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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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2017-12-18 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 |
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2017-12-18 21:41:05
easy BIT...int gives WA..use long long |
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2017-12-18 10:00:24
came to know a creative of using fenwick tree! |
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2017-12-01 05:07:40
Try Fenwick Tree Random testcases Input 5 6 1 2 3 4 5 6 8 5 1 4 2 6 2 6 2 5 4 1 2 4 2 7 1 4 2 5 2 56 2 1 1 output 0 11 4 6 0 |
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2017-11-13 12:41:27
BST approach using STL gives TLE! but mergesort works :) Last edit: 2017-11-13 14:54:21 |
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2017-10-23 21:40:05
tags are clickbait |
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2017-10-16 11:35:56 harkirat
1 Last edit: 2017-10-16 11:36:08 |
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2017-10-05 00:27:52
Use Long long int if you don't want WA! |
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2017-10-04 20:30:53
Mergesort does the job |
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2017-09-30 20:20:16
If you want to do it in orthrodox manner do it by merge sort.Its just modified mergesort.It can be solved using BIT concept too.There is a bit mask approach too but it gives TLE.AC in 4th go.... |