correction-->>"Then C loses two of its matches to(one to A and one to
C). " to "Then C loses two of its matches to(one to A and one to B)
".
On May 27, 7:44 pm, vishwakarma wrote:
> so here we go
>
> Let A loses two of its matches to (one to B and one to C
correction---it was typo mistake ...
Team C loses to(one to A and one to B)
On May 27, 7:44 pm, vishwakarma wrote:
> so here we go
>
> Let A loses two of its matches to (one to B and one to C).
> Let B loses two of its matches to(one to A and one to C)
> Then C loses two of
n May 27, 7:27 pm, Arpit Mittal wrote:
> @Vishwakarma
>
> it is now ok that 11 should be the answer, but why any 4 teams cannot win 12
> matches in total...
>
> for that they have to score 12*4 = 48 points out of 56. then wats the
> problem.
>
> i know how it is coming 11 no
@Arpit
Any four team cannot win 12 matches in total.
...Rishabh is wid right answer that is ( " 11 " ).
Hence any team winning its any 11 out of 14 matches ensures its entry
to semis.
But not below 11 its entry to semi will depend on other team
performance.
On May 27, 7:11 pm, Arpit Mittal
sorry !!! correction-->>..i misread the problem
My solution gives "what is the lowest possible number of matches won
by a qualifying team ".
On May 27, 6:37 pm, vishwakarma wrote:
> Correct me if i m wrong !!!
>
> Number of matches of each team = 14.
> Let team A
Correct me if i m wrong !!!
Number of matches of each team = 14.
Let team A,B,C,D qualify for semifinal.
1.maximum number of matches A can win = 14 (all played )
2.maximum number of matches B can win = 12 (all played except played
with team A)
3.maximum number of matches C can win = 10 (all play
wait !! wait !!
@Anshuman sir...u have left few case ;
I corrected ur code and got accepted at spoj.
Here is the correct code;
//here F[] is array which got 3 fields ( element value(same as ur D[]
array) , its index in original input(same as ur IN[] array) , number
of elements less than it till
@Anshuman sir
There is mistake in above code :
u wrote -->> " ans[IN[i]] = curtime + (D[i] - last - 1) * rem + (i -
L[IN[i]] + 1); "
I think it should be -->> "ans[IN[i]] = curtime + (D[i] - last - 1) *
rem + (IN[i]- L[IN[i]] + 1);
correct me if i m wrong
On May 26, 9:17 am, anshu mishra
u can use BIT(Binary Indexed Tree) to count inversions.
I prefer this becoz it is much easier to code BIT than merge sort.
On Apr 24, 4:25 am, سهراب ابوالفتحی
wrote:
> Let A[1..n] be an array of n distinct numbers. If i < j and A[i] > A[j],
> then the pair (i,j) is called an inversion of A. Give
Here i proposed an algorithm. correct me if i am wrong !!
int main()
{
int N; //number of boards
int W; // number of workers
int smax = 0;
cin >> N >> W;
int *A = new int[N];
for(int i=0;i> A[i];
smax += A[i];
}
int
occurence is greater than n/2 than u ll
always get an unique value left after step 1 , it wont depend on the
way u select 2 different element n removing them.
On Apr 15, 12:43 am, vishwakarma wrote:
> Let A be the input array;
>
> Now algorithm is follows;
>
> struct
elp .
>
> On Fri, Apr 15, 2011 at 12:52 AM, vishwakarma
> wrote:
>
>
>
> > I can post solution of this complexity if you want !!
>
> > On Apr 15, 12:19 am, vishwakarma wrote:
> > > complexity : O(n) + O(nlogn)
>
> > > Sweety wrote:
> >
I can post solution of this complexity if you want !!
On Apr 15, 12:19 am, vishwakarma wrote:
> complexity : O(n) + O(nlogn)
>
> Sweety wrote:
> > Question :Let A[1..n] be an array of integers. Design an efficient
> > divide and conquer algorithm to determine if A contains a
complexity : O(n) + O(nlogn)
Sweety wrote:
> Question :Let A[1..n] be an array of integers. Design an efficient
> divide and conquer algorithm to determine if A contains a majority
> element, i.e an element appears more than n/2 times in A. What is the
> time complexity of your algorithm?
>
> Answ
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