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] <= 10^7). 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
hide comments
polkerty:
2017-09-05 04:13:19
So my solution is really weird (respect to everyone who saw the merge sort). Basically I used 3 vectors and some carefully chosen constants to get the answer by sorting progressively smaller pieces of the array, and then using a bunch of binary searches to find the values I wanted. |
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jayharsh:
2017-08-25 16:36:54
Merge sort implementation in a innovative way............. |
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diaaeddin:
2017-08-21 13:00:29
AC in one go using trie |
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jarvis_1997:
2017-07-30 17:01:52
Comments are like GOT spoilers. Never look at them unless you are done ;)
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studyfather:
2017-07-24 13:50:35
use merge sort and AC!!! |
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ani_geek9654:
2017-07-01 13:37:46
AC in 1
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hzshang:
2017-06-30 18:26:07
Binary tree, easy :P |
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kartiks22:
2017-06-26 09:40:03
took me 2 hours to figure out long long mistake !!! |
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plab0n:
2017-06-19 21:54:15
I am getting TLE in my merge_sort approach. Plz help me how can I make my code efficient.
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lord_poseidon:
2017-06-19 07:24:22
Easy one AC in one go
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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 |