**Div2A**

**Div2B/Div1A**

**Div2C**

**Div2D/Div1B**

**Div2E/Div1C**

**Div2F/Div1D**

**Div1E**

**Div1F**

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Tutorial of Технокубок 2020 - Финал

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Nice editorial :)

thank you for editorial

so unlucky with problem div.2 E ;( thanks for editorials. can someone check my code and find my bug? i was debuging for 2 hours and found nothing. my code

The test on which your code is failing is small in size(n=m=p=10) so you can use it as a counter-testcase and debug..

I have a problem with 1321C - Удаление соседних. I don't really understand how the answer to the first test case is $$$4$$$

According to the problem, $$$ 1 \leq i \leq |s| $$$

Let

s= $$$bacabcab$$$$$$1$$$) We remove $$$s_2$$$ = $$$a$$$ and we get $$$s$$$ = $$$bcabcab$$$

$$$2$$$) We remove $$$s_2$$$ = $$$c$$$ and we get $$$s$$$ = $$$babcab$$$

$$$3$$$) We remove $$$s_1$$$ = $$$b$$$ and we get $$$s$$$ = $$$abcab$$$

$$$4$$$) We remove $$$s_3$$$ = $$$c$$$ and we get $$$s$$$ = $$$abab$$$

$$$5$$$) We remove $$$s_3$$$ = $$$a$$$ and we get $$$s$$$ = $$$abb$$$

$$$6$$$) We remove $$$s_2$$$ = $$$b$$$ and we get $$$s$$$ = $$$ab$$$

$$$7$$$) We remove $$$s_2$$$ = $$$b$$$ and we get $$$s$$$ = $$$a$$$

Which cannot be shortened further. So we removed a total of $$$7$$$ characters. Shouldn't that be the optimal answer?

'a' can never be removed as it has no previous letter (see problem statement)

You cannot remove "a" because it was clearly mentioned that only those letters can be removed whose neighbor is previous letter of the alphabet.

I think you misunderstood the meaning of this problem. You should read it again carefully.

Look at the problem statement!'a'has no previous letters!

We can't remove 'a'.it's mentioned in the problem.

Actually i got very confused so according to question b is previous element of a, or y is previous element of z,but a is not previous element of d.I wasted a lot of time in this

'a' can never be removed bro

I'm confused by the editorial for Div2F/Div1D. When I tried to implement it I seem to always get WA buried somewhere in test case 6, and I don't really know how it handles the case of odd number of consecutive ones, or the case:

$$$11010 \rightarrow 01110 \rightarrow 01011$$$

Edit: Oh, I think you meant deleting the $$$1$$$ s from the string altogether. I turned out to be converting them into $$$0$$$ s instead, which is clearly wrong

Edit 2: And now I fail at test 10. This is tough.

Edit 3: Got it. Had to add special handling for "empty" strings.

I don't understand this explanation for Div2C.

Can somebody explain it please?

For me that just proofs that for the lexicographically last letter (ie 'z') there is no letter after that one. Then it states that if such one is removed, all other possible removals still exist. So we can remove letters in reverse lexicographically order. Kind of induction.

Please, add author solutions too.

Hey, is there anyone here knows other problems that need to use the tree compression trick like D1E? I learned it before but I forgot :(

Can someone explain div1D please? I just can't get the point of what the tutorial is saying...

In short, the tuition is like this. Since you have the transformation is to change "011" to "110" and vice versa, the parity of the number of "1" between any two consecutive "0"s remains. Therefore, if there is any "11" in our original string, the fact that you delete it would not change anything. Just delete all "11" in the original string, and you obtain a shorter new string. To answer queries, just pick the substrings of the new string that correspond to the substrings in the original, and compare if they are the same.

Thanks a lot bro! A lot nicer explanation!

In div2E as per my understanding we need to implement segment tree to support max query over cumulative sum. Am I right?

For those like me — This problem requires segment tree with lazy update

In problem div2 D how is the minimum rebuild in example 3 0? why is not 1?

If you make the transposed graph (basically all edges in opposite direction) and calculate the shortest distance from the last node, you'll see that the given path is indeed the shortest path from 8 to 1. Therefore, if I take this path on node 8, I can continue with this. Question asked rebuilds not builds so first build is excluded.

Can anyone help me find which test case my code for DIV2-B fails.

You have an error in overflowing values. The maximum value for n is 2*10^5, for each number 4*10^5. Multiplying these numbers we get 8*10^10. While int can store data up to 2*10^9. Replace the int with long long. P.S. And in general, I would recommend doing this task the way it is editorial, because on big tests you will get TL.

Tanx alot.

I couldn't understand the editorial of Div2C .It will be great if someone explain it to me. Moreover,I cannot figure out which case i am missing.here is my code:https://codeforces.com/contest/1321/submission/72240578

It is something as follow: We first try to remove occurrences of 'z'. Now if there any 'z' remaining that can't be removed. Then we remove occurrences of 'y' wherever possible. and so on.

And here in each pass you can scan entire array. As there are only about 100 characters.

Now, you may ask why we are not starting with 'a' then 'b' then 'c' and so on...?

Because suppose you have following string — 'cbca'. Then if we had tried to remove 'b' then it would not be remove after it's pass gets over. But after removing 'c' string becomes 'ba'. So, we find out that we can also remove 'b'.

And if you think a little bit than you will see that in our approach of removing characters in reverse alphabetical order once we find that we can't remove given character then we will never be able to remove that character in future.

Can somebody explain why my soln to div 2 D is giving WA ??. Submission Link

Why I time out on 1320A?I think I only use time O(N)!

## include<bits/stdc++.h>

using namespace std; typedef struct{ int val,id; }node; int main(){ int i,n; long long s,k,m=0; scanf("%d",&n); vectorb(n),a,c; for(i=0;i<n;i++){scanf("%d",&b[i].val);b[i].id=i+1;m=m>b[i].val?m:b[i].val;} while(b.size()){ a=c;//清空 s=0; k=b[0].id-b[0].val;//按id-val分组 for(i=0;i<b.size();i++){ if(b[i].id-b[i].val==k)s+=b[i].val; else a.push_back(b[i]); } b=a;m=m>s?m:s;//更新最大值 } printf("%lld\n",m); return 0; }

I am posting a bit unique solution for div2C

remove adjacent. Hope you find it useful. https://codeforces.com/contest/1321/submission/73099847My code is running on my system and many online compilers but giving different answer with codeforces compiler. Please help and tell me why am I getting different output with codeforces compiler. My code is Here.

The problem statement is Here

Can someone help me out? My code for div 2 D(Navigation system) showing memory limit exceeded on test case 14. My submission link is https://codeforces.com/contest/1321/submission/73115697 . I am unable to debug my code.

That is because while adding elements in queue during bfs, the elements which are already present in queue may get added again in further iterations. For e.g, in the third test case given in the problem statement, 5 and 4 will be added while traversing adjacency list of 6, and again while traversing adjacency list of 7, which makes repetitive entries in queue and this can lead to exceeding of queue size which is approx 10^7. This is leading to MLE.

To avoid it, keep an already_present_in_queue array for each element, and add only those elements in queue which are not already pushed in queue.

Div2C also has a $$$O(n^3)$$$ dp solution https://codeforces.com/contest/1321/submission/74736147

can you see what's wrong with my solution I tried to solve using dp but failed https://ideone.com/JGqJIU

can you explain it's states?

$$$dp[i][j][c]$$$ is whether we can use some sequence of moves to reduce the substring $$$s[i..j]$$$ to the character $$$c$$$ or not. After that $$$dp2[i]$$$ is the minumum length we can obtain on the prefix after some sequence of moves

Is it necessary to use lazy propagation in div1 C? Thanks

Apologies, for div2D, this is failing on test 7 and I can't figure out what's up, could anyone look through and maybe see the logic? My # of possible cases is more than that of the test's.

https://codeforces.com/contest/1320/submission/76050423

You need to add a break at line 39.

https://codeforces.com/contest/1320/submission/79646922

Thank you!! I'd stopped debugging this and it was so helpful!

Can't figure out what's wrong:

https://codeforces.com/contest/1321/submission/77013730

Basically, I compute shortest paths in the transposed graph in a table

`dist`

.`dist[X]`

is the shortest path length from node`X`

to node`P[K]`

.Then, looping

`I = 2 to K`

(i.e 2nd to last node):I check if

`p[I - 1]`

to`p[I]`

is an edge in some shortest path from`P[I - 1]`

to the last node (i.e.`P[K]`

. I check this by checking`dist[P[I]] == dist[P[I - 1]] - 1`

.If the

check fails, then both`max`

and`min`

increase by 1, because in both cases, we need to rebuild because we transitioned to a non shortest-path edge.If the

check succeeds, then I check if how many outgoing edges does`P[I - 1]`

have that's part of some shortest path:there's more than one, then I increment`max`

only because the worst case is when the navigation system suggested a different path.there's only one, then max and min doesn't change.Maybe my bfs is wrong

Lol yep, my bfs is wrong.

In my implementation, it's possible that's nodes can be pushed twice in the queue. This should not be the case when counting paths. I got AC after fixing this.

Can someone tell me whats wrong in my code ? Question : Div 2C

Codeforces Round 625 div2 D wa4。 I can not make sense. https://codeforces.com/contest/1321/submission/82265166

I am already solved. thanks~.

Can anyone please explain me the editorial of div2E/div1C?

Let's represent all the monsters as the points at the coordinate plane: a monster with $$$x$$$ toughness, $$$y$$$ power, and $$$z$$$ coins is represented by the point $$$(x,y)$$$ with value $$$z$$$. For each pair of a sword with power $$$x_0$$$ and a shield with toughness $$$y_0$$$, the result is the sum of values over the points which belong to the rectangle $$$0 \leqslant x < x_0, 0 \leqslant y < y_0$$$ minus their costs.

Let's now apply the scanline technique: we will process the events of two types: a new monster and a new sword. Let's build a segment tree over the (sorted by toughness) list of shields and define the current set of monsters (initially empty) as $$$S$$$. For each shield with toughness $$$y_0$$$ and cost $$$c$$$, a leaf value in the segment tree stores

Let's sort an array of events by $$$x$$$. If some sword has the same power as some monster's toughness, then the sword is processed before. For each event in the sorted order:

- If it's a monster with power $$$y$$$ and value $$$z$$$, then for each shield with the higher toughness add $$$z$$$, which means adding on some suffix segment of the segment tree. That monster is added to $$$S$$$;

- If it's a sword with power $$$y$$$, then you need to find the shield with the biggest value in the segment tree. The final answer is the biggest value among the values of these pairs.

Thank you. I got it.

can anyone tell me why i am getting TLE in div1-A. Here's my submission 87744808

Same here

DP solution for journey planning is giving TLE. O(n^2) isn't fast enough even for 2 second time limit?

Is there any accepted recursive dp solution of div2A ? Is it possible in the given time limit?

1320D — Reachable Strings can be solved in O(n) with hashes without a segment tree. I have put Submission for reference.

Anyone can explain why this $$$O(nlog(n))$$$ submission in Div1 a is getting WA? submission