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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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2018-03-16 19:52:03
use long long int for the count |
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2018-03-16 16:24:21
Last edit: 2018-03-16 19:52:08 |
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2018-03-16 12:15:54
hint : : use BIT or Merge seriously invested more than 4 hours to come up with sol.. nd debugged a lot of times to understand the flow of code nd finally ac in 1go !!! gys try something new rather than merge ..go for hacker-earth tut for this. Last edit: 2018-03-16 19:55:46 |
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2018-03-16 12:01:15
I'm getting WA!!! Somebody please help. <snip> Last edit: 2022-06-26 16:00:27 |
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2018-03-16 07:15:19
@addy1397 There are n distinct numbers in the problem, none of them are repeated. Some of your random inputs are invalid. |
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2018-02-26 22:01:11
Use BigInteger (number of inversions) for java else you will get WA. Last edit: 2018-02-26 22:02:24 |
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2018-01-14 14:14:47
my 100th :) using BIT |
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2017-12-31 17:32:15
Make sure to use long long in c++ |
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2017-12-28 11:14:17
I used segment tree. But gives TLE... |
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2017-12-18 23:39:12
Although I solved it using merge sort but can anyone help me in its bitmask approach? Last edit: 2017-12-21 13:01:48 |