11532 - Simple Adjacency Maximization
Posted: Fri Jan 09, 2009 6:48 pm
I try to solve this problem
But i get WA.......
can any one help me????????
Code: Select all
CUT after acc..........................
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Re: 11532 - Simple Adjacency Maximization
Posted: Fri Jan 09, 2009 8:24 pm
Try some test cases..
My output is..
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6
0 1
1 0
5 1
1 5
7 2
2 7My output is..
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0
1
47
1
367
5Re: 11532 - Simple Adjacency Maximization
Posted: Sat Jan 10, 2009 11:52 am
Thanks helloneo
now i can solve it........
Re: 11532 - Simple Adjacency Maximization
Posted: Fri Jan 16, 2009 2:32 pm
I think this problem can be solved by using DP, but i can't find the recursive relationship... 
could someone give me a hint?
could someone give me a hint?
Re: 11532 - Simple Adjacency Maximization
Posted: Fri Jan 16, 2009 3:36 pm
I didn't solve this problem with DP.. Just try to find a pattern..f74956227 wrote:I think this problem can be solved by using DP, but i can't find the recursive relationship...
could someone give me a hint?
If there are enough '0's, the best representation would be like this..
...000000101101101101
If there are not enough '0's, you have to insert '0' in smarter way..
Re: 11532 - Simple Adjacency Maximization
Posted: Sat Jan 17, 2009 11:28 am
I got AC!!! It is just a simple simulation problem 
Thank!
Thank!Re: 11532 - Simple Adjacency Maximization
Posted: Sun Apr 26, 2009 8:49 pm
ACC
It was wrong for 3 1
It was wrong for 3 1
Re: 11532 - Simple Adjacency Maximization
Posted: Mon Apr 27, 2009 4:52 pm
I think you might get a precision error here..
Remove your code after AC
Do not use floating point operationaliahmed wrote:Code: Select all
sum+=pow(2,q);
Remove your code after AC
Re: 11532 - Simple Adjacency Maximization
Posted: Sun Jul 05, 2009 1:51 pm
What does this mean ? Why isn't 0101011 (43) better than 0101101 (45) in case of 4 3 .The number of 1s adjacent to one or more 0 in the binary representation is maximized.
Also i don't understand in case 3 2 , why is it 1101 (13) instead of 10101(21) ?
Re: 11532 - Simple Adjacency Maximization
Posted: Tue Jul 07, 2009 3:37 pm
43 -> 0101011 has three '1's which are adjacent to '0'
45 -> 0101101 has four '1's which are adjacent to '0'
So, 45 is better than 43
For input
3 2
1101 and 10101 both have three '1's adjacent to '0'
But, problem statements say that
45 -> 0101101 has four '1's which are adjacent to '0'
So, 45 is better than 43
For input
3 2
1101 and 10101 both have three '1's adjacent to '0'
But, problem statements say that
So, the answer would be 13 (1101)Find the smallest integer N ..