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Thanks for fast editorial! :D An easy trick for D: Once you have calculated the sum of minimum in all subarrays, you can just make all the numbers negative ( multiply each of them by -1) and then again calculate the sum of minimum in all subarrays. The sum of maximum will just be the negative of that.

Thank u sir for enlightening us with such a good method!! Hats oFf!!

C problem also solvable with DP (27810451) works in 18

^{3}* 9^{2}* 2.D problem also solvable with D&C approach(27810683) which works in

N*log(N).Can you explain your dp states for Problem C ?

dp[position in decimal representation][equal or not][sum of digits]

what is the D&C approach?

Divide and conquer.

why did this DC solution 27820818 get MLE? could you explain it to me please?

hehe inline??????? http://www.cplusplus.com/articles/2LywvCM9/ passed my solution

Can you please elaborate what is your approach, it is interesting.

D problem D&C aprroach link

This implementation is same as of Kerim.K , just some comments added for ease of understanding.

Thanks to Kerim.K for sharing the approach...

There is also another solution for problem C.

Since we know that the maximum value for

nis 10^{18}, then the maximum sum of digits we can get is forn= 999999999999999999, which is 162. Using this observation, and the proof mentioned in the editorial, we can be sure that all integers in the interval [s+ 162,n] arereally bignumbers. So, what is left now is to check the interval [s,s+ 161] with a simple for-loop.Source Code#include <bits/stdc++.h>

## define ll long long int

using namespace std;

int main() {

}

Amazing observation :D

Observed the same thing. Also while calculating the sum of digits, you can store the parts uptil the thousand's digit(excluding it as it can change). And only calculate the sum for the last 4 digits of a number. OR we could just store the four digits in an array and whenever any one of the digit goes above 9 make update the current to 0 and add 1 to next one.

Can i know the difficulty difference between an educational round and a normal round.How does difficulty of each problem in Normal round map to educational round?

It's usually like a standard div2 round + one task slightly harder than div2e.

E and D are good so that people can learn something new from the contest. Though they were standard trie and stack questions . I felt they were educational.Hope u introduce more advance concepts and data structures through educational rounds.

I did not understand the editorial to problem B. please explain. thanks :)

A very simple approach to problem B: Sort the array containing the numbers. Pick the first three. Now, we can prove that the minimum product will always contain only these three elements and nothing else. Thus, we next count occurrences of these first three letters in the whole sequence. Let the first, second and third numbers be a, b and c. And their repetition in the whole sequence be countA, countB, countC, and let they all be different. Then, ans is countA*countB*countC. But, if b repeats, then b is equal to c (sorted order). Then ans is countA* (ways of choosing 2 elements from countB elements). Similarily if a is repeated two times. Ans is countB only(any other repetition of a will cause it to be equal to c). If all three are same. Then ans is (no. of ways of choosing 3 elements out of countA elements). Here, the no. of ways of choosing 3 elements out of n elements = n*(n-1)*(n-2)/6 Here, the no. of ways of choosing 2 elements out of n elements = n*(n-1)/2

I did the same thing for B, but I keep TLE-ing on test 43. I'm convinced there's some kind of error in the judger or the data, because every other test takes less than 200 ms, including the 100000-length ones. I even bit the bullet and used BufferedReader instead of Scanner...no difference.

Take a look at my submission. Can you see some weird degenerate case that would cause my solution to not be O(NlongN + N)? I'm going crazy over here!

Can you please explain why the number of ways to choose 3 elements out of n is n*(n-1)*(n-2)/6 ?

I am not getting problem A ,I mean why does not (x2-x1)%x==0 and (y2-y1)%y==0 work?Please explain.

because let's say one is even positive and other is odd positive, then the answer would be NO.

it works but you also have to check if it's possible to reach the point doing movements in L. like a horse on chess. It will be reachable if the number of moves performed in the x axis + the number of moves performed in the y axis is even. This happens because when you move, you always do a move in both axis.

this is fine,but no of moves can't be given by (x2-x1)/x and (y2-y1)/y because there can be intermediate nullified moves in between too.

but also number of moves in x and y-axis can't be given by (x2 — x1) / x, (y2 — y1) / y because in 0 0 0 6 and x=2,y=3,no of moves in x axis is not(0-0)/2

So what? Yes, it isn't 0, but we only care of the parity of this number and (0 + 0)/2 + (0 + 6)/3 is even, so the answer exists.

Thanks.Got it.

can you please explain why we looking only for parity, and why not the number of moves given by (x2-x1)/x, (y2-y1)/y .plz explain using this case ( 0 0 0 6 and x=2,y=3).Thanks in advance

(x2-x1)%x==0 and (y2-y1)%y==0, only guarantees that we can reach x2 from x1 and y2 from y1 separately, but we need to reach them in the same number of steps, for that to be possible (abs(x2-x1)/x)%2 and (abs(y2-y1)/y)%2 should either be 0 or 1. You cannot reach both x2,y2 in same number of steps if one of abs(x2-x1)/x,abs(y2-y1)/y is odd and other is even.

this is fine,but no of moves can't be given by (x2-x1)/x and (y2-y1)/y because there can be intermediate nullified moves in between too.

but also number of moves in x and y-axis can't be given by (x2 — x1) / x, (y2 — y1) / y because in 0 0 0 6 and x=2,y=3,no of moves in x axis is not(0-0)/2

I think it's nature for most of us to use DP to solve Problem C…… But the prove is easy and important.....

How to approach a problem like 'B'? I mean how does the right technique strike? Are there any pointers(in the limits or similarity to a classic problem) to the right way? Any help would be appreciated.

Hmm. I don't really know :) I used the approach from the editorial in couple of other problems, though I can't remember any in particular. I guess that if this one doesn't seem intuitive then you can always try the other one. Where you just check some cases on the amount of the elements less than the third minimum of the array. Most people implemented this solution.

UPD: the author of this task told me that the similar approach is used here

Thanks a lot, PikMike.

i would request the author to provide the implementation described in the editorial Thanks for the help PikMike Also i have seen many people doing sorting will that not affect the ordering (i<j<k) as described in the question???

ordering will not be affected because if there is 3 indices in sorted array then there will be 3 indices in previous array. Isn't it .

Here it is.

Reorders will not affect answer. Why should they? You will just take the same triplets but indices in each will be in other order.

sorry my bad you are correct thanks for the explaination

I solved it differently. First sort array a. Then take three smallest values, let them be a,b,c in this order. Also, while doing input of array, keep array cnt[x], which tells how many times x appeared in input.

That covers all cases.

Can you explain how did you get these formulas?

SaynorPRO You may see these formulae as "n choose r". If we denote this value with the function c(n,r), c(n,r) = n!/(r!(n-r)!). So here, in case 1: It is simple product-rule of combinatorics case 2: It is c(cnt[a],2) that we need; because that gives us the number of ways to choose a(and hence, b). case 3: Like case 2. case 4: It is c(cnt[a],3). Hope this helps.

True. The condition i<j<k made me think that sorting would be of no help . This is may be because I assumed that a[i]<a[j]<a[k] which is not so. Need to practice more. Thanks

Hello, In problem D I don't know why did I get a Wrong answer on test 12. Can you please help me?

27814997

I guess integer overflow in the last for loop

How to solve the problem D when imbalance is defined as (max-min)^2 or (max-min)^k with k>=2 in better than O(N^2)?

WHat is the movement of

source: 2 -5 destination: -8 8

move: 2 1

. in Problem A ? How this is YES ?

0 -4, -2 -3, -4 -2, -6 -1, -8 0, -8 1, -8 2, .. .. .., -8 8

(-2,+1) (-2,+1) (-2,+1) (-2,+1) (-2,+1) (+2,+1) (-2,+1) (+2,+1) (-2,+1) (+2,+1) (-2,+1) (+2,+1) (-2,+1)

Thanks everyone :)

Why my MEX queries does not work? Here's my code:

Solutionaren't NlogN solutions supposed to pass for D ?? Mine got TLE because it uses insert and lower_bound in a set n times ! and I know their complexity is logN http://codeforces.com/contest/817/submission/27808174

I also have the same problem...

Here is my code.

NlogN solutions were supposed to pass but in your solution the constant factor is large (around 5).

In your code the complexity is :

Overall Complexity — 5*NlogN (Worst Case) = 5*(10^6)*(20) = 10^8.

And only those O(10^8) solutions pass in 2 sec which are highly optimised. In your case the hidden constant factor of set insertion also affects your code complexity.

Other NlogN solution which passed : 27810683 by Kerim.K

I solved Problem C by searching for the first 'Really Big Number' greater than 's' and less than or equal to n by running a loop.I considered the fact that if a number suppose i=25 doesn't satisfy the 'Really Big Number' condition then no number from 20-29 will satisfy the condition and thus I can jump from i=25 to i=30.By following this, if i found the first valid number less than or equal to n then our answer will be 'n-i+1' otherwise '0'. My question is whether my approach is correct or not because i got accepted but i couldn't proved why my approach is correct. Here is my accepted code.

Your approach is absolutely correct since you have used binary_search.

As you can observe from the value of num-sumD(num), it is always increasing.

And since the value of num — sumD(num) is strictly non-decreasing, take the lower limit to be 1, higher to be n + 1 and search for the appropriate num such that for values less than num (num — sumd(num)) is less than s and for values greater than or equal to num (num — sumd(num)) is greater than or equal to s.

So your approach is alright. For more clarification you can see my code : 27802867

Thank u priyanshupkm

I don't know where I went wrong in problem C please help I don't know much about dp and stuff I used the fact that difference of any number and its sum of digits is equal to nearest multiple of 9 of that number.

This is for itachi_2016's approach.Try the following test case and you will realize your mistake.

`109 91`

Thank you FazleRahmanEjazi I now understand where I went wrong..

Problem B reminds me of atcoder beginner contest 057 problem D (link). The idea of precomputing a combinations table can be also be used here since the problem is also to count how many ways to swap the last element.

My solution: here

Can anyone help me with sqrt decomposition solution for 817F - MEX Queries (According to editorial its possible).

I can't figure how this approach can be used to solve Problem F.

It's not much different from the segment tree one. Numbers are also compressed and the queries are the same.

You process assigning on a segment and xor of the segment by dividing original array into sqrt blocks, for each block you maintain its sum. Lazy propagation is used too. Just every query works in instead of .

We decided not to force participants to use solutions.

Lazy propagation in sqrt decomposition? Anyone has some resources for that?

Similar problem as D are :

http://practice.geeksforgeeks.org/problems/maximum-of-minimum-for-every-window-size/0

http://agc005.contest.atcoder.jp/tasks/agc005_b

Same approach can be use in these problem , next smaller , next greater using stack .

Can someone please tell why my code is getting WA ??? Link .Thanks in advance !!!

your code is giving wrong answer because suppose you are standing on a node where xbit=1 and ybit=0 then there xor is 1 and you are adding all the numbers that have that bit zero. but this is wrong because this does not gaurantee all the number in this group to be smaller then your leadership(l).this group till now are only equal, not became less. so you can't add them. if xbit=0 and ybit=1 then your case is correct. there may be case some of these number can latter become equal to or greater than leadership(l). link this is my code you can see . this might help you visualising what i am saying.

Problem B In the tutorial "Let's also store set of three elements which give minimal product of the array". In c++ how do you calculate the product if all the numbers are greater than 10^9?

Problem B In the tutorial "Let's also store set of three elements which give minimal product of the array". In c++ how do you calculate the product if all the numbers are greater than 10^9?

hi

we don't calculate product of three numbers

and question is "how many ways are there to choosing three minimal numbers"

because minimal numbers give us minimal product

There is also one more solution of problem C.

All numbers bigger than s+9*18 and not more than n are really big. So we can check numbers from s to s+9*18 and add max(0, n — s-9*18) to answer. Sorry for my English

t = s+10-s%10 while(t-sum(t)<s) t+=10 then if t < s there is none, else the answer will be n-t+1. sum(t) means the sum of digit this is accept. Is this right?

Problem F in such constraints can also be solved with coordinates compression and bitmasks in obvious way.

BTW, it's fastest solution at this moment. (171 ms without fastIO). 27832302

can you explain your approach . it is diificult to understand from code

Global idea as it is in the editorial:

Let's store this array of zeros and ones as bitset, it means that each byte of array will represent 8 values in 8 bit. Then you can perform first two queries on 32 values at once as simple assignment of int variable a[i] = 0 or a[i] = -1, and third query as a[i] = ~a[i]. To find the answer after each query you should find first value in array not equal -1 and find first zero bit index in it.

Can anyone explain me D&C solution for D? Because solution with stacks looks awful. I was thinking D&C on contest but couldnt solve it. Edit: got it. this solution helped me http://codeforces.com/contest/817/submission/27832753

Hello,

In problem E how do we insert pi in the trie? I didn't get the explantation given by the editorial. What is the criterion of insertion in binary trie?

Detailed approach for D???

Considering finding the sum of minimum values for all subsegments, the solution tries to know for each index

ihow many timesa[i] will be the answer. Ifxis the number of subsegments that contain indexianda[i] is the minimum value, thenans[i] =x·a[i]. The final answer will be .For the correctness of the algorithm, if a subsegment have more than one cell with the same value, we will consider the answer to be the leftmost cell.

For index

i, Let:L[i] be the rightmost indexj<ianda[j] ≤a[i]R[i] be the leftmost indexj>ianda[j] <a[i]Clearly, All cells in the range [

L[i] + 1,i- 1] have values greater thana[i] and all cells in the range [i,R[i] - 1] have values greater than or equal toa[i]. Now, we can say that indexican be the answer to any subsegment whose startpoint is in the range [L[i] + 1,i] and endpoint is in the range [i,R[i] - 1]. So,ans[i] = (i-L[i])·(R[i] -i)How to compute the array

L? Keep a stack with you and go from left to right. The stack is initially empty and will contain indices. Now, we are at indexi:L[i] = - 1.janda[j] ≤a[i], thenL[i] =jjanda[j] >a[i], then pop this element and do the same check for the new stack top.If

jis popped from the stack, then fork>i, canL[k] =j? The answer is no, becausea[j] >a[i] andiis closer tokfromj, so ifa[i] ≤a[k], thenL[k] =i. Otherwise,a[j] >a[i] >a[k]. That's why we removejfrom the stack because it will never be used again.After finding

L[i], pushionto the stack.You can compute the array

Rin a similar way from right to left.In editorial of 817A, why did we calclate cntx and cnty?

What does cntx and cnty signify?

These signify the number of moves in x-axis and y-axis respectively to reach dest from source.These are the number of moves to reach the destination from source assuming only one axis at a time, i.e. we can move

onlyin a straight line, or independent of other axis as constrained by the problem.As each move in original situation changes parity from odd -> even -> odd ... we can check if the parity of both are same, apart from checking whether the no. moves are integral or not.

Thanks shubham_827.I understood it now.

Hi, In problem E, why should we search for the numbers li(xor)pi when count the warriors?

Can you give me same problems like this one?

Can you give me a Tutorial about xor properties needed in programming contests.

Thank you in advance.

Can any one tell me why we are doing mod 2 in problem A.

hi

it is difficult to explain but

if there modes in 2 aren't equal then we can't arrive from x1 to x2 and y1 to y2 in the same time

and if there modes in 2 are equal then imagine that:

x2 > x1 and y2 > y1 and (abs(x2 — x1)/a) > (abs(y2 — y1)/b)

then you can simply do x1 + a, y1 + b and x1 — a, y1 + b k times and you will have

(abs(x2 — x1)/a) and (abs(y2 — y1) + b*2k)/b

and you can chose correct k to get (abs(x2 — x1)/a == abs(y2 — y1)/b + 2*k) because their modes in 2 are equals.

for another cases you can do like this.

if mod 2 is 1, then we can reach target with odd number of moves. Otherwise, we can reach target with even number of moves. So, the parity (odd/even) must be the same for both horizontal and vertical moves.

can you explain the logic of problem D. I have read it many times still it fails to strike me . I understand upto how to calculate left[i] and right[i] after that how he is finding no of segments and rest of solution. please help

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Can anyone explain Problem D?? I read the solution many times but not getting it.

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