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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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2018-07-12 12:42:30
Last edit: 2018-07-16 07:41:19 |
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2018-06-23 23:18:41
what will happen if i do it without bit mask ?? |
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2018-06-09 17:00:02
Use long in Java to avoid WA |
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2018-06-09 11:35:55
Merge sort and AC :) |
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2018-05-24 02:27:40
The test cases are very weak... I know my code is wrong still I went for a try and it got accepted. |
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2018-05-21 14:09:12
Its easy to think of a solution if one knows merge sort tree. Got to think more for BIT! |
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2018-03-30 19:43:28
Last edit: 2018-03-30 19:58:20 |
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2018-03-20 17:55:00
Segment tree ->0.28 :) |
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2018-03-18 08:37:26
BIT with long long is the key to success. |
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2018-03-16 19:57:39
how to solve this using #graph-theory #number-theory #shortest-path pls somebody share the idea . |