INVCNT - Inversion Count
Let A[0 ... n - 1] be an array of n distinct positive integers. If i < j and A[i] > A[j] then the pair (i, j) is called an inversion of A. Given n and an array A your task is to find the number of inversions of A.
Input
The first line contains t, the number of testcases followed by a blank space. Each of the t tests start with a number n (n ≤ 200000). Then n + 1 lines follow. In the ith line a number A[i - 1] is given (A[i - 1] ≤ 107). The (n + 1)th line is a blank space.
Output
For every test output one line giving the number of inversions of A.
Example
Input: 2 3 3 1 2 5 2 3 8 6 1 Output: 2 5
hide comments
dardev:
2020-07-03 16:48:44
The concept is covered very well here -> https://www.youtube.com/watch?v=7_AJfusC6UQ |
|
mukul202:
2020-05-19 17:39:01
Just solved this question using:
|
|
picchi_67:
2020-05-06 05:46:10
Can be Solved Easily with merge sort.
|
|
mt6170078:
2020-04-21 16:14:30
why tag of Bitmask? how to use Bitmask here?? |
|
yourmom__:
2020-04-21 07:40:01
I see the question. Okay. Thought of a simple O(n^2) algo. Then I submit and get TLE. Offcourse. Then I see the commenst section. People are commenting about Fenwick trees? Segment trees? Self balancing BST? Now WHAT THE FUCK are these? Spent 2-3 days learning their basics. People who are getting TLE, spend sometime learning these new data structures, ans dont give up.
|
|
offamitkumar:
2020-03-29 10:47:02
Solved it With
|
|
vishesh_2308:
2020-03-12 10:57:36
https://youtu.be/B29dVmpIpn0
|
|
terminator_pk:
2020-02-07 19:18:21
Last edit: 2020-02-08 12:55:30 |
|
savage_19:
2020-02-04 08:30:10
1. Do it with merge sort first.
|
|
avsd_47:
2020-01-21 18:22:23
Caution:the answer you are going to print must be of type long long costed me 3 wa's. |
Added by: | Paranoid Android |
Date: | 2010-03-06 |
Time limit: | 3.599s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: PERL6 |