maintain a multiset of all zombie's health. at any point of time sum of x in first query become greater than any element erase it form multiset and subtract it from total sum. and along with that subtract mst.size() times x as well. The main point is how you are going to reduce the total sum at every step. I wrote these line of code I think you can get a idea out of it.

int n,k;
cin>>n>>k;
vector<int> vt(n,0);
multiset<int> mst;
int sum = 0;
for(int i = 0;i<n;i++){
    cin>>vt[i];
    mst.insert(vt[i]);
    sum += vt[i];
}

int td = 0;
for(int i = 0;i<k;i++){
    int t;
    cin>>t;
    if(t==1){
       int x;
       cin>>x;
       // td = x;
       while((*mst.begin()<=td+x)&&(mst.size()>0)){
         sum -= (*mst.begin()-td);
         mst.erase(mst.begin());
       }
       td += x;
       sum -= (x*mst.size());
       // cout<<sum<<" "<<x<<" "<<mst.size()<<endl;
    }
    else if(t==2){
       int y;
       cin>>y;
       sum += y;
       mst.insert(y+td);        
    }
    else{
       cout<<sum<<endl;
    }
}

Keep a priority queue in max heap containing the health of the zombies and keep a variable denoting total sum.For query type 1 ,add x to total damage and keeping popping elements untill top element <= total damage so far and subtract it from sum as its dead.The sum of health of alive zombies is sum of their initial health — no.of zombies alive * total dmaage so far.While adding a new zombie for query type 2 consider its health to be (y + total damage so far) and push this to priority queue

Code Link

Thankyou authors for such a nice problemset.

Will be released by Sunday. Good to hear you enjoyed the contest.

The editorial has been released now. Hope you like it :-)

Thanks!

Glad to hear that you enjoyed the contest and found the problems interesting

For E:

#include <bits/stdc++.h>
#define fast                                                                  \
    ios_base::sync_with_stdio (false);                                        \
    cin.tie (NULL);                                                           \
    cout.tie (NULL)
 
using namespace std;

signed main() 
{
    fast;  

    int n,q;
    cin>>n>>q;

    vector <pair<int,int>> pos[n+1];
    vector <int> s(n+1);

    stack <int> st;
    for(int i=1;i<=n;i++)
    {
        string str;
        cin>>str;

        if(str=="Push")
        {
            int x;
            cin>>x;
            st.push(x);
            pos[st.size()].push_back({i,x});
        }
        else
        {
            st.pop();
        }
        s[i]=st.size();
    }

    int last=0;
    while(q--)
    {
        int i,j;
        cin>>i>>j;
        i=(i+last)%n+1;
        j=(j+last)%s[i]+1;
        int loc=s[i]-j+1;
        auto itr=lower_bound(pos[loc].begin(),pos[loc].end(),make_pair(i+1,0));
        itr--;
        last=itr->second;
        cout<<last<<endl;
    }
    
    return 0;
}

Done

<p class="rated-user user-legendary">You</p>

Thanks! Glad you enjoyed the contest :-)

Great to hear that! We'll be posting the Editorial of the contest quite soon, and all the contest submissions will be made public after that. This is done so that those who wish to can spend a little more time pondering upon the problems :-)

What if the weight of gold is so high that replacing it with silver is not optimal. Think about any such case.

also we can consider that if 17*s>=g, that amount which we can remove (of s) to convert it to g would always need to be a multiple of the lcm of both g and s. So here's an easier way of understanding it (in my opinion).

then we can easily find the quantity of g and s and compute the answer.

The editoral is out now, and all contest submissions have been made public. Apologies for the slight delay.