@Shishir
cool! this seems to work!!
C[i] = (N-1)/D[i] ; //this is super cool
On Sep 21, 10:29 am, Shishir Mittal 1987.shis...@gmail.com wrote:
Let the denominations be D[] = {1000,500,100},
and amount be N.
Let C[] , denotes the count of each denomination.
for ( i=0 ; i 2 ; i++) {
@Channa
thanks for explaining for the benefit of everybody.
On Sep 22, 4:50 pm, Channa Bankapur channabanka...@gmail.com wrote:
@eSKay, @Ankur, et al.,
Please be aware that there are non-Indians too in the group.
Hi All,
Let me try and define the problem precisely (as far as I can).
The
ok i can tell one with complexity of O(log(n)) , u take a no. pow =
3,temp=3;
then
while(1){
temp=pow;
pow*=pow;
if(pown){
pow=pow/temp;
break;
}
}
now do , pow =pow *3; untill pow=n;
and check from here
am i clear?
On Sep 23, 2:40 am, Dave dave_and_da...@juno.com wrote:
He didn't
I disagee. Please don't force your personal opinions on everybody like
this.
Thanks.
On Sep 20, 4:39 pm, ankur aggarwal ankur.mast@gmail.com wrote:
i think there is no use of discussing this ques..
On Sun, Sep 20, 2009 at 2:25 PM, eSKay catchyouraak...@gmail.com wrote:
yes it is
Vicky, Yup! you got it there.
BTW you might like to do binary search again instead of sequential
search when you hit the break condition.
to indeed make it O(logn).
Infact if M=3^n is given then this algo would be of complexity O(log
log M)
_dufus
On Sep 23, 11:59 am, vicky mehta...@gmail.com
You can also do it in O(log2(log3(N))) by implementing binary search
for power of 3 !
but its slower than normal way for small N's !
On Sep 21, 8:22 pm, Anshya Aggarwal anshya.aggar...@gmail.com wrote:
how to find that whether the given number was of the form 3^n. ( i.e. 3 to
the power n).
isnt his algo same like modular exponentiation in CLRS
On Wed, Sep 23, 2009 at 5:09 PM, Dufus rahul.dev.si...@gmail.com wrote:
Vicky, Yup! you got it there.
BTW you might like to do binary search again instead of sequential
search when you hit the break condition.
to indeed make it O(logn).
yaa , that's right as then we are sure of having complexity O(n)
On Sep 23, 4:39 pm, Dufus rahul.dev.si...@gmail.com wrote:
Vicky, Yup! you got it there.
BTW you might like to do binary search again instead of sequential
search when you hit the break condition.
to indeed make it O(logn).
@ Minjie Zha ,
hey, how does these things clicks to u , as i thought it for 2 hrs.
and still couldn't find a completely correct sol..
On Sep 19, 2:04 pm, Minjie Zha diego...@gmail.com wrote:
Oh yes, I made a mistake.
Your are right.
On Sep 18, 12:02 am, ashish gupta ashish12.it...@gmail.com
Computing log base 2 is one multiplication simpler than computing log
base 3.
Dave
On Sep 22, 7:49 pm, Prabhu Hari Dhanapal dragonzsn...@gmail.com
wrote:
is computing log to the base 2 simpler than computing log to the base 3
?
On Tue, Sep 22, 2009 at 5:40 PM, Dave
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