@Dave : in your pseudo code for B(m,n)
you are not using m ... i guess there is a typo error.
this should be num[i]=m + 1 -i instead of this
num[i] = n + 1 - i
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Dave's solution works fine. I should have close look at it.
On Sat, Nov 5, 2011 at 12:17 PM, mohit verma wrote:
> @ Dave
>
> I think your solution wont work for the cases like (MAX_INT) C
> (MAX_INT-1).
>
>
> On Thu, Nov 3, 2011 at 10:20 PM, wrote:
>
>> correction: "if P is prime"
>>
@ Dave
I think your solution wont work for the cases like (MAX_INT) C
(MAX_INT-1).
On Thu, Nov 3, 2011 at 10:20 PM, wrote:
> correction: "if P is prime"
>
> VM
> NSIT, Dwarka
>
>
> On , vaibhavmitta...@gmail.com wrote:
> > Use Lucas Theorem is P is prime.
> >
> > VM
> > NSIT, Dwarka
> >
> > O
correction: "if P is prime"
VM
NSIT, Dwarka
On , vaibhavmitta...@gmail.com wrote:
Use Lucas Theorem is P is prime.
VM
NSIT, Dwarka
On , Dipit Grover dipitgro...@gmail.com> wrote:
> Thats really cool! Thanks Dave :)
>
>
>
>
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Use Lucas Theorem is P is prime.
VM
NSIT, Dwarka
On , Dipit Grover wrote:
Thats really cool! Thanks Dave :)
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Thats really cool! Thanks Dave :)
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@Dipit: Here's how you would do it by hand:
B(1000,10) = 1000*999*998*997*996*995*994*993*992*991 /
(1*2*3*4*5*6*7*8*9*10).
We know that the result is an integer, so all of the terms in the
denominator are factors of the numerator. Do some cancelling: 10
cancels 1000 to 100, 9 cancels 999 to 111, 8
You can do something like this:
( n* alpha) * ( m*alpha) *p
where 0 wrote:
>
>
> On Sat, Oct 22, 2011 at 9:04 PM, Mad Coder wrote:
>
>>
>>
>> Can anyone explain why ((n%p)*(m%p))%p will give wrong answer ?
>>
>> Lets say, n = 10^15 , m = 10^15 and p = 10^18
>
> So,
> n%p = 10^15
On Sat, Oct 22, 2011 at 9:04 PM, Mad Coder wrote:
>
>
> Can anyone explain why ((n%p)*(m%p))%p will give wrong answer ?
>
> Lets say, n = 10^15 , m = 10^15 and p = 10^18
So,
n%p = 10^15
m%p = 10^15
And the intermediate result (n%p)*(m%p) will overflow the long long range.
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Can anyone explain why ((n%p)*(m%p))%p will give wrong answer ?
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@Aamir: Something like this might work:
n %= p;
m %= p;
result = 0;
while( m )
{
if( m & 1 )
{
result += n;
if( result >= p )
result -= p;
}
m >>= 1;
n <<= 1;
if( n >= p )
n -= p;
}
What this does is similar to the elementary school meth
Use NTL's ZZ extended integer type.
Don
On Oct 21, 12:26 pm, Aamir Khan wrote:
> Lets say n,m and p are three long long integers.
>
> i want to calculate (n*m)%p.
>
> If (n*m) overflows the limit of long long then the answer would be wrong.
> Moreover if i do ((n%p)*(m%p))%p then also there is no
On Thu, Jan 13, 2011 at 12:06 AM, Aviral Gupta wrote:
> we can do it when hcf(b,m)=1 , in that case find inverse of b by
> extended euclidean mod m and then multiply it by a
>
Yes. And when m is prime, B(mulitplicative inverse of b) = b^(m-2)
As b^(m-1)mod m = 1 if m is prime.
>
> Regards
> Avir
we can do it when hcf(b,m)=1 , in that case find inverse of b by
extended euclidean mod m and then multiply it by a
Regards
Aviral
http://coders-stop.blogspot.com/
On Jan 12, 6:36 am, mittal wrote:
> Somehelp with (a/b)modm expression.
>
> http://en.wikipedia.org/wiki/Modular_arithmetic
> i visi
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