Submit | All submissions | Best solutions | Back to list |
IITKWPCO - Create Collections |
Little Feluda likes to play very much. As you know he only plays with numbers. So he is given n numbers. Now tries to group the numbers into disjoint collections each containing two numbers. He can form the collection containing two numbers iff small number in the collection is exactly half of large number.
Given n numbers, Find out how many maximum number of collections he can form ?
Input
T: number of test cases. (1 <= T <= 100).
For each test case:
First line will contain n : (1 <= n <= 100)
Then next line will contain n numbers single space separated. Range of each number will be between 1 and 10^6.
Output
For each test case, output maximum number of collections that can be formed.
Example
Input: 2
2
1 2
3
1 2 4
Output:
1
1
Added by: | praveen123 |
Date: | 2013-08-12 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
Resource: | IITK ACA CSE online judge |
hide comments
|
|||||||||
2015-01-24 23:47:37 Anmol Pandey
To those getting WA or Runtime Error 1)range is not in 10^6 it is around 10^7 2)try changing cin to scanf there isn't any trick involved. |
|||||||||
2014-08-18 10:16:30 ||N0VICE||
No need to think too much my n^2 soln got AC in 0.0s Last edit: 2014-08-18 10:17:13 |
|||||||||
2014-08-17 07:27:17 Akshay Kamthan
wooh! AC in one go for the second time ;):P |
|||||||||
2014-08-15 18:16:32 Naruto uzumaki
DIVISION might lead to confusion . 15/2 =7 and 14/2=7 |
|||||||||
2014-06-30 21:11:40 Diksha Jaiswal
AC in 1st attempt...5 mins code :D |
|||||||||
2014-06-30 21:11:40 sarelfeniel
Very easy... plus it's on Cube so insanely fast!! |
|||||||||
2014-06-30 21:11:40 salman
Range of each number will be between 1 and 10^7 . |
|||||||||
2014-06-30 21:11:40 who cares
woww!!! it means if you are using division your solution would directly rejected!!!! :( :/ finally accepted with same logic but with * |
|||||||||
2014-06-30 21:11:40 NISHANT RAJ
simple brute force work here. try http://www.spoj.com/problems/SPCJ/ |