular Bracket Sequence

Tutorial
Solution

1095F - Make It Connected

Tutorial
Solution
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 » 5 years ago, # |   0 In problem C, what if the constraints for n and k are 1018 ? Is there any other better solution possible since the given one will run out of space and time !
•  » » 5 years ago, # ^ |   0 since in worstcase you need to print k numbers... there is no way... if we change the output format to print tuples (2^x, count) it is possible
•  » » » 5 years ago, # ^ |   0 Yes ! Completely forgot about it (maximum allowed output constraint)! Thanks..
 » 5 years ago, # | ← Rev. 2 →   0 I would like to know if my AC algorithm for problem D could actually be hacked somehowhttps://codeforces.com/contest/1095/submission/47607217What the program does is: If the I kid remembers kids A and B in front of him, that means I is before A and B, so we draw two directed edges I->A, I->B. There is an order to be decided between A and B that we don't know of yet, so we also draw a pointed line A---B between them. When all the lines have been drawn, we iterate over each directed edge U->V, and if there was a pointed line U---V, then i assume the order U->V is indeed correct, so we add this directed edge U->V to an answer graph G. Note that there might be multiple solutions, so U->V and V->U might end up existing at the same time. When we finished generating this answer graph G, visit all the nodes, starting from 0 and going to a any non-visited neighbor that we can go to through an edge I think the go to any part might actually be hackable
 » 5 years ago, # |   0 In problem C, the for loop is from i=0 to i=30. As we know that 'int' is of 32 bytes,so I am unable to understand as to why the loop is not from i=0 to 31. Also I saw the codes of many programmers for the problem C and nearly everybody has done it from 0 to 30. So please explain the reason for this.
•  » » 5 years ago, # ^ |   +5 230 = 1073741824. You don't even need 30 in this cycle, 29 is pretty enough, because for numbers up to 109 the maximum power of two in their representation is 229 = 536870912.
 » 5 years ago, # |   0 Where is editorial for problem 1095C - Powers Of Two?P.S: "Unable to parse markup [type=CF_MATHJAX]" in spoiler.
•  » » 5 years ago, # ^ |   +12 Oops, sorry. Fixed now.
 » 5 years ago, # |   0 So my solution for D was completely different... I created a matrix in which placed the 2 neighbors for each kid. But how? Well, for example, if the next 2 kids after the first one are 3 and 5, 3 will be a neighbor for 5 and vice versa. Now i could just pick a number, go through this matrix and find the next one. But one more problem.. Sometimes the result was counter-clockwise, so i just added an extra condition : so for the first kid, i would go through that matrix and see which one of those is a 3 or a 5 (one of them), then pick it and do the same thing for it and so on. It was such a rushed solution, yet it worked and i am really please with it ;)
 » 5 years ago, # |   +3 anyone please explain Problem E..I am unable to understand "if conditions in loop" especially.
 » 5 years ago, # |   +5 Problem E can also be solved easily using a segment tree+lazy updates
•  » » 5 years ago, # ^ |   +2 actually just segment tree works, without the lazy updates.
•  » » » 5 years ago, # ^ |   +1 true, didnt think of that thanks
•  » » 5 years ago, # ^ |   0 Then i wonder how it is implemented
•  » » » 5 years ago, # ^ | ← Rev. 2 →   0 we convert the string to +1 and -1 (+1 when opening parens, -1 when closing parens) then we compute the cumulative sum array, and then we put the cumulative sum array in a segment tree. To check if changing the i-th parens "fixes" the sequence we do the following: if the i-th element in the string is opening parens we subtract 2 from me segment [i,n] and if its a closing parens we add 2. (this simbolizes changing the parens to the oposite one) then we do rmq(0,n) and rmq(n,n) if rmq(0,n)<0 or rmq(n,n)!=0 then its an invalid sequence
•  » » » » 5 years ago, # ^ |   0 the way i did it was in each node of the segment tree, I stored i>The number of valid pairs in the range ii>The number of unpaired opening brackets iii>The number of unpaired closing brackets While merging the nodes, observe that the unpaired opening brackets in the left can be paired with unpaired closing brackets in the right child.In the end, just check whether the root has any unpaired opening or closing brackets.The problem is similar to https://www.spoj.com/problems/BRCKTS/
•  » » 5 years ago, # ^ |   0 Please give me the segment tree idea for problem E
•  » » » 5 years ago, # ^ |   0
 » 5 years ago, # |   0 Is there any easy explanation of problem E.
•  » » 5 years ago, # ^ |   0 The problem is similar to this one SPOJ/BRACKETS
•  » » » 5 years ago, # ^ |   0 I am getting TLE with segment tree.
•  » » » » 5 years ago, # ^ |   0 Here's a nice article about the implementation with segment trees (https://www.geeksforgeeks.org/range-queries-longest-correct-bracket-subsequence/)
 » 5 years ago, # |   0 In E, can you explain your choice of indices for this part: for (int i = 0; i < n; ++i) { pref[i + 1] = pref[i] + (s[i] == '(' ? +1 : -1); okp[i + 1] = okp[i] & (pref[i + 1] >= 0); suf[n - i - 1] = suf[n - i] + (rs[i] == '(' ? +1 : -1); oks[n - i - 1] = oks[n - i] & (suf[n - i - 1] >= 0); } It seems arbitrary to me whether you'd do that or do something like pref[i] = pref[i-1] + (s[i] == '(') : 1 : -1). The code is actually different, but I don't understand the difference.
 » 5 years ago, # |   0 For Problem E with test case 8 )))((((( why is the answer 0? Can't I change it to ()()()() with modifications at positions 1, 3, 4, 6 and 8?
•  » » 5 years ago, # ^ |   0 Because you can and only can change exactly one bracket according to the problem description
•  » » » 5 years ago, # ^ | ← Rev. 2 →   0 one type of bracket or just any one bracket (i don't get the logic for above example) , also then how come answer to 6 (((()) is 3
•  » » » » 5 years ago, # ^ | ← Rev. 2 →   0 Just one bracket. Possible for (((()) : ()(()) ; ((())); (()())
•  » » » » 7 months ago, # ^ |   0 void void void.....
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 Hi aakash have you understood this testcase Please clarify this problem if you got it .
 » 5 years ago, # |   0 In problem C what is the function of map ? and what it is doing in program i don't understand
•  » » 5 years ago, # ^ |   0 The map stores the number of occurrences of various powers of 2. It comes in handy when we want to split a set-bit ( a power of 2 ) into two lesser powers of 2.
•  » » » 3 years ago, # ^ |   0 jatinmunjal2k, Resonite here, big fan bro....
 » 5 years ago, # | ← Rev. 3 →   0 In problem E, in the first example, why the answer can't be ()(())? this is a regular bracket sequence
•  » » 5 years ago, # ^ |   0 It can. You can flip positions $2$, $3$, and $4$ (1-indexed) while being valid in that test case, and your answer is a result of flipping the position $2$.
 » 4 years ago, # |   0 In Problem D, why is the counter clockwise not acceptable?