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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
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 |
hide comments
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2015-01-26 13:48:05 Fufo
Last edit: 2015-01-26 13:48:30 |
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2015-01-26 13:15:50 internom
Last edit: 2015-01-26 13:16:19 |
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2015-01-18 19:49:38 Rajat (1307086)
Divide and Conquer. |
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2015-01-05 15:58:36 Yashpal
AC using BIT,Segment Tree,Merge Sort, Trie !! Loved this problem !! ;) |
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2014-12-26 12:13:29 agaurav77
Simple cin/cout IO with the divide and conquer paradigm (remember to use long long), AC :) |
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2014-12-04 17:32:23 rishabh aggarwal
use long long to get ac... Last edit: 2014-12-04 17:32:47 |
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2014-12-04 00:49:37 Rafael Perrella
@benevides resp should be long long |
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2014-12-03 20:46:22 benevides
i dont get it,why doesn't it work <snip> Last edit: 2022-06-26 16:04:56 |
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2014-11-25 17:55:04 arun
Real please AC in 1st go!!! use BIT And take care of repeat element. Last edit: 2014-11-25 17:56:04 |
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2014-11-12 12:19:45 [Lakshman]
@Abi you have declared count as long long and your mergesort is returning int , So declare mergesort and merge long long instead. Hope this helps. Last edit: 2014-11-12 12:25:19 |