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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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2016-06-11 09:57:54
just make everything long long and leave a line after each test case by ur own |
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2016-06-01 11:21:19
use long long for no.of inversions counter (else it would result WA) |
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2016-05-22 16:15:21 Mallika Agarwal
Can an admin please help me out with some test cases on which problem is not working? Edit - Resolved now. Last edit: 2016-05-23 00:35:43 |
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2016-05-18 08:26:26
did first using o(n^2)....time limit exceeded ! :( now realised have to use merge sort |
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2016-05-16 12:08:59
did using modified merge_sort ..can anybody tell me how to implement using BIT??? |
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2016-05-13 13:43:35
a very nice question with the use of BIT...(use long long) |
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2016-05-13 10:56:39
solved by Merge Sort in 1st go :) But i want to solve it using BIT. Pls :( give me some hints to think about BIT for this problem |
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2016-05-12 21:43:24
use unsigned long long int for counting inversions. |
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2016-04-28 02:43:41
Modification of Merge Sort. Did the same question from CLRS :) |
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2016-04-17 10:43:09
Read BIT from ch2 of "CP by Steven Halim" and it will be a piece of cake. :) |