MikeMirzayanov's blog

By MikeMirzayanov, 4 weeks ago, In English

You can view PDF version of the tutorial by the link.

Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
 
 
 
 
  • Vote: I like it
  • +70
  • Vote: I do not like it

»
4 weeks ago, # |
  Vote: I like it +37 Vote: I do not like it

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.

»
4 weeks ago, # |
Rev. 2   Vote: I like it -18 Vote: I do not like it

Nice editorial

»
4 weeks ago, # |
  Vote: I like it -17 Vote: I do not like it

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

»
4 weeks ago, # |
Rev. 2   Vote: I like it -8 Vote: I do not like it

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]

  • »
    »
    4 weeks ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    i can not understand editorial for A. can you explain?

»
4 weeks ago, # |
  Vote: I like it -8 Vote: I do not like it

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 $$$p^q$$$.

»
4 weeks ago, # |
  Vote: I like it -38 Vote: I do not like it

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

»
4 weeks ago, # |
  Vote: I like it -12 Vote: I do not like it

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

»
4 weeks ago, # |
Rev. 5   Vote: I like it -21 Vote: I do not like it

Thanks for good problems

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

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

»
3 weeks ago, # |
Rev. 2   Vote: I like it +3 Vote: I do not like it

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

»
3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like 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?

»
3 weeks ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

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

  • »
    »
    11 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I notice that you are sorting the vector every time a monocarp or polycarp query is done. This is probably making it slow. You can instead try using sets as they are implemented using binary-search trees and will be faster (O(log n) vs O(n logn) per query). My submission using sets

»
2 weeks ago, # |
  Vote: I like it 0 Vote: I do not like 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.

»
2 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anybody share the code of C. Berpizza

  • »
    »
    2 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    See in my submissions

  • »
    »
    9 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it
    #include<bits/stdc++.h> 
    using namespace std;
    typedef long long int ll;
    typedef vector<ll> vi;
    typedef vector<vector<ll>> vvi;
    typedef vector<bool> vb;
    typedef pair<ll,ll> pii;
    #define fo(i,s,e_ex) for(i=s;i<e_ex;i++)
    #define Fo(i,k,n) for(i=k;k<n?i<=n:i>=n;k<n?i+=1:i-=1)
    #define endl '\n'
    #define MOD 1000000007//998244353
    #define setbits(x) __builtin_popcountll(x)
    #define pbb push_back
    #define mpp make_pair
    #define ff first
    #define ss second
    #define all(x) x.begin(),x.end()
    #define mset(arr,val) memset(arr,val,sizeof(arr))
    vi guest(500005,1);
    bool compForPair(pii a,pii b){
    	if(a.first==b.first){
    		return a.second>b.second;
    	}
    	return a.first<b.first;
    }
    void solve(ll caseno){
        ll i,j,q;
    	ll type,m=0,guestNo=1;
    	priority_queue<pii, vector<pii>, decltype(&compForPair)> poly(compForPair);
    	priority_queue<ll, vi,  greater<ll> > mono;
    	cin>>q;
    	while(q--){
    		cin>>type;
    		if(type==1){
    			cin>>m;
    			poly.push(mpp(m,guestNo));
    			mono.push(guestNo);
    			guestNo++;
    		}else if(type==2){
    			while(true){
    				ll gtorem = mono.top();mono.pop();
    				if(guest[gtorem]==1){
    					cout<<gtorem<<" ";
    					guest[gtorem]=0;
    					break;
    				}
    			}
    		}else{
    			while(true){
    				ll gtorem = poly.top().second;poly.pop();
    				if(guest[gtorem]==1){
    					cout<<gtorem<<" ";
    					guest[gtorem]=0;
    					break;
    				}
    			}
    		}
    	}
    }
    int main(){
    	ios_base::sync_with_stdio(false);
    	cin.tie(0);cout.tie(0);
    	ll t=1;
    	//cin>>t;
    	for(ll i=1;i<=t;i++){
    		solve(i);
    	}
    	return 0;
    }