That trick for pairs in problem M is just awesome... During the whole contest we were struggling to remove that logn factor but couldn't find the way.

Nice editorial

Thanks for this Tutorials, It was a very cool contest, But there were a lot of test cases, which caused the long queue.

I think that there is a mis-written part in problem A in editorial: a[posi−1]≤a[posi]≥a[posi+1] should be a[posi−1]≤a[posi]≤a[posi+1]

I think there is an error in tutorial for L. "If there are no more than 2 such integers" (second sequence) should be "If there are no more than 1 such integers" or "If there are less than 2 such integers". Because in next paragraph $$$k_i > 1$$$ implies that $$$p$$$ can be important if there are 2 numbers of form $$$p^q$$$.

MikeMirzayanov is there an alternate to these explanations , it is difficult for me to understand this editorial.

Please someone can explain similar sets(problem M) editorial.

Thanks for good problems

for problem c, i just used two different sets, one for each operation type, sorting elements as either (i, m) or (m, i), where m is the estimated money and i is the customer number. this makes finding which customer is served much easier, as you can take from the front of the first set for the 1st operation and from the back of the second set for the 2nd operation

For M, how is finding all pairs (x,y) not time complexity (k^2)? How exactly do you create a separate array for each integer x and then add all values y to it

For L, I'm getting wrong answer on test case 99 (because I'm only able to find a sequence of 994 items, not 1000). Is this happening to anyone else?

In problem C Berpizza why I am getting tle on test case 12 My code please let me know and also how to resolve it

In problem M's tutorial, does it mean that: For every small sets, we find all pairs, which is O(k^2) where the k is the size of the sets. There are M/k sets so that their are O(M*k) pairs in total? How should I implement it? I tried 2 dimension unorderedmap but TLE.

Can anybody share the code of C. Berpizza

In problem C, can anyone tell me why my code is failing in the seventh test case that too by swap of two digits(136 in place of 133 and 133 in place of 136)? source: 105199970

Approach: I have used two priority queues with pairs to store customer id and price sorted with respect to id in ascending order for monocarps queries and sorted with respect to price in descending order for polycarps queries. And I have used unordered_map to check which id's have already been served.

My code for question c is giving tle on test case 17 plz help!!!!!

Can someone help me with problem K. I am not able to figure out where I am doing mistake?

Your code here...
#include<bits/stdc++.h>

using namespace std;
#define ull unsigned long long
#define ll long long
#define f first
#define S second
#define mod 1000000007


int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(0);
    // freopen("input.in", "r", stdin);
    // freopen("output.in", "w", stdout);
    int t;
    cin>>t;
    while(t--)
    {
      string s;
      cin>>s;
      int c = 0;
      for(int i=0;i<s.length()-1;i++)
      	if(s[i]!=s[i+1])
      	{
      		c = 1;break;
      	}
      if(c==0)
      	{
      	if(s[0]=='L')
      		cout<<-1<<" "<<0<<"\n";
      	else if(s[0]=='R')
      		cout<<1<<" "<<0<<"\n";
      	else if(s[0]=='U')
      		cout<<0<<" "<<1<<"\n";
      	else if(s[0]=='D')
      		cout<<0<<" "<<-1<<"\n";
      	continue;
      	}
      	vector<pair<int,int>>left(s.length()+1);
      	vector<pair<int,int>>ryt(s.length()+1);
      	for(int i=0;i<s.length();i++)
      	{
      		left[i+1] = left[i];
      		if(s[i]=='L')
      			left[i+1].f -=1;
	      	else if(s[i]=='R')
	      		left[i+1].f +=1;
	      	else if(s[i]=='U')
	      		left[i+1].S+=1;
	      	else if(s[i]=='D')
	      		left[i+1].S-=1;

      	}
      	for(int i = s.length()-1;i>=0;i--)
      	{
      		ryt[i] = ryt[i+1];
      		if(s[i]=='L')
	      		ryt[i].f-=1;
	      	else if(s[i]=='R')
	      		ryt[i].f+=1;
	      	else if(s[i]=='U')
	      		ryt[i].S+=1;
	      	else if(s[i]=='D')
	      		ryt[i].S-=1;
      	}

      	vector<int>inx(s.length());
      	for(int i=0;i<s.length()-1;i++)
      	{
      		if(s[i]!=s[i+1])
      			inx[i] = i+1;
      		else
      		{
      			if(i!=0 && inx[i-1]>i)
      				inx[i] = inx[i-1];
      			else
      			{
      				int j = i+1;
      				while(j < s.length() && s[i]==s[j])
      					j++;
      				inx[i] = j;
      			}
      		}
      	}
      	int b = 0;
      	inx[s.length()-1] = s.length();
      	for(int i=0;i<s.length();i++)
      	{
      		if((left[i].f + ryt[inx[i]].f==0 && left[i].S + ryt[inx[i]].S==0))
      		{
      			if(left[i+1].f!=0 || left[i+1].S!=0){
      			cout<<left[i+1].f<<" "<<left[i+1].S<<"\n";
      			b = 1;
      			break;}
      		}
      		
      	}
      	
      	
   

      	if(b==0)
      		cout<<0<<" "<<0<<"\n";







  

    }

    return 0; 

}

"Now, we can note that each a[i] is either minimum in LaIS or i = nxt[j] for some other element a[j]".

Can someone give me a proof of that?