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 <dave_and_da...@juno.com> wrote:
> @Karthikeyan: Is this supposed to relate to the question of
> determining divisibility by 5?
>
> Dave
>
> On Jan 21, 9:25 am, karthikeyan muthu <keyankarthi1...@gmail.com>
> 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
> > <aaron.nar...@gmail.com>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 <dave_and_da...@juno.com> 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 <the.um...@gmail.com> 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 <dave_and_da...@juno.com>
> 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 <anshumishra6...@gmail.com> 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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