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How is (a|b)-b == a&(~ b) ?
you may see it with an exemple :
f(5,3) = 5|3 — 3 = 7 — 3 = 4
binary representation : 101 &(~011) = 101 & (100) = 100
A | B has all the bits where A or B have 1, set to 1.
A | B — B has only those bits set to 1 where A has a 1 but B doesn't have a 1. If B had a 1, then A | B would have a 1 in that position and — B would set that bit to 0.
So it is equivalent to A & ~ B
i got it (a|b)-b=a&~b,but how to approach towards the answer from there
Only the number which you put first is taken as it is and the others have ~ added to them. So maintain prefix and suffix of ~ &. Answer would be maximum of pref[i] & $$$a_i$$$ & suff[i] for all i. pref[i] is the value of all the elements from 1 to i-1(using the ~ & operation) and suff[i] is the value of all the elements from i+1 to n(using the same operation).
The editorial explains it better so you might wanna read that :)
Check this submission for reference.
When you put INT_MAX as pref[0] and suff[n-1], how come it does not affect the final output? pref[i]&arr[i]&suff[i] would probably be bigger when pref[i] is INT_MAX?
https://codeforces.net/contest/1300/submission/78506793 not necessary to take int_max as pre[0] && suff[n-1] it will not make any sense,check my solution(^_^)
Of course it can be done without it, my question was, why the INT_MAX didn't affect the final result.
int max is 2,147,483,647 in binary it is (1111111111111111111111111111111) so if it is( & with ~ar[0]) the result will be ~ar[0],think over it
Nice. Got it. Thanks
Understood this one rather than the editorial. Thanks!
Nice explaination
a|b can be written as: a + b — a&b
subtract both side by b =>
or a|b — b = a + b — a&b — b
or a|b — b = a — a&b
or a|b — b = a&(1-b)
or a|b — b = a&(~ b)
Edit : Note that 1 here represents all the bits where a is set as any set bits apart from it is out of our concern
In 2nd last statement, you have used the identity that a&b — a&c = a&(b-c). How do you prove this?
Hint : a — b = a&( ~ b)
Although, the common element I took out was U or INT_MAX Edit : Just to clear out confusion here ,
I didn't prove a&b — a&c = a&(b-c) ,
rather a&U — a&c = a&(U-c) , where U is considered as all bits set
How is a-b = a and (not b)? Take a=5 and b=3. a-b=2 and a and (not b)=4
I am mentioning this again, Note that b is a subset of a in my cases all over
Back to real one: a&b is a subset of a
Why we are taking a common element and why are we taking it as INT_MAX? Please explain this part.
https://codeforces.net/contest/1300/submission/78506793 not necessary to take int_max as pre[0] && suff[n-1] it will not make any sense,check my solution(^_^)
Can you refer to any good resource to study bitmasking?
Apart from practice sites which you can easily google,
I really like the content of this website for bitmanipulation tricks :BitHacks
CSAPP's Datalab
For (a|b), let's define those bit positions which is 1 in
bas POS, what (a | b) does is to make those POS inabecome 1 (0 — > 1, 1 — > 1), and what (-b) does is to make those POS inabecome 0 (1 -> 0), so as a result, what (a|b)-b does is to make those POS intabecome 0, which is equal to a&(~ b).Corrected Solution :-
(a|b) — b
Now, a-b = a and (not b)
So
a|b — b = (a or b) and (not b)
= (a and (not b)) or (b and (not b))
= (a and (not b)) or (0)
= (a and (not b))
= a & (~ b)
thanks! helped me so much!
(a|b)-b = a^b — b = ab' + a'b — b = ab' + a'b — b.(a + a') [since a+a'=1] = ab'+ a'b — ab — a'b = ab' — ab
= a.(b' + b') [since b' = -b] = a.b'
[here, a' means complement of a]
hence, the proof.
You can imagine it using Venn Diagram if you have heard about it.
Please add a reference to this editorial in the round announcement post so that people can find it easily. Thanks.
Thank you for the wonderful contest. I really liked problem C of div2. The trick was to see what the operation is exactly doing!
I hate when the code is not included.
Finally became a specialist after veryyyy long time ＼(^o^)／
So happy for you dude! I'll hopefully reach specialist in the next contest ♥
Congratulations! Keep going.
Congratulations!!
What is the logic behind Div2/E solution?
I am not able to get it. Someone please help....
orz!!
Guys, does anyone have a non-geometric solution for Water Balance?
Is not it slow to get AC?
Yes. You can see all the water tanks as a succession of groups. Initially all groups have size 1. While you can find 2 adjacent groups, such that the left group has greater average volume than the right group, merge them. Here is my solution. It's not very well written and might be hard to read.
Here is my O(n) solution based on deque.
Using non-decreasing queue to solve this problem is a very simple way. Monotonic queue and monotonic stack both classical algorithms. Thank you for providing such a simple way.
I like your solution. Can you give more insight on usage of 'eps'. What might happen if we will not use it.
I just wanna make sure it will be true when tmp == Q.back().val.
I think it is necessary to use 'eps' only when calculate whether two double number are equal.It mean we can't use == to calculate whether two double number are equal.If we want to calculate, we can use follow way. assume a and b are double.
In my opinion, I don't think it is necessary to use 'eps' when use < > <= >= operator. Sorry my poor English
here is my solution I used stack
Yo should notice the fact that the each number must be smaller then the next numbers
Wish I'd noticed the simplification in C... Ended up using a segment tree on |
I solved Div1_C with CHT.
We want to know minimal value of a[1] so we can binary search it. When we get it we should get minimal value of such a[j + 1] that (a[1] + a[2] + ... + a[j]) / j = a[1] and so on.
Now we must understand how we can efficiently binary search. Let x be value that we want to check. So this inequality should be performed:
(pf[j] — pf[i — 1]) >= x * (j — (i — 1))
Where pf[i] = a[1] + a[2] + ... + a[i] This can be represented as:
pf[j] — x * j >= pf[i — 1] — x * (i — 1)
Left part nothing else than line equation. So we can store lower CHT. When we check x for corectness we should also binary search for line at CHT that contains x.
Thus we achieve complexity O(n * log n * (log 10^6 * ε ^ -1)) where ε is needed precision.
But this solution will get TL. So we need further optimizations.
Note that answer is non-decreasing sequence. So needed x only increase. We can use pointer technique for searching line at CHT that contains needed x and than binary search at some interval.
Thus we achieve complexity O(n log (10^6 * ε ^ -1)).
This is my solution https://codeforces.net/contest/1299/submission/70688494.
I'm interested in your approach. Could you tell me what is CHT?
CHT — convex hull trick. You can read about it here https://cp-algorithms.com/geometry/convex_hull_trick.html
Thanks.
Really fascinating !!. But there is a question I want to ask:
In your code, I saw this :
So this is where you binary search for the minimum answer x. But $$$ps$$$ seem to remain the same during this loop (equal to $$$st$$$). So you always work with line $$$K[st] * X + B[st]$$$ with every $$$x$$$ from 1.0 to 1e6. I have just learned CHT and I think each line should hold the max/min in an only interval, so different $$$x$$$ will have the max/min in different intervals, right.?
Could you clarify this please? Thanks so much.
Yes, each line should hold the max/min in an only interval. But I search for needed interval (such interval where our x lays) with pointer technique and when I found it I start binary searching.
So every x from 1.0 to 1e6 get max/min at that same line? Am I missing something?
No. To this interval belongs only values from [fr[st], fr[st + 1]) (or from [fr[st], +oo) if our line is last). But it makes no difference because value more than right bound we can't achieve (I think it's clear) and value smaller than left bound we can't achieve (because optimal value belongs to other interval that doesn't fit (we check this while moving the pointer)).
Thanks so much! I understood it. And actually we don't need to binary search for x at all. x is the x value of cross point for line -(i — 1) * X + pf[i — 1] and line K[st] * X + B[st]. I test this with your solution and get accepted. Link
I understood your optimization. Thank you too!
About 2E/1C, there is an arguably simpler "greedy" solution using a stack in $$$O(n)$$$
As given in the samples, the solution to
is
Let me write this like
Now suppose we want to solve the following problem (same thing but with an extra 2 at the end)
One possible solution would be
But we can see that this isn't optimal since the 7.5 and 2.0 can be combined making the 7.5 smaller. Combining them gives
Now we see 5.0000000 < 5.66666667 so we wouldn't gain anything from combining again. So we are done! (The proof that this gives the optimal solution is left out to the reader =P)
Here is my implementation 70666728
Awesome!
it is TLE with py3
Not really. This problem is heavily IO bound, and in Python3 a lot of that stuff is slower because they changed to using unicode strings (I'm using Python2 just to easily get around that). But there are ways to make Python3 fast as well.
For example I took your code and added my fastIO template to it 70718663, and now it runs in 1887 ms.
very thanks, it`s awsome
Isn't it more or less equivalent to the editorial solution? Since this is basically how you find the convex hull anyway.
Hahaha, it definitely is xD. These 78 upvotes and "Awesome!" comments xDD
As stated in the editorial, the solution to the problem is a lower convex hull. So any algorithm solving this problem will find a convex hull, and in particular this stack algorithm can be interpreted as creating a convex hull.
However I would argue that it is possible to get an intuitive understanding of the stack algorithm without thinking about it as finding a convex hull. It's this intuitive understanding that I'm trying to highlight.
One stupid question, is it always possible that any problem that has solution using CHT,can also be solved by stack,if yes or no what is the reason.Thanks in advance.
Awesome, Helped a lot !!!
i couldn't find 'so mean' and 'around the world' problems during the contest. what are these problems and why are they in the editorial
Thanks for the useful tutorial <3
Does anyone have a submission to div2E / div1C that runs in time on C++11?
I swear I'm so pressed rn:
By far the majority of the runtime comes from the IO. For example see 70688591 by neal running in 249 ms. He has a custom template for printing doubles. Narut tested to remove neal's fast IO template in 70694340 and now it runs in 1902 ms.
As pajenegod mentioned problem here is IO. Try changing cin/cout. I'm sure printf/scanf with c++11 runs faster. E.g. -
Printf/Scanf gets ac whereas same soln got tle on cin/cout
I thought the ios:: thing at the beginning made cin/cout the same performance as scanf/printf?
Still cin/cout cant match printf/scanf. We are printing more than 15e6 here. Here in c++17 printf/scanf was ~600ms faster than cin/cout on submissions I made during testing.
dang it, thank you very cool
It's true for ints but cin/cout is very slow on doubles.
I read it in as an int though
Love u dori
You didn't output it as an int.
<3
Checking for central-symmetry is equivalent to checking if each pair of opposite sides are equal and parallel. You can do this easily using integers only: no doubles needed.
Check my code here: 71015952
HINT: To prove that central symmetry is equivalent to opposite sides being equal and parallel, we can use congruence of triangles.
C could be solved by repeating the greedy We are going to keep track of the segment sum, and size Make N 1-lengthed segments. Starting from an element, make a new segment by adding elements. If the element is bigger than average, stop and make a new segment. If we complete whole thing, do this again from the newly formed segment array. If it doesn't change, print it. It looks like N^2, but it somehow passes.
Yes,look at mine.Mine is also like a O(n^2) algorithm,but it passed. https://codeforces.net/contest/1299/submission/70672559
and this https://codeforces.net/blog/entry/73763?#comment-579507
Believe N^2
Isn't it the case that the isosceles trapezium after joining accordingly will make a hexagon but is following the property of symmetric around the center. ( Div 2 : D, Aerodynamic)
Isosceles trapezium does not have central symmetry. A central symmetry is when a line passing through the center of symmetry has equal length on both sides of the polygon.
My bad.Thanks though.
i understood a|b-b is equivalent to a&-b but then what we have to do. i was unable to understand editorial. Can anyone explain me in detail. Thanks in advance.
Consider 4 no.s :
a b c d, Now,f(x,y) = x&~yThe question requires:
f(f(f(a,b),c),d)which is essentiallyval = a&~b&~c&~d.Notice that you have to maximize
valby positioning any element at first position and remaining at any position after and for that,you can iterate and for any element
ifind~a[0]&~a[1]...&a[i-1] & a[i] & ~a[i+1]...&~a[n-1].Find when this value is maximized and put the element at i at front. That's your answer
Sample submission : Jiangly
thanks a lot for this clear explanation. it always helps a lot for beginners like us
for array
9 1 2 5the output of Jiangly this code is9 1 2 5which has function value 8 .. but isn't maximum function value 9 is possible with array reordered as5 1 2 9?5 1 2 9gives 4 tho (Think of it like the value cannot be bigger thana[0]because of&property)I don't know the problem of my code.
Help me, please.thanks.
The problem is RUNTIME_ERROR.
https://codeforces.net/contest/1300/submission/70701084
Your array size is 2e4, the input size is 1e5.
Even if Runtime error is removed, your solution will give TLE nevertheless
Use better sorting algorithm
Very thanks.
I use the quhanks.
I use the quick sort.but still TLE
This is my new code,what's the problem?
https://codeforces.net/contest/1300/submission/70725391
Why not use the available sort function, instead implement your own?
I think the quick sort is the fast methed.
The function , I have used.But have too many problem.
I just do not understand why ,one group of 100000 data , the code is right. but five group of 100000 data , the code is wrong.
your sort function is not a O(nlogn) sort when the input array is a strict increasing array like 1 2 3 4 5 6 7 ....
Thanks！
I have a DP solution for 1C .
dp[i].FIR means the smallest average number of an array ai~ar (i<=r<=n)
dp[i].SEC=r
now we can update the dp in "O(n^2)" ,LIKE THIS:
then if we know the smallest prefix average number from each position.The problem become much easier!
We just need to perform the operations from left to right by "Greedy".
You can also see my code in my "submissions".
Although it is like an algorithm of O(n^2),but in fact it is very fast.
Update it is O(n)
It's linear O(n) because in the inner loop you merge some ranges so you need to merge n times in total
Yes.thank to beacon_cwk.I have confirmed that it is O(n)
Please elaborate a little if possible!
EDIT: Actually, never mind.
Your argument proves that for any $$$(x_0, y_0)$$$ in $$$T$$$ you can embed the vector $$$\left(\frac{x_0}{2}, \frac{y_0}{2}\right)$$$ inside $$$P$$$. It doesn't prove that the point $$$\left(\frac{x_0}{2}, \frac{y_0}{2}\right)$$$ actually lies inside $$$P$$$, which is what you really need. And this is false in the general case, so your proof has to somehow involve $$$P$$$ being centrally symmetric.
You're right, my mistake. Thank you!
Для проверки того, что выпуклый многоугольник центрально-симметричен, достаточно проверить, что противоположные стороны параллельны. Доказывается через проведение прямой, параллельной рассматриваемой стороне, яерез центр симметрии. Это немного проще, чем пересекать прямые, соединяющие противопложные точки. Так как точки целочислены, то проверять параллельность можно в целых числах.
UPD. Необходима не только параллельность каждой пары противоположных сторон, но еще и равенство их длин.
Дядя Петя, ты дурак? Возьми правильный шестиугольник и параллельно сдвинь одну сторону, не меняя направления других сторон, он не будет центрально симметричен, хотя стороны всё ещё параллельны
Я думаю, что он забыл добавить, что заодно с параллельностью проверить равенство противоположных сторон, т.е. просто равенство векторов.
Можно рисунок параллельно сдвинутой стороны в правильном шестиугольнике? Ну или 6 точек как контр-пример.
UPD. Понял, необходима не только параллельность каждой пары противоположных сторон, но еще и равенство их длин.
А ничего, что пересечения должны быть в серединах отрезков? Так что даже не надо писать код для пересечения прямых
I think Dishwashers should never prepare contests again
Div1D definitely should have been posed without this stupid condition about no cycle passing through 1 of length >3. It can be solved as noted by tfg here: https://codeforces.net/blog/entry/73664?#comment-579444 (that's not the best description, but it can be made working). First thing is that this condition is super ugly and unnatural so that it would be much nicer without it and second thing is that finally there would be something to think about in this problem making it significantly more interesting to solve as well.
Sorry mr grandmaster but I don't understand tfg's solution
Yeah, I understand this description might be quite cryptic, so let me explain how I think of this (or should I say "sorry ms grandmaster, I should have expected this"?)
Delete 1 and for every connected component calculate subspace spanned by cycles in it. Now let's focus on adding edges to one of these connected components and then merge partial results (which are mappings from 374 subspaces to number of ways we can get them in this particular component) to get overall one. This merging introduces *374 overhead to complexity, but that's not something huge (and actually for smaller components we will not get all 374 possibilities, so it will be significantly less in practice).
So how to solve it for one particular components with some edges from 1 added to it? For a fixed component we divide edges into ones from spanning tree and ones outside of it (each such gives one vector to basis). Let's assume we fix our favourite edge from 1 to that component and add it to spanning tree. Every following edge that we decide to add from 1 we treat as non-tree one and it gives us one cycle whose cost we are able to calculate and add it to the subspaces dp. So we are able to solve it if we add *n to complexity by fixing first added edge. Key observation here is that edges actually can be partitioned into 32 equivalence classes. Instead of calculating cost of the cycle directly we can fix some favourite vertex in that component r and calculate cost of new cycle as xor of costs of paths from 1 to r using first added edge and from 1 to r using new added edge. There are only 32 possibilities for cost of path from 1 to r using first added edge what adds *32 factor to complexity for particular component.
Overall complexity for original problem was $$$O(374n)$$$ while now it is $$$O(374 \cdot 32 \cdot n + 374 \cdot 374 \cdot n)$$$, where first part comes from dp for each component and second one comes from merging results from different components (and that part should be actually significantly faster).
Understood, thanks. It's sad that this possibility is wasted
Actually that's not exactly what I had in mind, you can have dp with states [i-th edge][current subspace][height of the first edge you've chosen in the current connected component or if you haven't chosen it yet]. This way it's actually O(374 * 32 * N) because you don't have an O(374) transition to merge stuff, just an O(1) transition but with an additional state.
Ah, right, makes sense, thanks!
Sorry but I have a problem about Div1.D that how to caculate the number of subspaces?
Your text to link here...
The most challenging part is the edges between nodes which are connected to 1; I don't know how for each subspace count the number of ways to delete edges adjacent to 1 such that remaining edges form this subspace
Hello in 1299B, how can i avoid precision errors checking whether the midpoints of segments connecting the opposite vertexes coincide? I am receiving WA test 53
https://codeforces.net/contest/1299/submission/70740871
Don't divide by 2.0
Can someone pls explain the solution to aerodynamic? I couldnt quite get it asto why a figurewith central symmetry would be an ans.
Not sure if anyone discussed it already, but I find the following solution for problem E (Div 2) easier and more intuitive. It is based on the fact that if the prefix is computed optimally, when you are at position i, all you have to do is to iterate from i to 1 (lets say using j) and see if you can obtain a better solution by applying the operation on the segment [j, i]. If you can, then move j, if not, then stop (since prefix is optimal, it means that it is sorted, so you dont need to check smaller indexes since they also have smaller values). Now this is just a brute force, but it can be optimized by representing a subsequence having the same value as a pair of (sum, number of elements). So you end up with O(n) complexity. 70909016
Here how I solved
Div 1 B - Aerodynamic (Div 2 D)in contest:Step 1: Use GeoGebra to plot test 3 and realized that the opposite sides are parallel.
Step 2: Go back to the statement and realized that it's also true for test 1.
Step 3: "Coincidence? Nah, I don't think so. Let's code it"
My submission: 70663347.
I still don't know how to prove it. Could someone help me?
equal and parallel sides imply the central symmetry of the polygon
In 1300C - Anu Has a Function i have used this:-
what will happen if we are unable to find a1? How to take a1 for the solution ?
My approach to C:-
If any bit occurs more than 1 times in the array then it will always be 0 in the final answer so we just need to take in account the bits having count 1 in each number and then sort in descending order each number according to this new value..
My submission 86092229
thanks for the easier explainantion
you are doing this to check the jth bit
and it works for(i=0;i<n;i++) { cin>>a[i]; x=a[i]; for(j=0;j<32;j++) p[j]+=(x>>j&1); }
can you please explain why for(i=0;i<n;i++) { cin>>a[i]; x=a[i]; for(j=0;j<32;j++) p[j]+=(1LL<<j)&x; } this code of mine is failing ![user:dhananjay_049]
The statement p[j]+=(1LL<<j)&x means
shift 1 j bits to the left and then if x also has jth bit set then add 2^j to the p[j]. but here you only need to check if jth bit is set or not .
you can make your code work by just adding a conditional operator like this.
p[j]+=( (1LL<<j)&x ? 1 :0)
thanks d_evilal
Here is a much easier solution for C.
I have gone through all comments/editorial for Problem C(div 2)/ Problem A (div 1). But Not able to understand it completely. Can anyone please explain it in more detail.
What does Z(2,5) mean? (Problem D)