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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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2015-12-18 21:20:46 Shubham Gupta
Only keep the counter as long long. Take care of the formatting,newline after the no. of testcases(t) and then after every set of numbers. |
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2015-12-17 17:48:21 archit saxena
omg took 10 hours to solve use mergesort to get O(nlogn)time.bubblesort might look attractive at first but gives O(n^2)time |
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2015-12-11 15:35:28 Sarin Nanda
advice : use long long instead of int i was getting WA before that |
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2015-12-09 14:30:18 computer science
nice segment tree |
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2015-12-08 09:59:58
Stupid mistake costed 3 wrngs!!! |
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2015-12-03 12:20:22
i can't understand what the judge says wrong answer on test 1 but when i run in my pc it is correct |
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2015-12-02 19:09:18
there should be 7 inversions in case -2 instead of 6 Confused!!!! |
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2015-11-15 12:23:59 Priyank
Key Learnings --- Data Type. Use of Merge Sort. Remain careful in setting bound on array. |
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2015-11-05 00:26:40
@Anh Quan Thanks for the tip about using long long. That did the trick. |
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2015-11-02 20:37:04
superb use of divide and conquer ;) |