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] <= 10^7). 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
adityaseth1011:
20180623 23:18:41
what will happen if i do it without bit mask ??


dennislo:
20180609 17:00:02
Use long in Java to avoid WA 

futile55:
20180609 11:35:55
Merge sort and AC :) 

a000:
20180524 02:27:40
The test cases are very weak... I know my code is wrong still I went for a try and it got accepted. 

devarshi09:
20180521 14:09:12
Its easy to think of a solution if one knows merge sort tree. Got to think more for BIT! 

aditya12legend:
20180330 19:43:28
Last edit: 20180330 19:58:20 

a2j007:
20180320 17:55:00
Segment tree >0.28 :) 

shauryauppal:
20180318 08:37:26
BIT with long long is the key to success. 

jmr99:
20180316 19:57:39
how to solve this using #graphtheory #numbertheory #shortestpath pls somebody share the idea . 

mranderson:
20180316 19:52:03
use long long int for the count 
Added by:  Paranoid Android 
Date:  20100306 
Time limit:  3.599s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: PERL6 