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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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2015-08-18 05:15:53
long long 1WA |
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2015-08-17 08:33:10
used mergesort even then it shows time limit exceeded... |
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2015-08-11 13:26:21 Himanshu Garg
Learn MergeSort & GO AHEAD...YOYO :) |
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2015-07-30 18:04:49 Anh Quan
use long long u will get AC :) if use int u will get WA out! |
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2015-07-29 12:52:05 SUBHAJIT GORAI
use long long to store the result ... |
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2015-07-25 05:15:36
stop putting useless comments to confuse....!! |
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2015-07-23 21:33:09 anando_du
used BIT + Binary search (for compression) got AC . long long caused one WA :v |
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2015-07-21 09:26:53 aman mahansaria
finally AC :) pls use long long instead of int |
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2015-07-19 11:51:23 Komal
for the number of inversions do use a 64 bit integer(long in java) Last edit: 2015-07-19 11:51:55 |
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2015-07-14 17:55:54
i got it correct for one test case but i dont know how to apply it for t test cases , pls give a hint . p.s. i am using c++ and just started coding and i am 16. |