Shivansh wants to give chapo to Arpit, but only if he is able to solve the following problem. As you are a dear friend of Arpit, you decide to help Arpit solve the problem. Given an array Arr of N numbers. You have to process Q queries on that array. Queries can be of following types :
(1)add x to index y
(2) given y compute sum of (Arr[i]*(y%i)) for i from 1 to N.
Input
First Line of input contains two space separated integers N and Q. Second Line contains N space separated integers Arr[1] Arr[2] ..... Arr[N] Next Q lines describe one query each and can be one of the two type:
-> 1 i X — Add X to index i.
-> 2 Y — compute sum of (Arr[i]*(Y%i)) for i from 1 to N.
Constraints
1 <= N,Q <= 100000 1 <= Arr[i], X, Y <= 100000 1 <= i <= N
Output
For each second type query output one integer per line — answer for this query.
Sample input
4 3
3 2 3 4
1 2 1
2 10
2 12
Sample output
11
0
I think it can be solved by segment tree, but I'm not able to figure out how to handle 'y'?
Auto comment: topic has been updated by ajraj27 (previous revision, new revision, compare).
Auto comment: topic has been updated by ajraj27 (previous revision, new revision, compare).
What's the source? We don't want to accidentally help you in the upgoing contest. :)
Its from the contest called "newton challenge" which got ended. Link — https://my.newtonschool.co/course/ldrj7tqzlt/timeline/
Can you please explain now ?
I see only $$$O(q \cdot \sqrt{n \cdot \log(n)})$$$, but with rather low constant factor.
Whats the approach you're taking?
Bruteforce small $$$i$$$ for each query and maintain Fenwick tree over possible values of $$$y$$$. When a value of $$$Arr[i]$$$ for big $$$i$$$ changes, then you have to update $$$O(\frac{10^5}{i})$$$ intervals in the Fenwick tree.
For each query, we then bruteforce small $$$i$$$ and read from Fenwick tree for big $$$i$$$. I think that with a sqrt decomposition instead of Fenwick tree we can achieve even $$$O(q \cdot \sqrt{n})$$$.
Will segment tree be a feasible option ?
I'm not sure if $$$O(q \cdot \sqrt{n \cdot \log(n)})$$$ will be enough. If you want to squeeze it, I definitely do not recommend a segment tree.