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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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2016-11-16 10:34:27
use long long to get AC. Use ORDER STATISTIC BST algorithm. Last edit: 2016-11-16 10:35:25 |
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2016-11-10 18:57:11
good one to do with BIT .... will give more understanding of BIT |
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2016-11-10 16:49:21
For INPUT {5,2,3,8,6,1} : OUTPUT must be 8 but answer is 5 in sample input! Is it Wrong?????? |
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2016-09-23 20:18:01 munjal
long long long !!!!!!!!!!!!!!!!!!! makes me 3 WA. |
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2016-08-16 05:59:44 Romil Jain
Got several WA because of int. Use long long int. Also, a little weak test cases. It accepts A[i] == A[j] also as an inversion, whereas in question it is clearly mentioned A[i] > A[j]. |
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2016-08-05 11:06:06 kunal todi
used mergesort in c++..but getting TLE.. |
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2016-07-22 11:53:18
I ll do this some day!! Done ! Too much WA after long time! ^_^ BIT is little tough to understand at first. Last edit: 2016-07-29 15:36:55 |
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2016-07-22 11:53:18
Last edit: 2016-07-22 11:53:42 |
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2016-07-20 11:10:11 Bruce
Using mergesort and Python. Generated a test case with n = 16000 and response was instant. But when submitting I'm getting time limit exceeded. Are further optimisations required? |
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2016-07-17 15:16:16 liuxueyang
@praval_singhal Thanks for your remind. I shouldn't use int for the ans. :-( use Fenwick Tree. |