@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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