@dave: thanks for that..but i just wanted to know as to how u thot of
converting n to a-b in the iteration. when u say 4a +b is a multiple of 5
iff a-b is a muliple of 5 i was able to get that only when i tried an
example...if they ask divisbility by 3 or 6 or 7 how wud the logic change??

On Sat, Jan 21, 2012 at 9:34 AM, karthikeyan muthu <
[email protected]> wrote:

> check the last char ... it should be 0 or 5 , int to string without mod
>
>
> On Sat, Jan 21, 2012 at 10:05 PM, Dave <[email protected]> wrote:
>
>> @Karthikeyan: Is this supposed to relate to the question of
>> determining divisibility by 5?
>>
>> Dave
>>
>> On Jan 21, 9:25 am, karthikeyan muthu <[email protected]>
>> wrote:
>> > @dave
>> >
>> > int no=10;
>> > char ans[100];
>> > sprintf(ans,"%d",no);
>> > cout<<ans;
>> >
>> > On Fri, Jan 20, 2012 at 10:29 PM, Arun Vishwanathan
>> > <[email protected]>wrote:
>> >
>> >
>> >
>> > > @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 -
>> >
>> > >> > > --
>> > >> > > You received this message because you are subscribed to the
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>> >
>> > >> > --
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>> >
>> > >> > - Show quoted text -
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