HORRIBLE - Horrible Queries
World is getting more evil and it's getting tougher to get into the Evil League of Evil. Since the legendary Bad Horse has retired, now you have to correctly answer the evil questions of Dr. Horrible, who has a PhD in horribleness (but not in Computer Science). You are given an array of N elements, which are initially all 0. After that you will be given C commands. They are -
* 0 p q v - you have to add v to all numbers in the range of p to q (inclusive), where p and q are two indexes of the array.
* 1 p q - output a line containing a single integer which is the sum of all the array elements between p and q (inclusive)
In the first line you'll be given T, number of test cases.
Each test case will start with N (N <= 100 000) and C (C <= 100 000). After that you'll be given C commands in the format as mentioned above. 1 <= p, q <= N and 1 <= v <= 10^7.
Print the answers of the queries.
0 2 4 26
0 4 8 80
0 4 5 20
1 8 8
0 5 7 14
1 4 8
Segment tree is way easier to comprehend than BIT, especially because you can view it as a node-based tree, no need for array index tricks. Most of the pain comes late in the implementation w/ lazy propagation. For BIT (if you actually try to understand why it works), everything is painful. Especially for this problem, you have to go through 3 iterations, PURQ (what Fenwick did originally), then RUPQ (to help get a feel for what's next), and finally RURQ.Last edit: 2016-09-28 08:44:59
this may save your life ((long long int)r - l + 1)*v
make sure that 1) use long in java.
The limits given are correct.
you have to increase the size of the segment tree , if you'r using one ,
Last edit: 2016-08-23 20:45:15
if you trying using BIT do this first
only BIT :v
For each test case update whole lazy array to zero(2*100000)