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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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2021-08-16 11:15:07
IDK but i am not able to solve it using merge sort, it says SEGMENTATION FAULT. |
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2021-06-30 06:33:50
Getting TLE, even after using merge sort algorithm and scanf/printf. Any idea. Why it is so |
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2021-06-23 16:15:35
Can be easily solved using the Merge Sort function. Just add some checks while merging and you are good to go. |
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2021-06-04 10:02:16
Similar problems using Ordered_set: Points on Line(CF) Sliding Median(CSES) Sliding Cost(CSES) Inversion Count(SPOJ) |
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2021-05-15 10:21:11
A good problem to learn PBDS if you use C++ https://www.geeksforgeeks.org/ordered-set-gnu-c-pbds/ |
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2021-04-09 10:29:22
Ac in one go using segment tree |
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2021-03-31 07:07:42
modified merge sort on passing by reference and using printf/scanf (instead of cin/cout ) gets accepted. Apparently differences are too big between cin/cout and scanf/printf in SPOJ. |
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2021-03-29 14:22:52
Solved using Fenwick Tree ! |
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2021-03-16 08:18:55
Finally get AC by using binary search tree |
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2021-02-24 12:25:00
Return type long long vs int tripped me up. Last edit: 2021-02-24 12:25:12 |