For particular values of p we might be able to do better, but for
unknown values of p, I can't think of anything better than this:
int g(double p)
{
int n = 0;
for(int i = 0; i < 30; ++i)
n += n+f();
return n > (int)(p*1073741824.0);
}
On Sep 12, 9:55 am, JITESH KUMAR <jkhas...@gmail.com> wrote:
> Hi
> You are given a function f() that returns either 0 or 1 with equal
> probability.
> Write a function g() using f() that return 0 with probability p (where 0<p<1
> )
>
> --
> *Regards
> Jitesh Kumar*
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