We will hold AtCoder Beginner Contest 173.

- Contest URL: https://atcoder.jp/contests/abc173
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20200705T2100&p1=248
- Duration: 100 minutes
- Number of Tasks: 6
- Writer: evima, gazelle, kyopro_friends, sheyasutaka, ynymxiaolongbao
- Rated range: ~ 1999

The point values will be 100-200-300-400-500-600.

We are looking forward to your participation!

Participate, we must.

Hope that C is not the killer this time :p

you jinxed it (at least for me) :(

this is third time in a row that,I'm able to solve A,B,D but not C ):

My Solution

can anyone please check my solution and correct me?????

same bro :(

i used backtracking.

really bro

Starts in 15 mins :)

See you all on the scoreboard!

Video editorial and screencast are being uploaded to my youtube channel now. There are lots of nice solutions described in the comments here (and the atcoder editorial is usually quite good), but if you would like to see stuff explained more visually, feel free to look out for that :)

cant find atcoder 173 in ur channel... is it in the process of being uploaded as of now?

Yeah it is being processed by Youtube right now.

I hate to say this because it sounds so cliche, but there is a bell icon you can click if you subscribe if you want to get an email when YouTube decides to finish processing it.

subscribed !! upload more and more tricks, tutorials and solutions of cf and atcoder rounds

i guess u r the only cp channel in youtube with the clearest english accent!

keep up the good work

Looks like Youtube is finally done! Here’s it is :)

The video quality will improve as it processes. Usually it is only 360p or something for a bit, even though it is recorded in 1080p.

orz that F solution was so damn cool!!

not able to solve C again!!!

How to solve D?

always try inserting new comer into two maximum existing values! Submission

That was a good idea.

Should have thought of that.

Greedy, keep track of which element contribute to the answer and how many times, like the 1st one contribute only 1 time to the answer, rest of the numbers does it twice. So just take the (sum of largest n / 2 element — 1st largest).

I tried this approachI maintained a priority queue, if anyone is interested I can explain the solution , because I think people have posted better solutions.

Mine was a O(nlogn) solution by sorting the given friendliness of each player in decreasing order. And Taking into account the cases:

So just take sum of them in the order mentioned till you reach exactly n-1 numbers.

I also did using same approach but My Solution is failing some test case. Can you please point out the error and link your solution.

Your code is not considering the case that other elements will contribute twice. Try to change your approach.

Please link your solution and how every element will contribute twice??

Yeah Sure! CODE

What the hell with me . can solve D but not c ..everytime stuck on c..

Consider all 2^h subsets of rows and 2^h subsets of cols and then try every pair of subsets that is brute force for all 2^h*2^w subsets

what is "Meet in the Middle"?

Meet in the middle is a standard technique to divide the problem into two halfs and brute force over all the possible subsets of the problem. SOme problems on Meet in the middle: if you want to practice. Meet in the middle practice problems

Thank you so much. I heard this technique for the first time.

This is not at all meet in the middle, but just basic brute-force. Meet in the middle would be something like: iterate through all subsets of rows, all subsets of columns and somehow combine result from them in time complexity closer to $$$O(2^W + 2^H)$$$ then $$$O(2^{WH})$$$

Ya you are right , I have to edit it as when I was writing my mind was out of my body, sorry if it disturbed someone.

What if the constraints were somewhat bigger. What would be the efficient approach then?

My first thought would be: iterate through all $$$2^H$$$ subsets of rows, for each column find number of black cells at intersevtions of given column and one of chosen rows, then use some basic dp to count subsets of those numbers adding up to $$$K$$$. Total complexity is $$$O(2^HWK)$$$

Quite a nice idea. Thanks!

I am sorry but where's the Meet in the Middle part in C, it's just plain brute force.

Ya While writing it I was thinking of some other thing! Sorry !

Yeah, exactly, LOL. This is a simple bruteforce solution.

But try extending the problem to H<=20 and W<=20, I think then we have to go for that? or may be some other heuristic.

how can be solved using Meet in the middle ?

Same here Solved D but not C

C is just BruteForces My submission

Can you please explain??

Iterate over all subsets of rows. Now for each subset of rows, iterate over all subsets of columns. Iterating over the subsets can be easily done by

`for (i = 0; i < 1<<N; i++) { ... }`

It will give us: $$$000, 001, 010, 011, 100, 101, 110, 111$$$ in binary if $$$N = 3$$$.This is the basic idea

since it was only 6. I did the Bruuuuuuuute force. Generated all sized subsets then ran a loop for every row and every column and for each combination checked whether it's valid or not.

My Submission-C

It's better to be general, rather than doing all combination do all subset. To find all subsets just use integer from 0 to 2^n — 1.

Btw felt the need to tell this because your code was a bit lengthy.

What's the approach for F ?

You can consider the nodes and edges independently if you count nodes as +1, and edges as -1

Uhmm, can you elaborate a bit ? Thanks!

If you have a forest, the total number of components is nNodes — nEdges.

You can easily count the number of ranges a node is part of with combinatorics, and same for an edge (it is just nLeftRangeEndpoints * nRightRangeEndpoints). So you can count this contribution of each node and each edge as if it were the only thing in the graph.

Oh wow, quite elegant a solution!

Can you tell me whats wrong with this submission?? https://atcoder.jp/contests/abc173/submissions/15045823

When N = 1, your code gives 0 as output instead of 1.

It is similar to this question of codeforces

In fact it is exactly this question asked.

Not exactly, the tree in codeforces question is linear. But we can convert a[i — 1] , a[i] logic to parent, child logic and it works same.

Yes, you are right.

Brief Editorial :

AJust print $$$\left \lceil n/1000 \right \rceil * 1000-n$$$

BYou can use a "map<stirng,int>" .

CJust try all the states.in $$$O(2^{(H+W)}*W*H)$$$

DFirst you can easily find that all the player arrive in a fixed order,that is "Decreasing Order".But how to prove ,it is also very easy.Because if i-th player arrive earlier than j-th player,and ai<aj,just swap the i-th player and j-th player ,the answer won't be worse.

Than consider how many times the biggest one is added to result?Yes,exactly once.Because when you let the second the one arrive,the first one is added to the result.

But how about the second biggest ,third biggest and so on...It seems not to be very clear.We just know at most twice!Because after you add aj and ak next to ai, ai won't be added to the result because aj and ak is smaller than ai.

That's enough !!! We can make the second biggest one ,third biggest one and so on...be added to the result exactly twice.

Try to come up a way to achieve it .

EIf k=n just print the product of all the numbers and mod 1e9+7

If there isn't a positive integer and k%2=1 just print the product of smallest k numbers and mod 1e9+7

If there isn't a positive integer and k%2=0 just print the product of biggest k numbers and mod 1e9+7

Otherwise just try how many negative integers are used,clearly the number of used negative integers must be even.

We can use to vector to store all the positive integers and the (negative integers * -1) ,and sort them in decreasing order.

Suppose we use i negative integers,i%2=0.Then the i negative numbers are the first i integers in the negative integers' vector. And first k-i integers in the positive integers' vector.We can use prefix product to calculate quickly.

But when it is the biggest ? We can try i from the 0 to the k.The present result = the last result * the product of i-th and (i-1)-th negative integers / ((k-i+1)-th * (k-i+2)-th positive integers).

So if the product of i-th and (i-1)-th negative integers > (k-i+1)-th * (k-i+2)-th positive integers the present result is bigger than the last one.Otherwise end the loop.Because the product of the negative integers is decreasing , but the product of the positive one is increasing.

FSuppose ui<vi.

Just consider different pair of (L,R) ,there are total (n)*(n+1)/2 .

For each case , imagine a graph of some nodes,and some edges.

And the i-th node appears in i*(n-i+1) graphs .

The ui and vi both appear in ui*(n-vi+1) graphs.And if ui and vi both appear in a graph ,ui and vi belong to the same connected component.

so the answer =

https://atcoder.jp/contests/abc173/submissions?f.Task=&f.Language=&f.Status=&f.User=Gary

F nice formular, but not understandable for those who did not solve it.

Sorry I will update it later ,I have something to do now.

Use this until then https://codeforces.com/blog/entry/79733?#comment-656818

Problem D:I have implemented it in the exact way that you mentioned, but it just happened to fail at only 2 test cases, can you help me in finding out the bug in my submission ?

Than consider how many times the biggest one is added to result?Yes,**exactly once**.Because when you let the second the one arrive,the first one is added to the result.

I added the biggest element exactly once!

You can compare yours with mine.

OK, got it, thanks!

if n = 1 then answer will be 0 not a.front().

Is N ever less than 2?

can you explain F?

The idea is to count the edges, somehow... Because the count determines the number of components.

In solution F: My understanding:

X= contribution of all the nodes in total summation

Y=Contribution of each edge in the total summation

ans=X-Y

Kindly explain your recurrence(I understood Y ,kindly explain X)?

Can you tell me what's wrong with my submission?? https://atcoder.jp/contests/abc173/submissions/15045823

check your code when N=1.

E

Isn't this unfair, giving a question which is already available online, I mean the exact same question, most of the participants who googled something similar to find hints might have ended up copying the exact same code without any efforts of their own.

The code provided doesn't have the mod 1e9+7 part, and that's a little bit tricky for negative numbers.

In the last 2 contest I have seen happening this for C and now E.

How to solve E and F? Thanks in advance:)

E: case out n==k, and when everything is negative.

Otherwise, you can repeatedly remove the smallest two nonegatives and replace them with the largest two negatives as long the product of the two negatives is bigger.

F: You can consider nodes and edges independently. Each node contributes +1 to the answer, each edge -1. You just need to count the number of ranges each node and each edge is part of.

In E why after sorting both positive and negative numbers then repeatedly removing pair from the end which has maximum value(product) and if k is odd and there are only 2 positive numbers left then only removing pairwise from negative numbers didn't work?

For odd K when i first removed the largest positive number separately and solve the rest for even k worked, why?

WA AC

Consider this testcase:

The first method will select 3 and 1 since 3 x 1 > (-2) x (-1). Then it has to select 1, the only positive number left. However, 3 * 1 * 1 < 3 * (-2) * (-1)

That problem can be remedied by ensuring that whenever two numbers are selected together at either end, another two numbers will be selected at either end unless no more element will be selected. It is easy to show that strategy cannot be wrong.

That is why the second method works, because for even K, the maximum product, if it is greater than 0, must come from even number of positive numbers and even number of negative numbers.

Your solution for E is just brilliant. Thank you

For F for every segment think that every element is disconnected, and their contribution to answer is the length of segments. Now for each edge see in how many segments it can occur and subtract that number from answer.

Another Series :)

Multiplication 1

Multiplication 2

Multiplication 3

Multiplication 4

Ok! can anybody tell me how to solve C? :)

brute force, check all possibilities

just look up all possible way to select rows and columns either by bitmasking or dynamic programming and check number of blacks for each combination

Brute force

Check for all possible subsets, if we exclude these then if the number of black cells is k or not.

Max number of rows = 6

Max number of subsets of rows = 64

Max number of columns = 6

Max number of subsets of columns = 64

Total number of subsets = subsets of rows * subsets of columns

Max total number of subsets = 64 * 64

Complexity to check each subset = O(nm)

Total complexity = O(2^n * 2^m * n * m)

Thanks man!

You can just brute force all the possible choices and for each choice count the number of remaning black squares. Check my submission for details: https://atcoder.jp/contests/abc173/submissions/15012234

E was on codechef as MMPROD.

Wow, E took me forever (and a lot of WA and lots of messy casework). Wondering if anyone has a cleaner solution.

(My messy solution: https://atcoder.jp/contests/abc173/submissions?f.Task=abc173_e&f.Language=&f.Status=&f.User=AnandOza)

I was too getting wrong ans on 1 test case . Then I checked the modular operation I was doing and found the error in it

I think it is less code if we separate the numbers into three groups, counting also the 0.

https://atcoder.jp/contests/abc173/submissions/15010921

My solution with comments

I think mine is much cleaner. The idea is that you case out n==k and cases in which all numbers are negative.

Otherwise, you can start with the positive numbers, and replace two positive numbers with two negative number repeatedly.

That code is just:

Nice solution!

Ah, that is a lot nicer. Thanks! I'll try implementing it later.

btw congrats anand finally made it to red!!

Thank you!

I was also approaching the problem in the similar manner. I was checking the two largest negatives not taken with two largest positives and whichever product is greater I was taking those two elements.But I got confused when the case when we will have to take odd number of negatives.So how to handle the product in such a case(i.e how to take the modulo in that case) or there won't be any case like that?

UPD:Got it there was just a small mistake in implementation.I was printing the wrong value:(.

https://www.geeksforgeeks.org/maximum-product-subsequence-size-k/

Here is good implementation

E also makes me frustrated. :'(

Defeated by Mod ;____;

Same

can we see others submission? if yes then how?

click on Results, then All Submissions

If you want to see submission of a perticular person then, go to standing hover over the username and click on the searching symbol (magnifying lens). You can also do that by writing the username in all submissions. But I like the first one this way I can see the fastest submissions.

Can somebody guide me to an article(or anything for that matter) which has good theory on bitmasking? Couldn't solve C this time! (easy to read)Code for A, B, D, E @ https://atcoder.jp/contests/abc173/submissions/me

https://atcoder.jp/contests/abc173/submissions/14996063

I can explain you if you need help for C

Thanks, that'd be great!

For D, getting WA on 9 test case , please help, what's wrong in my code. https://atcoder.jp/contests/abc173/submissions/15012383

For Problem E

First

Second

First code is not working for 2 testcases so i added one if for k==n that is second code and Second code is not working for 3 testcases. I can't understand how it is possible can anyone help me?

because your if(k==n) part is wrong

How Can you explain i can't get it still?

Why are you doing mul=-mul? You are anyway multiplying the actual numbers.

thanks got it.

I tried E by sorting all the numbers. If all the numbers in the array are positive the max product of k numbers is last k numbers. otherwise, we need to find the maximum of the first k numbers and the last k numbers and print the maximum. What is wrong with this logic.

Consider this testcase:

The product of first k numbers and the product of last k numbers are 0, but the correct answer is 4.

Yes, got it. Thanks

Suppose the array is-

-4 -3 -2 -1 0 1 3 4 and k = 4

Product of first k numbers = 24

Product of last k numbers = 0

Your Answer = 24

Actual Answer = (-4)(-3)(3)(4) = 144

Answer should be(-3*-2*-1*1)-6%1e9+7=1000000001 no?

i got it, thanks though

can someone please help out with C?

It can be solved with a simple brute force approach. Notice that

`h,w <=6`

, so just iterate over all possibilities and check.I didn't even understand the problem statement of C.. xD

Anyone's comparing summation of log2()s failing for E?? can anyone explain how to overcome it??

What is wrong in this :

Try to take even number of negative numbers starting from descending order and multiply with even numbers left , again in descending order.

If the above is not possible, then check if a zero exists.

If a zero exists, ans is zero else answer has to be negative.

SO, try to take odd number of negative numbers in ascending order and choose remaining numbers as even, again in ascending order ?

https://atcoder.jp/contests/abc173/submissions/15018728

A%mod > B%mod does not imply A >B

I know, I haven't used that.

Apparently,the mistake is overflowing of the product.

Thanks anyway.

Can someone kindly explain the E problem https://atcoder.jp/contests/abc173/tasks/abc173_e

Here is a link to my solution and also give a test case where it fails. https://atcoder.jp/contests/abc173/submissions/15013047

Anybody used gfg's code for problem E?? If yes,please share your sol, i want to see my mistake!

The Python code had an incorrect range for CASE II. Also, to avoid TLE, the mod had to be applied throughout.

Yes, but got 26 AC and 11 WA. Maybe has some issue with modulo

Can anyone explain how to approach C ?

May somebody help me debug my code? Getting WA on one testcase. https://atcoder.jp/contests/abc173/submissions/15021212

You can view all of Atcoder's testcases here

My Solution

can anyone please check my solution and correct me?????

can anyone explain the approach of Problem C , I was able to solve A, B, D but not C. https://atcoder.jp/users/aniketakgec/history/share/abc173

It is brute force .

you can store count of no of black cell in rows and columns in an array . And then simply brute over all the possibilities of selecting row and column (i used bitmasking for it) . And can use above array to count number of black cell in each case ,if it is k , do ans++;

Notice that the constraints for this problem are noticeably little, so all you need to do is to consider all possible combinations of rows and columns. Then, check how many of them have k remaining black cells.

My Submission for this problem

Simple brute force, try all possible combinations of rows and columns that are to be painted and simply count the no of black cells left in O(h*w) for a particular combination, total time complexity — O(2^h * 2^w * h * w)

Someone please help , why my logic is wrong in task D:

CodeFor n=1 your code gives the result a[n] but it's 0.

In problem constraints n>=2 , btw i got my mistake , i forgot to put f=0 inside f==2 condition , Can't believe , i wasted 40+ minutes to find my bug , but was not able to observe it while in contest .

Question says N >= 2

in constrains

`2 ≤ N ≤ 2×10^5`

It was a nice contest. Thanks for sharing the link.

What is wrong with this code of E-

SolutionNot sure if this is all but that block outputs negative numbers, the mod calculaton is not correct:

this code is from g4g

Initially, problem C had the constraint H, W <= 13 and needed DFS and pre-calculation :(

Well, today's C was easy i.e. just briteforce. But was definitely much harder than regular ABC's

How to do if H,W <=13, Can someone please explain the approach a little bit. I solved for the low constraints using bitmasking.

can c be solved using recursion if yes can you please explain it? I am trying to solve it using recursion for hours but it's not working.

Here is my DFS solution (O(2^{H + W} + H * W * 2^H)): https://atcoder.jp/contests/abc173/submissions/15031206

Actually, the DFS part of the code can speed-up by using dynamic programming (like the knapsack problem) and solve the problem in O(H * W^2 * 2^H).

Link is broken, could you correct it? I am very interested in reading it.

Sorry, It’s now available.

In problem E-Multiplication 4 I was checking whether they have this kind of test case

`5 4 1 2 3 -1 0`

in their test set or not.so, I removed one condition in my code and got AC instead of WA.

Output produced by AC code is 1000000001 ( -6 % (10^9 + 7) ) you can see here. Where the correct answer is 0.

OMG I misread problem E as maximum of all possible k length subsequences (a1*a2*a3*....ak)%(1e9+7). How to solve this task?

Got WA in B due to writing capital 'X' instead of small 'x'

My submission for E is giving WA for 5 tests. Actually I tried only limited cases:

First I sort array of absolute value of numbers descending order

If first k have even number of negatives, we got our answer

Otherwise we remove smallest negative and search for a positive among remaining n-k elements. If found positive, then this is an estimate.

Another estimate is dont take one non negative among first k elements and search for a negative in remaining n-k elements. If negative is found then this is another estimate.

If none of 2, 3 occurred then we know our answer must be negative so we chose last k elements as estimate. Otherwise we chose biggger of 2, 3.

Is this wrong? While writing this comment I realised that I havent checked which one of 2, 3 yields greater value. ;p submission link

I had a problem with E, most of the test cases passed but I don't know why I was not able to have AC on E. can anyone help me ? here is my the link to my code https://atcoder.jp/contests/abc173/submissions/15005580

It is wrong logic. Among other things you totally ignore that there can be 0 in a[].

Thanks for the reply. I tried the geeks for geeks solution which included the case when there is a 0 in a[] but still got WA verdict. here is my code https://atcoder.jp/contests/abc173/submissions/15012483 Can you have a look at it? Thanks!

English editorial may be helpful for all coders.

Can someone tell me why this solution is failing on certain cases.

It is wrong logic. Consider the third and fourth being added to the circle.

The third contributes min(a[0], a[1]), and the fourth, too. Because the fourth is placed on "the other side" of the second, which is between second and first, too.

How third and fourth will contribute min(a[0],a[1])?? In the given example third is contributing min(arr[1],arr[4]) and fourth min(arr[3],arr[2])

After third

`a[0], a[2], a[1]`

, after fourth`a[3], a[0], a[2], a[1]`

a[2] and a[3] where both placed between a[0] and a[1]. Its a circle.

I have submitted several times and still getting WA on three specific cases. https://atcoder.jp/contests/abc173/submissions/15025193

Is anyone stuck in the same cases to give me hand, because I have really no idea what is wrong. My main logic is to get the max K numbers and apply a replacement when the parity of negatives is an odd number.

Thank you !

In Atcoder the correct order must be A, B, D, C. Thrice in a row I was not able to solve C. Can anyone help pls. :(

ll solve(){ ll n; cin>>n;

} why this code is giving WA for D

please explain solution of C in easy language..

Basically we have to pick a subset of rows and a subset of columns

Then count the number of # character in those selected rows and columns say cnt if the (total # in matrix)-(cnt) equals k then increment the answer

Do this for all combination of subsets of rows and columns i.e. 2^(W) * 2^(H) combinations

Can there be cases where we have to choose 3 rows or 4 rows to get the desired number of black colors??

I mean is there any restriction on how many rows or columns we can choose in a single time to get the desired number of black colors??

No there is absolutely no restriction that way.

Thank You......and sorry for the very late reply

oh now i understand what is going on.. thanks bro

Since the constraints are small, we can use brute force. Notice that there are h rows and w columns. If we represent each row and each column by a bit, then we can use a bitmask of the form (1<<h+w) For eg: If h=4 and w=3, then 1101001 means we are considering the first row, the fourth row, the second column and the third column. Similarly consider every possibility, and subtract the number of # you encounter from the total. If this is equal to k increment your answer by 1.

You can check out my submission here

Submission

In D question, why can't the max element be added more than once? For example, consider the case :

5

2 2 3 3 3

Won't the max element be added more than once here??

UPD: I understood. Ignore this comment.

My code passes all the tests except one. I can't figure out any mistake. Please help. Link to my submission

Edit: I got my mistake. Here is my new submission

What is wrong with this approach for E? I covered the other cases such as only negatives and k=n. So when both positives and negatives are there, we take biggest elements by absolute value in our answer and count the number of negatives. If number of negatives is even, this is our answer. Otherwise, I am trading the least positive value in my taken numbers with biggest negative value among left out numbers or trading least negative value with the biggest positive value among left out numbers, whichever is better. Any counter test case for this ?

Please help me its failing on 2 testcases. Submission

Can anyone help me understand why this code fails for a few testcases https://atcoder.jp/contests/abc173/submissions/15031526

i inserted the element as pair where first = absolute value and second value = 0 if positive else 1 and sorted the elements using abs value.In case of all negative elements i picked first k elements.Otherwise i picked the last k elements and noted the number of negative values(say nneg) and also kept track of the first negative value i picked(say firstneg).ifnneg is even then i print the product else in the first n-k elements i first look for maximum positive element and failing to do so i pick the least negative value and multiply it by product and multiply it by inverse of the firstneg.

vkm23061998, your code fails for this TC. Maybe, this helps.

Input:

8 6

1000000000 1000000000 -1000000000 -1000000000 -999999999 999999998 -999999997 999999996

Correct Answer:

192080

(By taking the first 4 numbers and -999999999 and -999999997)

Your Output:

237699

(By taking the first 4 numbers and 999999998 and 999999996)

Also, I have done in a similar way, as yours. My code is passing all the testcases.

You may like to view it : Link to my submission

Can anyone provide proof for the problem D.

Same here. the proof in editorial is not sufficient imho.

Here is my proof not depends on people arrived in the decreasing order. Consider all the n-1 comfort we can get. I will prove: If there are k numbers >= x, others < x, then among all comfort we can get, there are at most 2*k-1 comfort >= x.

We note these k person as nice person, and the position between 2 nice person as nice position. also we note comfort >= x as nice comfort. 1. There are at most k-1 nice comfort we can get when these k nice person arrive(first nice person can not get nice comfort). 2. when a nice person arrive, nice position will increase at most one, when a not so nice person get a nice comfort, he must decrease a nice position. so not so nice persons can get k nice comfort at most.

so we can get at most 1 comfort >= a[0], 3 comfort >= a[1], and so on...

Brother i got you. Thanks

Think greedily

Since we have to maximize the total comfort, so we'll try to get the maximum comfort possible every time.

Maximum comfort using some i friends=Maximum comfort using some i-1 friends + maximum comfort from any remaining friend.

By exchange arguments,step 2 will end up in maximum total comfort.

can any one explain me why this wrong in case of problem E ans=((ans*a)%mod+mod)%mod but this gives me AC ans=((ans+mod)%mod*(a+mod)%mod)%mod WA:https://atcoder.jp/contests/abc173/submissions/15038118 AC:https://atcoder.jp/contests/abc173/submissions/15038907

For E, what is the case 'after_contest_01.txt'? My code is getting 500 as it passes all the tests from the contest but fails on this one.

Note: It's also not in the TestCase Dropbox.

I don't know if it will solve your problem but my code was failing for the same case only. Check this case.

Input :7 5

35 25 5 -25 -25 -25 -25

Output :13671875

I guess, maybe you've made pairs of positive/negative numbers and multiplied them. And if K is odd, then you multiply another positive number to the answer. I tried this as well and it got a WA on the case 'after_contest_01.txt'. You can multiply the only (not paired) positive number first if K is odd. It did solve my problem, hope it'll be helpful to you too.

BTW, you can check the code since my English isn't good:

AC: https://atcoder.jp/contests/abc173/submissions/15069072

WA: https://atcoder.jp/contests/abc173/submissions/15068710

Can anyone check my submission for problem E : https://atcoder.jp/contests/abc173/submissions/15041696 it fails on just single case.Not able to figure out.

Test: random_mz01.txt

The Output is 0. Expected is 649491334

`prod=prod * a[i]%MOD * a[i+1]%MOD;`

Won't this overflow?

ok... I was returning a 0 if the max element in array is 0 immaterial of k(even or odd).That was the problem.Thanks for the test case.It works now :)

random_mz01.txt? Are the OJ inputs freely available somehow? I don't see them linked anywhere from AtCoder.jp…

Yeah they are: AtCoder Post

Yup, thank you!

Ha! Found it. https://atcoder.jp/posts/21 :)

Please help in question E, why does this give WA????

Can someone look at this approach for C? Link to the post

What's wrong with my E submission? Submission

What wrong in my submission in E ,it's failing on 6 test cases https://atcoder.jp/contests/abc173/submissions/15042752 ?

arpit_pro, vkm23061998, your code fails for this TC.

Input:

8 6

1000000000 1000000000 -1000000000 -1000000000 -999999999 999999998 -999999997 999999996

Correct Answer:

192080

(By taking the first 4 numbers and -999999999 and -999999997)

Your Output:

237699

(By taking the first 4 numbers and 999999998 and 999999996)

Hope, this helps.

Can anyone give a proper explanation of problem of C ? I didn't get the editorial.

Someone knows What's in "after_contest_01.txt"? （Problem E）

I don't know if it will solve your problem but my code was failing for the same case only. Check this case.

Input :7 5

35 25 5 -25 -25 -25 -25

Output :13671875

thanks,bro~

Nuclear launch codes

https://atcoder.jp/contests/abc173/submissions/15051664

Please help in E

For some (like me) who still couldn't figure out F

g(l,r): be graph of vertices[l,r]

e(l,r): count of edges in g(l,r)

f(l,r): count of connected components in g(l,r)

f(l,r) = count of vertices in g(l,r) — count of edges in g(l,r)

$$$\displaystyle\sum\limits_{l=1}^n \displaystyle\sum\limits_{r=l}^n f(l,r) = \displaystyle\sum\limits_{l=1}^N \displaystyle\sum\limits_{r=l}^n ((r-l+1) - e(l,r)) $$$ $$$=\displaystyle\sum\limits_{l=1}^n \displaystyle\sum\limits_{r=l}^n (r-l+1) - \displaystyle\sum\limits_{l=1}^n \displaystyle\sum\limits_{r=l}^n e(l,r) $$$

First part is easy O(n). For second part Let c(u,v) be number of ranges edge(u,v) contribute to

$$$c(u_i,v_i) = u_i*(n-v_i+1) $$$

Hence, $$$\displaystyle\sum\limits_{l=1}^n \displaystyle\sum\limits_{r=l}^n e(l,r) = \displaystyle\sum\limits_{i=1}^{n-1} c(u_i,v_i)$$$

PS: My first attempt at an explanation. I think there should be line-spacing option.

https://atcoder.jp/contests/abc173/submissions/15093804

Can anyone please check my solution? I am not getting anything wrong with my approach but getting wa continuously.

If anyone need Detail explanation (Not a video tutorial) for E Here