Yup I got it man thanks!

Hi .Can someone please explain the lazy propagation part from the codeforces post "Efficient and easy segment trees" .I have understood the recursive one but I am having difficulty understanding the iterative one as it is given there.So can some one please explain a bit so that it may become a bit easier for me to understand?

Can you explain your code for segment trees?

OH!! That was easy .I wonder why it did'nt strike me.

I cannot figure out how to check if a difference of diff can be achieved or not efficiently.Can you give some more hints?

I have a doubt .My solution got accepted.If I am not wrong it works in O(n^3).So if max. n is 300 then O(n^3) means 27*10^6 operations.How come this gets accepted in 2 seconds? Is'nt the maximum number of operations that we can perform in 1 second around 10^6? So for 2 seconds should'nt it be 2*10^6?

Thanks a lot! Those were three very good points.I was making some big mistakes.

This is my code .Please check this out. 

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef vector<int> vi;
//vector<long long int> ar;
//const ll mod=1000000007;
//map<int,int> mp;
int ar[1000005];
int dp[1000005];
int dpb[1000005];
int main()
{
int n;int k;
 cin>>n>>k;
int i;
for(i=1;i<=n;i++)
{
     cin>>ar[i];
}
if(n==1)
{
     cout<<0<<endl;

I dont know why I am getting some tests as incorrect.I calculate the forward case as dp[i]=max(dp[i-1],dp[i-2])+ar[i] ,this loop I run from k to n. For backward case I do the same from 1 to n. and then I find the index where I get max(forward+backward) .Please help me out.Can you tell me what is wrong with this approach

Can you please help me out?What I am doing is finding out the index where I'll get the maximum score from k , using dp. And then I am finding out the maximum score I can get at that particular index if I start from 1.Is this approach correct and if not can you tell me what is wrong?