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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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2016-07-14 08:20:08
Great problem for learning application of Fenwick tree! :) |
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2016-07-12 19:18:18
Got WA because of INT. |
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2016-07-12 09:18:03
good application of BIT |
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2016-07-03 12:39:13
why is the problem tagged as shortest path? |
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2016-06-28 06:05:41
You would need to fully understand Merge sort/BIT to even start with this ... if you choose it to do via merge sort , this video will finally tell you the logic you need to use to count the number of inversions https://www.youtube.com/watch?v=hqeoAIryJOc |
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2016-06-26 20:07:32
Use long for number of inversions Costed me so many WAs -.- |
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2016-06-22 15:18:30
Don't use STL set (gives TLE) use MergeSort |
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2016-06-21 22:43:19
CLRS rocks :-) AC in one go ! |
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2016-06-14 20:01:23
inversions are hidden in mergesort |
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2016-06-11 09:57:54
just make everything long long and leave a line after each test case by ur own |