You can try like this
#include stdio.h
#include string.h
#define N 3
void copy(int A[N][N],int B[N][N])
{
int i,j;
for(i=0;iN;i++)
for(j=0;jN;j++)
A[i][j]=B[i][j];
}
void mul(int A[N][N],int B[N][N],int M[N][N])
{
int i,j,k;
for(i=0;iN;i++)
time complexity is (cost of multiplication)*log(n)
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Yes its running fine at gcc 4.3.2 . And it might show warning in that case
just change the name of the function exp().
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To
Yes sure ,
we know if A B C are matrices of same dimension then Ax(BxC)=(AxB)xC
So now what SAMM is saying i.e normal exponential of any number can be
applied in case of matrix expo too and the same thing I too did but using
loop without recursive fashion.
I hope its clear now .
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You
Better go through following link , its explained nicely their.
http://www.ihas1337code.com/2011/01/find-k-th-smallest-element-in-union-of.html
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I don't think so that probability would be exact 1/1000 .
suppose number is ( a9 a8 a7 a6 a5 a4 a3 a2 a1 a0) where a0 is least and a9
is most significant bit then you can generate each of the bit ai using given
bit generator
but if but at a time (a9 , a8 , a7 , a6 , a5 , a3 ) and any other bit
but suppose you have to generate numbers between [1,513] then also it would
generate numbers them each with probability 1/1024 which could make a big
difference and I feel the above solution to be incorrect . May be we could
think of a better one .
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@Surender Your solution in correct one .
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It can be done in O(2^n)
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For more options,
suppose bottom 4 teams have won least matches and upper 4 teams have won
equal number of matches ...
1 - x
2 - x
3 - x
4 - x
5 - 6
6 - 4
7 - 2
8 - 0
total matches are 56
and let upper four teams have won x matches each
so x = (56-(6+4+2+0))/4
x = 11
so in this way to ensure qualification
No , you are wrong .. problem statement says how many matches should a
teams win to ensure its qualification , their no word like minimum or
maximum ...
8 gets wrong if following situation arises
1 - 9
2 - 9
3 - 9
4 - 9
5 - 8
6 - 6
7 - 4
8 - 2
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