@dave or anyone: can u pls explain the logic of n&3 in dave's solution? why
is it subtracted from n(which is divided by 4 using >>2) and what does n& 3
indicate?

On Sat, May 7, 2011 at 9:38 AM, Dave <[email protected]> wrote:

> @Umer: Do you suppose that you can convert an int into a string
> without using division or mod, either directly or indirectly?
>
> Dave
>
> On May 4, 1:12 am, Umer Farooq <[email protected]> wrote:
> > I'm surprised to see that why are you guys making this problem so
> complex.
> > This problem can be solved in two steps only.
> >
> > 1- Convert the given int into string
> > 2- Check if the last character is 0 or 5. // it yes, then return true
> else
> > return false
> >
> > for e.g.
> >
> > 125 (last character is 5 ... therefore it is divisible by 5)
> > 120 (last character is 0 ... therefore it is divisible by 5)
> > 111 (last character is 1 ... therefore it is not divisible by 5)
> >
> > The pseudo-code has been written in my above email.
> >
> >
> >
> >
> >
> > On Wed, May 4, 2011 at 1:49 AM, Dave <[email protected]> wrote:
> > > @anshu: Spoiler alert... I was thinking of something more along the
> > > line
> >
> > > int DivisibleBy5 (int n)
> > > {
> > >    n = n > 0 ? n : -n;
> > >    while( n > 0 )
> > >        n = (n >> 2) - (n & 3);
> > >    return (n == 0);
> > > }
> >
> > > To see that it works, write n as n = 4*a + b, where 0 <= b <= 3. Then
> > > the iteration replaces n by a - b. Consider (4*a + b) + (a - b), the
> > > sum of two consecutive values of n. This simplifies to 5*a, which is a
> > > multiple of 5. Thus, n is a multiple of 5 before an iteration if and
> > > only if it also is a multiple of 5 afterwards,
> >
> > > It is clearly log n because n is replaced by a number no greater than
> > > n/4 on each iteration.
> >
> > > Examples:
> > > n = 125. The sequence of iterates is 30, 5, 0. Ergo, 125 is a multiple
> > > of 5.
> > > n = 84. The sequence of iterates is 21, 4, -1. Ergo, 84 is not a
> > > multiple of 5.
> >
> > > Dave
> >
> > > On May 3, 3:13 am, anshu <[email protected]> wrote:
> > > > algorithm:
> >
> > > > if any number(a) is divisible by 5 it can be wriiten as 4*b + b -->
> > > > this cleary shows the last two bit of a & b will be same.
> >
> > > > lets understand by an example (35)10 = (100011)2
> >
> > > >  xx1100
> > > > +   xx11
> > > > ---------
> > > >  100011
> >
> > > > now this clearly shows we can calculate the unknowns(x) by traversing
> > > > right to left
> >
> > > > code:
> >
> > > > int main()
> > > > {
> > > >         int n, m;
> > > >         cin >> n;
> > > >         m = n;
> >
> > > >         int a, b;
> > > >         int i=2;
> >
> > > >         a = (m&3)<<2;
> > > >         b = (m&3);
> > > >         m >>= 2;
> >
> > > >         bool rem = 0,s,r;
> >
> > > >         while (m>3)
> > > >         {
> > > >                 r = a&(1<<i);
> > > >                 s = r^(m&1)^rem;
> > > >                 b = b|(s<<i);
> > > >                 a = a|(s<<(i+2));
> > > >                 rem = (r&s)|(s&rem)|(r&rem) ;
> > > >                 i++;
> > > >                 m >>= 1;
> > > >         }
> >
> > > >         if (a+b == n) cout << "yes\n";
> > > >         else cout << "no\n";
> >
> > > >         return 0;
> >
> > > > }- Hide quoted text -
> >
> > > > - Show quoted text -
> >
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> > --
> > Umer- Hide quoted text -
> >
> > - Show quoted text -
>
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