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Hi! Just a small note about problem D.

In the editorial you said to subtract

cnt_{ai}*a_{i}, but then you are addingcnt_{ai}*a_{i}again in the last sum, so you could just omit those two terms.I know it adds the same, but I think is more natural for the understanding to just don't include them.

what is cnt?

cnt_{x}is the count of ocurrences ofvaluex (at the left of i).When you process the

i-thnumber, you then add 1 tocnt_{ai}Isn't the time complexity for E gonna be O(n * n * n * k) ?

n *n -> generating all the possible string 't'.

n *k -> checking it will all the strings s1, s2 ... sn

If this complexity is right, then won't the above solution give TLE ?

It would have been really nice, if the author would have attached a sample submission.

Why it will be

O(n^{3}·k)? Authors solution generatesO(n) stringst, for each string checking will beO(n·k) -> Complexity isO(n^{2}·k). And my solutionYour solution is not accessible.Please correct it.

Fixed, sorry :)

There's a simple observation that can make the E solution a lot faster and less likely to get TLE. When you are determining the indices where

s_{i}≠t_{i}, if there are more than 4 such indices, then the answer is -1 (sincesandtare too far from each other for there to exist a "base string"). Now you'll only haveO(1) candidates for the "base string", and the rest of the approach is quite fast.Oh my god, i use it in my solution but forgot to write it in editorial. Sorry!

Another solution for F:

Use DP state (i, mask) to mean the minimum cost to make everything before column

idots, and the columnsithroughi+ 3 having asterisks in the positions indicated by mask (which ranges from 0 to 2^{16}- 1). The transitions are to clear a square in the mask (1x1, 2x2, 3x3, or 4x4), or, if the bottom 4 bits of the mask are empty, to shift to columni+ 1.There are 1 + 4 + 9 + 16 = 30 transitions of the first type, and 0 or 1 transitions of the second type, so the total time is

O(n·2^{16}·31). Implemented in a top-down (recursive) fashion, this doesn't run in time (submission: 33204536). The worst case is with a 4x1000 matrix, all asterisks, which takes about 5 minutes on my machine.To speed it up, we note that if the bottom 4 bits are empty, the shifting transition is always best: we don't need to check all the square-clearing options. Furthermore, unless we are at

i=n- 4, it is always best to clear squares that border the left edge of the square under consideration. This reduces the possible transitions by about a factor of 2, but it also drastically reduces the number of states visited: in the maximum case described above, our recursion only visits 106371 states, and the submission (33204643) passes in 61ms.You need only 3 columns for DP.

Why are there only those 30 transitions? What if the last 4x4 square looks like this:

Which transition would be used to use 4 1x1 stamps to clear the leftmost asterisks?

Nevermind, I misread your solution.

Can someone explain the logic behind the solution of problem D please?

D can also be solved using a Fenwick tree. Go through the sequence of numbers from least to greatest, updating the tree from 0 to 1 as you go. For each new value, update all previous values more than 1 less than it in the Fenwick tree, then add val*query(0, ind-1) — val*query(ind+1, MAX_N-1) to the final answer. Then do the same thing again, but in reverse sorted order.

I cannot understand your solution my bad,can you explain using an example...

I will do the sample case:

The initial BIT is like this:

`0 0 0 0 0`

We go through the elements in sorted order. So we start at 1, which are at indices 0 and 3. Before we look at 1, we have to update the BIT with all elements that are less than 1-1 = 0. There are none of these, so the BIT is still at

`0 0 0 0 0`

. For the 1 at position 0, we add 1*(0) — 1*(0) to the final answer. For the one at 3, we also add 0 to the final answer.Now we go to 2. The only 2 is at index 1. Before adding the 2, we update the BIT with all elements less than 1, but there are none of these, so we don't update it. As before, add 0 to final answer.

Now we go to 3. They are at 2 and 4. First, update all elements less than 3-1=2. There are 2 of these: the 1 at index 0 and the 1 at index 3. So now our BIT looks like this:

`1 0 0 1 0`

The 3's are here:For the first 3, we query for the sum of the first 2 elements of the BIT, which is 1. Then we query sum of last 2 elements, which is 1. So we add 3*(1) — 3*(1) to the final answer.

For the second 3, we query for the sum of the first 4 elements, which is 2. Then we query sum of last 0 elements, which is 0. So we add 3*(2) — 3*(0) to the final answer. Currently, the final answer is 6.

Now we go through and do the same thing in reverse sorted order. First, reset the BIT to all 0s.

For the 3s, we update all positions of elements greater than 3+1=4. There are none of these, so again add 0 to final answer.

For the 2s, update all positions of elements greater than 2+1=3. None, so add 0.

Now, the 1s. Update all positions of elements greater than 1+1=2. So our BIT becomes

`0 0 1 0 1`

. The 1s are here:For the first 1, add 1*(0) — 1*(2) to final answer. For the second 1, add 1*(1) — 1*(1) to final answer.

Final answer: 4

nice logic WhaleVomit

I too taught of Fenwick tree but as array values are in order of 10^9. how to handle that much memory. Or i m doing something wrong here ?

The Fenwick tree only has order of 10^5 elements. You just have to process the elements in the correct order, so their max size doesn't really matter.

i really didn't know how to follow you way. what i did is index compression.

HERE

The answer for problem C is just the mode (the highest frequency among all numbers) of the array.

nope, number of times mode appears

Can someone explain author's approach to the problem F in more detail? What is the bit mask representing, and how is he using it to form dp states?

very nice solution for problem G, thinking the min-cut way is much easier :-)

In problem D test 29's correct answer is -9999999990000000000. Isn't it exceeding 10^18 by it's absolute value?

Read the clarification carefully.

903F - Замена подматриц What does mask mean? I am visualizing it like this.

~~~~~ mask = 1

## for i = 3 and j = 4

## |*|.|.|

## |.|.|.|

## |.|.|.|

## |.|.|.|

mask = 3

## for i = 3 and j = 4

## |*|*|.|

## |.|.|.|

## |.|.|.|

## |.|.|.|

~~~~~

Is this correct.

Its better to visualize it like

for j = 4

for j = 1

Then doing mask = mask >> 1 shifts it one cell further

I still don't get it. :-(

Can you give recurrence relations between state domains in 903F - Clear The Matrix.

Thanks in advance.

I can share my solution. I believe that it's pretty easy to understand with editorial given.

can you please understand the transitions i didn't get it from editorial. please help !

In problem D how to calculate that the absolute value of the answer will be up to 10^19?

My estimation was actually - 2·10

^{19}.. 2·10^{19}:`a = 10^9; n = 2 * 10^5; a * n * (n/2)`

please correct me if i'm wrong

let's make y-x value maximum so y must equal to 10^9 and x equal to 1

let's approximate maximum value of y-x to be 10^9

now let's make y-x occurs alot

consider this case : (n=4)

1 , 10^9 , 10^9 , 10^9

so y-x happens 3 times

now let's make n=2*10^5

so let's divide the array into two equal parts ones and (10^9) part

the array will be like this 1 1 1 1 1..... 10^9 10^9 10^9..

so for every one y-x happens 10^5 times

and we have 10^5 ones

so our value will be (10^5)*(10^5)*(10^9) = 10^19

sorry for poor english

I suspect -- though I have not tried to prove --

that 65 bits (64+sign) are

enough to represent an answerto this problem.We need an extended range of integers (under 2

^{66}different values) when we do summation.We probably do

notneed all 66 bits to represent an answer.True or not, there are no test cases which would break my demo 33243686 solution (it uses an unsigned 64-bit integer when printing the result in BN::show() method).

Can someone explain problem D please?

In my opinion,

there are

three major difficultieswhich could be encountered in Problem D:d(x,y) would beunconditional(justy-x).transform a solutionof the above problem into a solution for this specific sort ofd(x,y).large numbers(they do not fit into a 64-bit integer).The below is my attempt to explain the first two of the difficulties.

I do not know how to resolve the third one fast and bugless in C++ (I implemented a solution in practice mode but it takes too much time to be practical, i.e. to write it in a contest from scratch).

Thus,

Solving a simpler problem...I suggest to start with a simpler unconditional difference function:

d'(x,y) =y-x.I found it useful to imagine the problem with summation of

d'(a_{i},a_{j}) as a summation of elements of matrixD'where

D'_{i, j}=d'(a_{i},a_{j}), which means:D'_{i, j}=a_{j}-a_{i}.(It is convenient to write down

a_{i}values as both a header column and a header row of a table representing theD' matrix, thus filling the matrix witha_{j}-a_{i}values becomes a simple process if you want to do it on paper.)An important condition is

i≤j, which means that we consider the bottom-left triangle ofD' to be filled with zeros.We only need to work with the

upper-right triangle of the matrix.Note also that we can

exclude the diagonal elementsofD' from consideration (a_{k}-a_{k}≡ 0).Summation over

D' is now straightforward but requiresO(n^{2}) iterations which is too much forn= 2·10^{5}.So the point is how to optimize it.

Observe that for any

k, thea_{k}value is added to (or subtracted from) the final result many times.If we could calculate the number of times in

O(1) then we can replace multiple additions with a multiplication which means that we would needO(n) operations (addition, subtraction, multiplication) in total.Calculating the number of times in

O(1) is easy if we observe the following:k-thcolumnof the upper-right triangle ofD', weadda_{k}to the sum.k-throwof the triangle, wesubtracta_{k}from the sum.a_{k}influences the final answer.Now, the number of times is easy to calculate for any

a_{k}, so the summation overD' could be done inO(n).Fixing the result of the simpler problem...This is the point where we

return to the originald(x,y)(and, thus, to the corresponding matrix

Dinstead of simplerD')...Dcould differ fromD' only in those elements where .Actually, we can only consider because

D_{i, j}=D'_{i, j}= 0 ifa_{j}-a_{i}= 0 (no need to fix).So, we are interested

how to fixso that it would become equal toD'D...For any

a_{i}, we could try to count how manya_{j}can be found (j>i)so that

a_{j}=a_{i}- 1 (let's call itcnt_{i, prev}).Likewise, how many

a_{j}can be found (cnt_{i, next}) so thata_{j}=a_{i}+ 1.Note that:

a_{j}=a_{i}- 1, thenD'_{i, j}= - 1,a_{j}=a_{i}+ 1, thenD'_{i, j}= 1,which means that we need to

subtract 1·from thecnt_{i, next}D'-based resultand to

subtract ( - 1)·from it.cnt_{i, prev}This could be implemented in multiple ways.

All of them (which I could think of) require a

mapwhich would answerhow many times ainvalueis encountereda[] sequence (i.e. how many distinctiexist so thata_{i}=value).In my demo implementation (33250280), I modify this map while iterating by

itowardsnso that on each step

cnt_{i, prev}andcnt_{i, next}can be easily taken.Thank you very much for helping me understand the problem.

Can someone explain the solution to F?

For problem, A is this logic is correct or not

if(x%7==0 || x%3==0)_ print("YES");_else if((x%7)%3 == 0)_ print("YES")_else_ print("NO")_It's not correct, because when you did (x%7) you divided x by 7, so you lost values that may be divisible by 3. For example, and as your logic says: x=19 -> (x%7)%3=2!=0 -> the answer is

NO(here you made two meals by 7 and there was 5 remained), but we can make one meal by 7, and four meals by 3, so we obtain 1*7 + 4*3 = 19, and the answer for this sample isYES.means the else if block has the wrong logic

Yes.

By the way, for any

x> 11 the right answer isYES.Wow, it's very good that by at least four 3's, it's guaranteed that we can make one 7 by adding one, and two 7's by adding two, and by adding 3 we come up with a number that's divisible by 3, and so on :D, Thanks.

You are welcome :)

I'm not sure I understood your logic, unfortunately.

I proved that property after trying to

greedily constructall possibleYES-numbers, in ascending order. I ended up with the following:x=a·3 +b·7, thenx+ 3 = (a+ 1)·3 +b·7This means that, starting with three

consequtiveYES-numbers, all larger numbers are alsoYES-numbers.The smallest three consequtive

YES-numbers are:P.S. I had not noticed this property myself. I found a solution which is based on this idea and wanted to check if I can hack it :)

Firstly, I'm sorry for these untidy comments... (

UPD: now this is tidy :) ).Another "Wow!" for this prove which I was looking forward one like it.

My logic was that the number

`12`

is the first number that contains at least four 3's`3 + 3 + 3 + 3`

.Then the next number will contain

`3 + 3 + 3 + 3 + 1`

and this additional`1`

can be merged with two 3's obtaining`3 + 3 + 7`

which gives us the answerYES..And the next number will make another single

`1`

which also can be merged with the remained two 3's obtaining`7+7`

->YES..The next number is divisible by 3 ->

YES, and so on.It's guaranteed that this is true for all the next numbers because in all these numbers we have that four 3's which we need in merging, and after merging we will have some 7's and some 3's (may be 0).

P.S. I had not have this prove before I read your comment, Thanks.

Now I understood. Thanks for explaining!

can anyone please explain me the logic of Problem D almost difference

Have you seen my earliest comment on this subject?

You could also take a look at my Python code.

In problem E, rather than swapping

swith any other character of_{i,diffpos}s, Can't I swap it with any other character at a position in array_{i}pos?A simple, intuitive solution for Problem D :

About a dozen lines in Python.

See my virtual contest code here.