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
dragonemperor:
2014-07-13 09:58:16
My BIT implementation is giving SIGSEGV. I am not accessing any array out of its bounds.
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kelaseek:
2014-07-01 07:32:16
BST 1.07
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excursionist:
2014-05-11 14:05:03
very easy :D AC ;) time(.c) << time(.cpp) |
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Mr Tambourine Man:
2014-05-10 13:36:33
use long long int..that cost me two WA |
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californiagurl:
2014-03-03 09:18:07
didn't do anything special for duplicates.....but got AC in 2 sec :)
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Ajay Prasadh:
2014-02-20 13:12:06
Check for duplicate numbers finally ac ! :D
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Anubhav Balodhi :
2014-02-04 07:52:44
good Inversion problem... AC :-D |
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rahul kumar:
2013-12-24 09:40:32
did it using merge sort...easy one |
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Achmet ibn Rashid:
2013-12-12 16:51:56
Tried both merge sort and binary indexed (aka fenwick) trees; got AC with both well within bounds, but merge sort was a lot faster:
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Achmet ibn Rashid:
2013-12-12 16:02:28
Last edit: 2013-12-12 16:56:11 |
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 |