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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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2023-03-11 09:54:46
This problem is solvable using: 1- Divide & Conquer (Merge Sort) 2- Segment tree 3- Fenweick tree (BIT) 4- Self-Balancing BST (indexed_set in C++) 5- Tries (Idk how does this work) |
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2022-11-07 14:44:15
I solved it using a binary tree :) |
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2022-08-08 17:59:27
Can this problem can be solved using Segment tree ? |
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2022-01-11 20:03:21
why is my solution accepted in CPP 14 but giving WA in C++ ? |
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2022-01-03 15:17:13
you have given that arr[i-1]<=10^7 , but it causes int overflow, when i used long long it got submitted, please correct it. Last edit: 2022-01-03 15:17:27 |
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2021-12-27 07:23:11
AC in one GO Using ordered set |
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2021-11-13 08:37:16
<snip> [NG]: Quit posting code / solutions / spoilers or you'll see your commenting privilege removed. Last edit: 2021-11-13 09:42:38 |
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2021-11-03 13:52:57 4444
@sabkx thanks a lot for the long long int comment |
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2021-10-03 07:06:22
EZ AC in second go! use long long instead of int. Merge sort works with 0.11 second |
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2021-08-19 06:33:52
Fenwick Tree :D, return type long long int |