Suddendly, all ideas of this contest are mine. And I'm very sorry about one missing small test in the problem B. This is the biggest mistake I made during the preparation of this round.

1324A - Yet Another Tetris Problem

**Tutorial**

Tutorial is loading...

**Solution**

```
for i in range(int(input())):
n = int(input())
cnto = sum(list(map(lambda x: int(x) % 2, input().split())))
print('YES' if cnto == 0 or cnto == n else 'NO')
```

1324B - Yet Another Palindrome Problem

**Tutorial**

Tutorial is loading...

**Solution**

```
for i in range(int(input())):
n = int(input())
s = list(map(int, input().split()))
ok = False
for i in range(n):
for j in range(i + 2, n):
if s[i] == s[j]: ok = True
print('YES' if ok else 'NO')
```

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
vector<int> pos;
pos.push_back(0);
for (int i = 0; i < int(s.size()); ++i) {
if (s[i] == 'R') pos.push_back(i + 1);
}
pos.push_back(s.size() + 1);
int ans = 0;
for (int i = 0; i < int(pos.size()) - 1; ++i) {
ans = max(ans, pos[i + 1] - pos[i]);
}
cout << ans << endl;
}
return 0;
}
```

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n;
cin >> n;
vector<int> a(n), b(n);
for (auto &it : a) cin >> it;
for (auto &it : b) cin >> it;
vector<int> c(n);
for (int i = 0; i < n; ++i) {
c[i] = a[i] - b[i];
}
sort(c.begin(), c.end());
long long ans = 0;
for (int i = 0; i < n; ++i) {
if (c[i] <= 0) continue;
int pos = lower_bound(c.begin(), c.end(), -c[i] + 1) - c.begin();
ans += i - pos;
}
cout << ans << endl;
return 0;
}
```

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
bool in(int x, int l, int r) {
return l <= x && x <= r;
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n, h, l, r;
cin >> n >> h >> l >> r;
vector<int> a(n);
for (auto &it : a) cin >> it;
vector<vector<int>> dp(n + 1, vector<int>(n + 1, INT_MIN));
dp[0][0] = 0;
int sum = 0;
for (int i = 0; i < n; ++i) {
sum += a[i];
for (int j = 0; j <= n; ++j) {
dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + in((sum - j) % h, l, r));
if (j < n) dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j] + in((sum - j - 1) % h, l, r));
}
}
cout << *max_element(dp[n].begin(), dp[n].end()) << endl;
return 0;
}
```

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
vector<int> a;
vector<int> dp;
vector<int> ans;
vector<vector<int>> g;
void dfs(int v, int p = -1) {
dp[v] = a[v];
for (auto to : g[v]) {
if (to == p) continue;
dfs(to, v);
dp[v] += max(dp[to], 0);
}
}
void dfs2(int v, int p = -1) {
ans[v] = dp[v];
for (auto to : g[v]) {
if (to == p) continue;
dp[v] -= max(0, dp[to]);
dp[to] += max(0, dp[v]);
dfs2(to, v);
dp[to] -= max(0, dp[v]);
dp[v] += max(0, dp[to]);
}
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n;
cin >> n;
a = dp = ans = vector<int>(n);
g = vector<vector<int>>(n);
for (int i = 0; i < n; ++i) {
cin >> a[i];
if (a[i] == 0) a[i] = -1;
}
for (int i = 0; i < n - 1; ++i) {
int x, y;
cin >> x >> y;
--x, --y;
g[x].push_back(y);
g[y].push_back(x);
}
dfs(0);
dfs2(0);
for (auto it : ans) cout << it << " ";
cout << endl;
return 0;
}
```

Can someone suggest some basic dp problems like E (same or little higher difficulty) please.

You can always look at the problemset and turn on the

`dp`

tag and sort by rating or number of solves in ascending order. I'd recommend looking around a rating of 1600-2000 for some basic (or a bit harder) dp tasks.DP tasks from 1600-2000 rating

I'm aware of that feature but in practice, I found that most 1600-2000 problems tagged with dp don't have their intended solutions with dp. Rather they are some greedy problems which have in their editorials "This problem can be solved using dp also", which is not helpful when I'm unable to solve the problem. This is why I'm asking :)

I see. In that case, here are some USACO Gold problems I would recommend that aren't just greedy.

Thanks very much

Good DP problems for all rating range.AtCoder Educational DP

I think this might be helpful. DP PROBLEMS AND TUTORIALS

Thanks.It really has good quality of DP problems

https://www.youtube.com/playlist?list=PL_z_8CaSLPWekqhdCPmFohncHwz8TY2Go

You can go refer this if you know Hindi.

DP is a good choice

E: A different dynamic programming approach, with runtime`O(n*h)`

.Let

`dp(i,j)`

be the maximum number ofgoodsleeping times if Vova slept the`i`

to`n`

times with the starting time`j`

. Then, the value`dp(1,0)`

will be the answer. Initially, all dp(i,j) = 0.Transitions(to be done from i = n to i = 1, with j being all possible values of time (0-h)):Let

`t1 = (j + a_i) % h`

and`t2 = (j + a_i - 1) % h`

Then,

`dp(i,j) = max(good(t1) + dp(i+1, t1), good(t2) + dp(i+1, t2))`

are all the cases covered in such? can you elaborate? thanks in advance

I think my solution will be clear to you. Just have a look, basic recursive function with memoization. https://ideone.com/i53rk9

can you explain your approach a little bit ??

Since, we know that we can either select

'a[i]'or'a[i]-1'. So, I called the recursive function two times, one with 'a[i]' as sum and the other with 'a[i]-1' as sum and hence returns the maximum of that. Do the same for allifrom1 to n-1. It is just the same as calling 'fib(n-1)' and 'fib(n-2)' in fibonacci series.2-d DPworks because 'n' is in the range [1,2000] and every time we are doing'sum%h'the range of value will be [0,h) and h ranges from [3,2000]. I recommend you to make a recursive tree of the given sample input (take the help from my code). This will help you to see the repeated calculation which I am storing in array(dp).https://codeforces.com/contest/1324/submission/73769097

Can you please see to my code . why my code fails for testcase 40

Your code fails for input when

Lcontains0i.e, whenL,Ris [0,..]. This happens beacause at the time of calling you are passingsumas0. For example try for input (1 44 0 14 23), the answer should be0as neither23nor22lies betweenL,Rbut your code gives1. To remove this problem, you just need to call the recursive function two times : (1. index=1,sum=arr[0]) (2. index=1,sum=arr[0]-1). And finally returns maximum of1and2. Hope it helps !!!Thanks.

Thank you so much man. I was completely frustrated as my code was not working but your comment was here to help me. :-)

What does the dp[i][j] state signify ?? Thanks

dp[i][j]signifies the total number of good sleeps from indexiton.Thanks

Plzz can you explain about the state transitions.

Hi I used similar approach getting WA on test 40 can u please help me https://codeforces.com/contest/1324/submission/73656520

Read the previous comment. Your code also behaves similar.

Essentially, what my state is how many good sleeps Vovu can have by starting from this hour and disregarding the previous i-1. Now, we can only disregard the first 1,2,3,...n and starting hour can only be in 0,1,2,...,h-1 so we have n*h states which cover all possible cases.

I did the same using memoization table.

Hey, i have solved question E. but i have done it in different way !! can you please explain, what you have done? i can't understand what you are trying to say !! please reply, i am right now focusing on DP.

Well I don't get what exactly u didn't understand, so I am copying the main logic of my code which got AC in hopes you can get the algorithm I used:

hmm, got it...you have done push dp and i have done it in reverse way.

Simpler way to solve C is to calculate the longest contiguous sequence of L's and then output it plus one.

Or you can find out farthest 2R's and print the difference in their positions. Edit : The editorial says the same . Sorry I hadn't read editorial before commenting here . XD

Still though, there is no need to remember the whole array of distances between R's. Instead, it is enough to remember the max distance.

This is what I did..but by seeing others solution I thought it was wrong lol

Alternate solution to task E:

Considering all times of the day modulo h, let

`dp[i][j]`

be the maximum number of good sleeping times assuming that Vova has slept the first`i`

times and that the current time of day is`j`

.Also, let's maintain boolean array

`pos[i][j]`

such that`pos[i][j]`

is true IFF it is possible to end up at time`j`

after sleeping`i`

times.Now, transitions can be calculated by setting

`dp[i+1][(j+ai)%h]`

to`max(dp[i+1][(j+ai)%h, dp[i][j] + ((j+ai)%h >= l && (j+ai)%h <= r ? 1 ; 0)`

and applying a similar transition for`ai-1`

.Now, the answer is just the max

`dp[i][j]`

such that`i`

=`n`

.I noticed a comment denoting a similar solution although my submission was in-contest unlike the other comment.

i thought exactly the same but mine seems doesn't work

can you please help me

my submission

Thanks in advance

EDIT:

Now it is working but why do we need to maintain a boolean array of possibilites?

Shouldn't it be handled automatically by setting to zero the dp matrix initially

please explain.

During the contest, my intuition was thinking that not all

`(i,j)`

pairs would be possible during DP, so I included the boolean matrix as an extra precaution.I knew that the extra space wouldn't really matter and played it safe. IDK if it makes a difference but I wasn't worried in-contest because I wanted to make sure I just got the task right.

Can anyone suggest where can we study and practice rerooting question like F. Thanks in advance : )

You can solve the following problems using the rerooting idea:

Thank you ^_^

luciocf Can you tell how to solve the first one — Save Nagato!

Do the same as said, in the above editorial !

Instead of keeping the black and white count keep a multiset for each vertex storing the depth using each branch, now reroot for each vertex and then erasing the largest depth kept in the multiset,putting the largest depth into each children multiset and then finally erasing it from each children multiset & putting it back in afterwards in the vertex' multiset.

Have a look here at my submission : https://dmoj.ca/src/1961273

I am unable to see your solution. It is showing access denied

http://collabedit.com/qfhfs

you can see my code : https://dmoj.ca/src/1961373. I think i write a similar structure to the solution given by vovuh.

I am also unable to see your solution

https://pastebin.com/SVGg3f0H Sorry.

Yes I will dm you

ok will wait for your message

Kindly post it here, it will help others too. TIA

What was the case that wasn’t included for B? I got hacked and I just wanted to know what the case was.

I think your mistake is that you are keeping track of the last seen element rather than the first seen.

I just know a bit of cpp so I'm unable to run your code, but I can tell you the test case on which most of the codes where hacked It was: 1 3 1 1 1, try running this and the output should be "YES"

Incase you want the specific test case on which your code was hacked, you can check it at https://codeforces.com/contest/1324/submission/73037905, you will be getting an option of view test (right under Hacked written in red text)

Thank you, I'm new to CodeForces so I didn't know how to view the test that hacked my code.

To view the hack:

Your Profile -> Submissions -> Hacked Submission ->

`#`

Button -> View HackYour Hacked Submission

@vovuh I tried to solve F using two DFS but kept failing test case 3 and I couldn't figure out why it fails on this test case. Can you help me point out my mistake? My submission is 73111564 and I've added comments to make it easy to read.

the problem is on how you build the tree, i changed your buildTree method, check https://codeforces.com/contest/1324/submission/73114269

Oh, yeah, I definitely did not build tree correctly :). Thank you very much!

In the 1324D — Pair of Topics's Tutorial

at the beginning of the Tutorial is $$$ i < j $$$ , but then turn into $$$ j < i $$$

can someone explain it ? plz.

Edit ： i made a mistake , sorry

A can also be done by checking if difference of adjacent elements mod 2 equals 0 or not

Why 6 downvotes?? :(

Codeforces community is kinda weird and specific to it's liking. Don't pay attention to upvotes or downvotes on your comments. Do check the FAQ on this to know more.

Can somebody explain why the above solution for D works? Once we sort all

`Ci`

there is no track of their original indexes anymore.. so how is it correct?The indexes don't matter. The condition i<j is given just so that you don't count (1,2) and (2,1) as different pairs. Ex. If the array is a, b, c, Then the pairs are ab, ac and bc. But if it is b, a, c, Then also the pairs are ba, bc and ac.

Thanks got it now. The pairs will be unique anyways as per the above solution.

When will the ratings get updated... Generally how much time it takes after the contest ends .?

What type of dp in problem E. Editorial said that is a standard dp problem, but i can't recognize that. Help me plz.

Its a simple include/exclude concept used in dp/recursion problems, like subset sum, Coin change etc.

Thanks

thanks NoobCoder1998. That's also my question too.

Question D can be solved in O(n) after sorting also using two pointers, which seems a bit efficient.

Link to my solution

The sort itself takes O(nlogn)

I clearly said after the sort, it takes O(n). It is just an efficient way to do the same thing. Suppose the sorted array is given, my logic would take O(n), but the given would still take O(nlogn).

And yet, the solution still has a time complexity of O(NlogN)...

Lol , If the array is sorted you say,its the same as saying that you would have done it in O(1) time if they gave you a table with sum of pairs which is O(n^2) or an array which is sorted(which is what you are asking). But O(1) doesnot matter because of the O(n^2) process , similarly , The very thing that your O(n) approach depends on an O(nlogn) process (doesnt matter if it is sorting) is what makes your solution O(nlogn) .

Can somebody explain palindrome problem? How does it check the palindrome by removing characters in between?

the question says we need a palindromic sub sequence of length greater than or equal to 3, So the easiest palindrome that can be formed by sub seq is of length 3 which has 1st and 3rd elements same, the 2nd element won't matter . Simply search for 2 same elements in the array but they should not be consecutive (so that you can insert any 1 element between the 2 similar elements).

If the array has 2 same elements that don't occur together then the answer is "YES".

The solution given by vovuh for E is giving WA on tc 2. Any suggestions on what could be wrong in it?

I copied the code from local directory, not from the polygon, the local code contained one bug which was fixed on polygon. You can realize what is this bug yourself (there is some kind of hint in the tutorial). And now you can copy and submit the correct solution but I don't know why do you need it.

this contest be like:

so funny

Can anyone explain O(N) solution for problem B ??!

The only thing important is the indexes of repeated elements in the array. so what we could do was to store the min index and the max index where each element is in the array and then just check is the diff of min index and max index is greater than 1. If it is then we could include any element between these indices to make a palindrome of length 3.

Link to my submission — https://codeforces.com/contest/1324/submission/73029984

P.S this is my first post so please ignore the informalities.

Thanks!

$$$O(N*LOG)$$$

Yes , im sorry . its n*log(n).

You can maintain an auxiliary array whose index will be the numbers present in the original array. That auxiliary array may be initialized by -1. So when you traverse the array if the corresponding index in aux array is -1 that means it is the first occurence of that number in the main array and you can replace -1 with that first index of that number. Now if you encounter that number again you can simply take the difference between current index and first index(stored in the aux array). if that difference is >1 then just print yes and return. Or if you cant find any such index print NO

$$$O(N)$$$ 73087070

Perhaps I am very poor at parity-trick questions, but problem A is very non-intuitive to me, and I was very shocked to see the number of submissions, problems B,C,D were much more straightforward for me.

I still don't think I understand problem A fully.

OK, but this can prove that answer is "NO" when parity of any two numbers is different. But how do I prove that it has to be "YES" when all numbers have same parity?

It's somehow very non-intuitive to me :(

The way Tetris(the game) works is the bottom most row gets cleared if there is a block present in every column. Since the block we can add spans 2 rows and 1 column, the height increases by 2 when we add it to a column. If the parity of each column is same then you can add those blocks to make the height of each column equal to, say, x. Therefore, you can clear the bottom-most row x times to get a fully empty board. Lets say your numbers are 1 3 5 9 3. Then you can add some number of blocks to each of the columns to get 9. After doing that, you can clear the whole board. Hope this helps :)

When parity of all numbers is same, in one step you can increment the minimum element by 2 and then decrease all elements by 1. So the overall effect is that minimum element got increased by 1 and rest all elements were decreased by 1. Now it's easy to see why after a finite number of steps all elements in the array will become equal.

You are not alone.I spent much more time on A than B、C、D.

I don't know it will help you this way or not but here's how I looked at it. Consider the cases when when all the adjacent numbers differ by an even counts(which means all the numbers have the same parity). Like 3 5 7 5 3 1 or 2 4 2 16 4. The only motto of the question is to make all numbers equal(Obviously, after that you can turn them all to zero).You start by adding +2 to the minimum number possible and then decrease the numbers till at least one of the numbers reaches to zero. Let's say p q r s t (numbers differs by an even count) and let's say t is the smallest number , you add +2 to t and let's say (for simplicity) t+2 is again the smallest number in the array so what happens if you actually perform operation 2nd described in the question — you subtract (t+2) to all numbers because t+2 is still the smallest number (so it's the first number that will be turned to zero) so the numbers become p-(t+2) q-(t+2) r-(t+2) s-(t+2) 0. Now you add +2 to zero and repeat the process and now it's easy to see that you can turn down them to same number if and if only they have same parity.

Thank you all for your attempts. I think I have a somewhat better understanding now.

The solution provided for E is not running. I think the issue is that we are running inner loop from 0 to n which includes the case of dp(i,j) where j>i which is invalid. Please correct if I am wrong.

IN PROBLEM A, The two processes are given in the problem. Does the process 2 has always to be performed whenever process 1 is performed or we can perform process 1 as many times as we want and then perform process 2. Can someone explain it please. Thanks in advance.

I have an interesting approach to question D,Time complexity: O(n).

You can define c[i] = a[i] — b[i],then sort them,finally while(l<r){ if(c[r]+c[l] > 0){ ans+=r-l; r--; } else { l++; } I think many people will use it to pass;

Your approach is

`O(N Log N)`

$$$-$$$ sort functionoh,sorry I forgot it,haha!

Hi! I solved Problem F but it throws TLE on Test case 42. Unable to understand why, can someone please help? https://codeforces.com/problemset/submission/1324/73134886

Where can I practice tree dp problems like https://codeforces.com/contest/1324/problem/F. Thnaks!

there you go 1092 pF

I never saw that kind of lengthy system testing. Did it happen like this before?

Can anyone explain Problem A.Tetris problem?

Yes !! In this problem we can observe that if there all odd nor even numbers in the array then the ans is yes else it is no It is just simple observations nothing else

Adding block of size 2x1 will not change the parity (eg. if a number is odd after adding 2 it will remain odd).So, in order to complete tertris we must have all elements of array of same parity either odd or even.

After 100% it went to 91% .WTH

When will the ratings get updated ?

I did my first Problem E in life but after the contest myself. I'm happy. I used Memoization. This question was much like LIS to me.

My F question uses two DFS, the complexity should be no problem, why is it TLE?73153261

I tried the same approach with question D as given in tutorial, still i am getting a wrong ans. Can someone please spot where i might have gone wrong? It goes wrong in test case 12. Thanks in advance.

/* arr[i]= a[i]-b[i]; ->as in the tutorial

only that arr is sorted in DESCENDING order

and i have used a 2-pointer technique */

static int findPairs (Long arr[] , int n )

change the data type of result as long. Overflow is happening.

Wtf!!! Such a silly mistake. I wasted an hour almost to debug my solution. Thanks a lot

2 pointer is good approach ............

Can anyone help me in knowing https://codeforces.com/contest/1324/submission/73161536 is giving a TLE

I have been looking at it for a long time and unable to understand why.

Thanks!

I think problem E can be solved by greedy Did anyone solved it by greedy ? I have tried , but i failed

No, you can't solve E by greedy.

Can someone help me find the error in my solution of E https://codeforces.com/contest/1324/submission/73208155 It gives wrong answer verdict on test 40

Replace the line

`if(sleep>=l && sleep<=r)`

with`if(sleep>=l && sleep<=r && i>0)`

This is because both

`l`

and`r`

have lower limit 0 in the question. So you need to omit that case.See my solution for clarity.

P.S — Assuming you don't have made any silly mistakes elsewhere.

Thanks it worked.

P.S I love pizza as well

I am the first russian?

Someone please explain me the rerooting technique used in problem F. I am unable to understand it's editorial. Please help.

If you don't understand the editorial, just see others code, i think it is more straight forward, when you understand the code, then you can go back see if you can understand the editorial.

I hope this helps

Just solved E all by myself. I think it is an easier approach. my solution

someone plz drope solution for A by using eazy c++

here is my solution in c++

thanks but i couldnot understand anything i prefer this one

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

using namespace std; int x, t, i, j, n, l, k; int main() { cin >> t; for(i=1; i<=t; i++){ cin >> n; int a[n]; for(j=0; j<n; j++) cin >> a[j]; l=0; k=a[0]%2; for(j=1; j<n; j++) if (a[j]%2!=k) {cout <<" NO " << endl; l=1; break; } if (l!=1) cout << "YES"<< endl;

}

return 0; }

Why is this code of 1324D — Pair of Topics getting TLE?

Idea of code in short :- BIT / Fenwick tree is implemented using unordered_map (C++) to query smaller values, of type (b[ j ] — a[ j ]), than (a[ i ] — b[ i ]) for all j < i.

Any suggestions to optimize this code using the same idea?

what's wrong in my code(failed on test case 4)

## include <stdio.h>

## include <stdlib.h>

int main() { int T,n,i,j; scanf("%d",&T); while(T--) { scanf("%d",&n); int

arr = (int)malloc(sizeof(int)*n); for(i=0;i<n;i++) scanf("%d",&arr[i]); for(i=0;i<n/2;i++) { for(j=n-1;j>i+1;j--) { if(arr[i] == arr[j] && j-i >1) { printf("YES\n"); break; } } if(arr[i] == arr[j]&& j-i >1) break;}

someone plz drope solution for D by using eazy c++

https://codeforces.com/contest/1324/submission/74789853

can someone please explain that in

pair of topicswhy-c[i]is not working,why **-c[i]+1 **.where does my code goes wrong for PROBLEM E link of solution:-

solution here...

.thank in advance:)

1 1 2 2 3 3 4 4 5 5 14 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 6 15 15

the checker is saying this is a palindrome. can anyone explain how? this is test 5,1st problem

as you can see number 6 occurs 2 times => 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 6 so you can choose any number of those 6 {7 7 8 8 9 9 10 10 11 11 12 12 13 13 14} 6

thanks a lot. got it. I didn't know that I can delete elements between two 6's

Is there any other DP solving for E problem?(thanks in advance)

I did D in some different manner and got wrong answer on test 6 . can anyone find where i am doing mistake

my submission : 73524489

can anyone please help me uderstanding the editorial of problem D; we have to contruct the Ci array Now if ci+cj is greater than zero atleast one will be positive so as to ci+cj to be positive. I don't understand this part of the editorial---- "Now ci>0 and we need to calculate the number of such j that ci+cj>0 and j<i. It means that each cj≥−ci+1 (for some j<i) will be okay."

please help

Can anyone explains why my solution for the Problem-F is giving TLE. According to me its complexiety should be O(2*(n+v)) in worst case?

vector mv[200005];

int a[200005];

int dfs(int x,int p)

{

}

void solve()

{

}

This gives TLE : * Used Arrays A,B,C and 2 loops to input A and B and form C * Sorted C * This finally used lowerbound thing https://pastebin.com/icGaeuH8

This doesn't give TLE : * Used vectors A,B,C and 3 loops to input these * Sorted *Lower bount thing https://pastebin.com/9fajyGEi

Can anyone help me out ?

Tutorial Question E :

loop 1: i=0;i<n;i++ loop 2: j=0;**j<=n**;j++

-> i is number of sleeps -> j is number of sleeps he goes one hour earlier

So j <= i for all j Thus, j<=i rather than j<=n.

can anybody explain to me that in problem d -c[i]+1 logic? I didn't get it

I hope you got c[i] + c[j] > 0. Now for this inequality , either c[i] must be positive , or c[j] or both. So we consider c[i] to be positive for all cases. Now c[j] has to be greater then -c[i] atleast to be 0 or positive. However c[i]+c[j] > 0 and not equal to zero and thus c[j] >= -c[i]+1.

the problem is nothing we have to remove the duplicacy therefore we have neglected the negative terms...

can someone please explain me problem D. I am not getting after sorting how we implement it.

I finally figure it out. Help this can help you.

After we sort it, we make the $$$c$$$ list sort from smallest to biggest. Because we are using the lower_bound() function, which imply we only find the $$$c[j]$$$ is smaller than $$$c[i]$$$. That's why we could skip $$$c[i] <= 0$$$.

For problem D, I still can't understand why we can skip it when c[i]<=0? Can someone help me?

We sort c: c[i-1] <= c[i]

We need c[i] + c[j] > 0 (j < i). So, if c[i] <= 0 with all j = 0..i-1 we can't get c[i] + c[j] > 0 :)

Thanks! I can understand it now.

Question (B) says that we can solve it by $$$O(n^2)$$$ or by $$$O(n)$$$. I cannot understand the writer's explanation of how we will do it in $$$O(n)$$$. Can anyone please provide code for the same. Thank you.

Try this 74688485

10/10 Mr. Specialist a14789654. Thank you very much!

int problem E am getting WA on test 40 can anyone help please my code : code

can some explain how can we use binary search in C i saw it in its tags

In problem D, doesn't sorting the c[] array mean losing information about i < j?

For the sleeping schedule problem, there seems to be a logical problem as in the first and the second transition. For the ith time you go to sleep, you can't go to sleep earlier more than i times or you can go to sleep early at most i times & the code or the tutorial doesn't address that.

Hi, I am not able to figure out the error in my logic for the problem E. Start with range [0, 0] and keep updating the range by adding ai-1 to the left border of range and ai to right border of the range. So, basically the range tells you the possible sleeping times after i_th time. At each i, check if there is any intersection with [l, r]. If yes, then good sleeping time++. Following is the code-

iam not able to understand F problem please help

Check it out

Also, what does this even mean?

`The subtree of the tree is the connected subgraph of the given tree.`

I feel the english is bad and it should be:

`A subtree of the tree is a connected subgraph of the given tree.`

can somebody explain why in problem F the dp[v] = a[v] + max(0,dp[to]) as in subtree all nodes are included so it should be dp[v] = a[v] + dp[to] why we are taking max over 0 . can somebody explain 1st testcase how for node 1 ans is 2.

In E can anyone please point the mistake in dp equation :

`if(j>i) dp[i][j]=0;`

`if(j==0) dp[i][j]=dp[i-1][j]+is((sum[i]-j)%h,l,r);`

`else dp[i][j]=max(dp[i-1][j],dp[i-1][j-1])+is((sum[i]-j)%h,l,r);`

solved.. it may be an interesting task to find out bug in this solution.

Spoilerthere should be an else if instead of if in the second line of logic code mentioned.

The approach used in solving problem D- is it any algorithm or has it any name? I find the way really cool.

Sorry to bother you vovuh, but it seems like problem B is broken. Validator is crashing and solutions are not being evaluated. "Denial of Judgment" is being shown on evaluator

Can someone tell me why am I getting time exceed error?

Q 1324D Pair of TopicsI have done sorting by both ways i.e quick sort and also using heap. Still, I am getting this error.Please Help!

The code is written below.

import heapq

def countGreater(arr, n, k): l = 0 r = n — 1 leftGreater = n while l <= r: m = int(l + (r — l) / 2) if arr[m] > k: leftGreater = m r = m — 1 else: l = m + 1 return n — leftGreater

N = int(input()) l1 = list(map(int, input().split())) l2 = list(map(int, input().split())) diff1 = []

for i in range(N): diff1.append(l1[i]-l2[i])

diff = [] heapq.heapify(diff1)

for i in range(N): diff.append(heapq.heappop(diff1))

count = 0

for i in range(N): if diff[i] > 0 : a = countGreater(diff[0:i], i, -diff[i]) count = count + a print(count)

Can anyone tell me why this submission is going wrong on TC 12 for problem D? Submission Link. Its very odd, I see that the Editorial solution isn't too far off from my method (It uses binsearch while I use a ptr method).

Can anyone explain me why he took max(0,dp[child]) everywhere?

Assuming you're talking about problem F, we don't want to add $$$dp[child]<0$$$ to $$$dp[parent]$$$ since we want to maximize $$$dp[parent]$$$.

found an easier

O(n)solution for problem D using two pointers, code hereIn problem D, Why should i need -c[i]+1?

can i solve it without adding 1?

Why does this solution for Problem F get timed out ..The Time complexity is O(n).. I guess its just traversing the tree 3 times... (https://codeforces.com/contest/1324/submission/86646387)

Problem D: Using Policy based Data Structure:

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

## include<ext/pb_ds/assoc_container.hpp>

## include<ext/pb_ds/tree_policy.hpp>

using namespace std; using namespace __gnu_pbds;

typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update> PBDS;

## define ll long long int

## define inf 1000000000

int main() { int n; cin >> n;

}

can any one please tell me why we canot calculate problem e as some function of previously calculated values

I understood the solution for the Problem D explained in the editorial, but I have one issue with the solution ( or the logic with which I tried to solve first but couldn't get AC).

What I did,

After sorting :

__> take the element from the beginning of c[] one by one

__> if c[i] < 0, then search for position having value >=( -c[i]+1 ) from the very next element in c[] ( i.e lower_bound(c.begin() + i+1, c.end(), -c[i] + 1)-c.begin()),

__> if c[i] >= 0, then search for position of value >=( c[i]+1) from the very next element in the c[] ( i.e lower_bound(c.begin() + i+1, c.end(), c[i] + 1)-c.begin()).

__> then add (n-j) in answer.

But doing so gives me WA. I couldn't understand why my logic is not correct, tried but still can't.

Link of submitted code

can some one explain me the concept behind problem a??

If someone comes across this, can you please tell why is this solution failing at case 50.