Why are the video editorials (made by Shayan, who to my knowledge had no relationship with the problemsetting of today's round) being downvoted? Anyway, thank you Shayan for these video editorials. Always appreciate them.

All those downvotes are coming from the people who craved these good solutions during the contest.

Great explanation for problem E, very easy to understand. I think that if the problem statement was clearer with saying that the special balls were on the first k positions there would be more ACs.

thanks

Thanks for these video solutions Shayan . Helped a lot!

Hi Shayan,

For problem F, you said that, while counting the subarrays having xor value less than k, there will be a index j for index each index i, such that all the subarray's xor values will be less than or equal to k. can you prove this ?

Can Anyone please help me understanding why my code is failing as tle on tc 16 ,though its a nlogn approach and should pass according to the constraints

include

include

include

include

include

using namespace std; typedef long long int ll;

ll bin(vector a,ll l,ll r,ll target){

ll si=l+1;
ll ei=r;
ll ans=-1;
while(si<=ei){
    ll mid=(si+ei)/2;

    if(a[mid]-a[l]>=target){
        ans=mid;
        ei=mid-1;
    }else{
        si=mid+1;
    }
}
return ans;

}

int main() {

ll t;
cin >> t;

while (t--) {
    ll n;
    cin>>n;
    vector<ll> arr1(n,0);
    vector<ll> arr2(n,0);
    vector<ll> arr3(n,0);
    ll i=0;
    ll j=0;
    ll k=0;
    double tot=0;
    for(ll i=0;i<n;i++){
        cin>>arr1[i];
        tot+=arr1[i];
    }
    for(ll i=0;i<n;i++){
        cin>>arr2[i];
    }
    for(ll i=0;i<n;i++){
        cin>>arr3[i];
    }
    tot=ceil(tot/3);
    for(ll i=1;i<n;i++){
        arr1[i]=arr1[i-1]+arr1[i];
        arr2[i]=arr2[i-1]+arr2[i];
        arr3[i]=arr3[i-1]+arr3[i];
    }

    vector<pair<ll,ll>> ans;
    for(ll i=0;i<n;i++){
        if(arr1[i]>=tot){
            ll j=bin(arr2,i,n-1,tot);

            if(j!=-1){
                ll k=arr3[n-1]-arr3[j];
                if(k>=tot){
                    ans.push_back({0,i});
                    ans.push_back({i+1,j});
                    ans.push_back({j+1,n-1});
                    break;
                }
            }
            j=bin(arr3,i,n-1,tot);
            if(j!=-1){
                ll k=arr2[n-1]-arr2[j];
                if(k>=tot){
                    ans.push_back({0,i});
                    ans.push_back({j+1,n-1});
                    ans.push_back({i+1,j});
                    break;
                }
            }
        }
        if(arr2[i]>=tot){
            ll j=bin(arr1,i,n-1,tot);

            if(j!=-1){
                ll k=arr3[n-1]-arr3[j];
                if(k>=tot){

                    ans.push_back({i+1,j});
                    ans.push_back({0,i});
                    ans.push_back({j+1,n-1});
                    break;
                }
            }
            j=bin(arr3,i,n-1,tot);
            if(j!=-1){
                ll k=arr1[n-1]-arr1[j];
                if(k>=tot){

                    ans.push_back({j+1,n-1});
                    ans.push_back({0,i});
                    ans.push_back({i+1,j});
                    break;
                }
            }
        }
        if(arr3[i]>=tot){
            //cout<<arr3[i]<<" "<<tot<<endl;
            ll j=bin(arr1,i,n-1,tot);

            if(j!=-1){
                ll k=arr2[n-1]-arr2[j];
                if(k>=tot){

                    ans.push_back({i+1,j});
                    ans.push_back({j+1,n-1});
                    ans.push_back({0,i});
                    break;
                }
            }
            j=bin(arr2,i,n-1,tot);
            if(j!=-1){
                ll k=arr1[n-1]-arr1[j];
                if(k>=tot){

                    ans.push_back({j+1,n-1});
                    ans.push_back({i+1,j});
                    ans.push_back({0,i});
                    break;
                }
            }
        }
    }
    if(ans.size()>=3){
        for(ll i=0;i<3;i++){
            cout<<ans[i].first+1<<" "<<ans[i].second+1<<" ";
        }
        cout<<endl;
    }
    else{
        cout<<-1<<endl;
    }

}
return 0;

}

How about moving these wonderful video into China. They really benefits a lot!