The Challenge Day - CHADAY CodeChef Medium-Easy Question

Welcome back to my blog.

Here is the link to the PROBLEM .

So , the problem looks fairly simple , we just need to iterate on all players and need to find which all of compos will give positive scores .

Now , for each player , if its number does not lie on the boundary of l and r of any challenge(m) , then the answer of that player is the same as before .

This little trick helps us bring our answer in the expected time limits .

Now , if it is a boundary of l or r , then we go on to compute a matrix S whose ith element is sum of the scores upto (i-1)th challenge .

and the answer to a compo , is S[right]-S[left-1] , if its greater than 0 , we add it to the answer .

Do take a look at the following C++ code for better understanding .

Code:-

#include <iostream>

typedef long long int ll;

#include<vector>

using namespace std;

int main() {

    int t;

    cin>>t;

    while(t--)

    {

        ll l[100005],r[100005],L[100005],R[100005],S[100005],v[100005],vis[100005];

        ll i=1,j=1,k=1,n,m,q;

        cin>>n>>m>>q;

        

        i=1;

        while(i!=m+1)

        {

            cin>>l[i]>>r[i]>>v[i];

            i++;

        }

        i=1;

        while(i!=m+1)

        {

            vis[l[i]]=vis[r[i]+1]=1;

            i++;

        }

        while(j!=q+1)

        {

            cin>>L[j]>>R[j];

            j++;

        }

        i=1;

        vis[1]=1;

        ll ans=0;

        while(i!=n+1)

        {

            if(vis[i])

            {

            ans=0;

            j=1;

            S[0]=0;

            while(j!=m+1)

            {

                if(i>=l[j] && i<=r[j])

                {

                    S[j]=S[j-1]+v[j];

                }

                else

                {

                    S[j]=S[j-1];

                }

                j++;

            }

            j=1;

            while(j!=q+1)

            {

                if(S[R[j]]-S[L[j]-1]>0)

                {

                    ans=ans+(S[R[j]]-S[L[j]-1]);

                }

                j++;

            }

            }

            cout<<ans<<" ";

            i++;

        }

        cout<<endl;

    }

// your code goes here

return 0;

}

Do leave your doubts and queries in the comments section .