Actually BIT supports max/min queries and continuous sum( a[i] ~ a[i+L] ==> +x) queries.

We can implement by using 2 BIT.

Oops, should have reread my reply before posting, thanks for the correction.

'Almost every solution you can do with Segment Tree can also be done with Binary Index Trees.' Its true the other way round because BIT only supports sum queries

I'm guessing that by B, you mean the max degree?

Why is the solution O(NB)? Isn't it just O(N)? I'm trying to figure out why the max degree constraint is given in the prob...

    int N; cin >> N;
    int A[N]; for(int i=0;i<N;++i) cin >> A[i];
    

#include <stdio.h>
#include <iostream>
#include <climits>
#include <map>
#include <cmath>

I see. I like the Stack Overflow link Shreeyash provided. I also like this tutorial on Codeforces: Counting problems from A to B.

Here are my thoughts and observations on it:

• You should think of digit DP problems as j...

The best bit manipulation tutorials are Fun with Bits – topcoder and Bit Manipulation - HackerEarth. Although I haven't tried any of these problems myself, HackerRank seems to have a good collection of problems on bit manipulation.

I find bit manipulation to be useful quite rarely. B...

It is dynamic programming on the digits of a number. It involves problems like this one Codeforces problem 489C.

Also, this might be useful : http://stackoverflow.co...

However I get TLE in last 2 subtasks. O(N*log(N)*log...

Instead of taking the center B, can't we sort the co-ordinates and then take the leftmost co-ordinate A while sweeping from right to left.

Then, find the number of co-ordinates with y-coordin...

You can also compress the coordinates and use a normal BIT, which would be much easier.

Full Solution:

• At first get the shortest path tree rooted at S using Dijkstra.Now what we observe is if those edges w...

I am still unable to figure out the solution from Keshav's Hints. Would be grateful if someone expanded those hints.

Thanks.

I'll add hints to problems that I've solved (the few ones that I could manage :P).