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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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2019-08-15 12:14:19
solved it using PBDS |
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2019-07-09 00:47:06
AC in one go using merge sort, very nice problem |
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2019-07-08 16:00:46
Divide and conquer approach |
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2019-07-01 16:11:36
good problem. can be solved using segment tree, fenwick tree, and merge sort Last edit: 2019-07-01 16:11:56 |
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2019-06-25 19:18:46
just store the element and index in pair and after that sort pair in decreasing order with respect to first element and then we have to just find the values of indexes which are smaller then the ith indexe keep the count and sum it simple BIT AC in one go loved this question my first on BIT Last edit: 2019-06-25 19:22:45 |
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2019-06-25 12:26:45
Great problem, can be done easily using divide and conquer. |
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2019-05-28 10:17:07
can anyone tell,how to solve this by BIT? |
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2019-05-27 08:33:53
Great problem...! can be done using D&C, BIT or Segment Tree... |
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2019-04-24 00:27:33
AC in one go using divide and conquer technique :D |
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2019-03-27 23:41:09
Yes, output can be up to n^2 so O(n^2) should definitely pass. |