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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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2017-06-19 21:54:15
I am getting TLE in my merge_sort approach. Plz help me how can I make my code efficient. Here is my Code link: <snip> Thanks in advance :) Last edit: 2022-06-26 16:04:40 |
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2017-06-19 07:24:22
Easy one AC in one go |
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2017-06-12 18:05:30
AC in one go!! Only Merge Sort.. |
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2017-06-02 09:48:56
AC in one go)))) |
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2017-05-25 04:38:39
Core concept of sorting algorithm.. |
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2017-05-02 14:53:24
thanks @fane_faiz :) |
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2017-04-10 19:42:03
Merge Sort->"ACIN1GO" |
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2017-04-05 20:10:49
Segment Tree+Merge Sort AC in 1 go :) |
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2017-04-05 14:12:46
If BST implementation gets AC, then the test cases are weak because the running time for BST is O(nh), which is O(n^2) in worst case. Basically it should fail if the sequence is sorted. |
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2017-03-23 18:42:59
solved using binary search tree..AC in one go |