@Karthikeyan: Again I ask: Do you think that you have used neither
division nor mod indirectly? I.e., can you assert with certainty that
there is no division or mod in sprintf() or anything it calls? I
didn't think so. In fact, I am willing to assert with certainty that
division is used, since the algorithms for converting binary integers
into base 10 use division by 10.
Dave
On Jan 21, 11:34 am, karthikeyan muthu <keyankarthi1...@gmail.com>
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 <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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> > > >> > --
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>
> > > >> > - Show quoted text -
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