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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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2019-03-27 23:41:09
Yes, output can be up to n^2 so O(n^2) should definitely pass. |
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2019-03-27 21:50:29
original problem, I know how to solve it, is O(n ^ 2) okay? |
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2019-03-27 16:12:41
not bad task... |
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2019-03-07 13:47:32
SOLVED using BIT in one go |
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2019-03-06 15:55:46
solved this problem after months. really good question |
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2019-01-20 19:49:52
@marethyu thank u |
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2019-01-18 23:37:10
used bit to solve |
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2019-01-17 09:09:39
Try solving https://codeforces.com/problemset/problem/459/D after you solve this. |
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2019-01-01 20:58:36
Use long long when counting inversions, it cost me several WAs! |
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2018-12-23 15:58:33
why others is AC with int althought it has already big number but this one have 2 use long in c++ ?? why ?? |