awesome solution
On Tue, May 3, 2011 at 1:43 PM, 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;
> }
>
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--
Regards
Aditya Kumar
B-tech 3rd year
Computer Science & Engg.
MNNIT, Allahabad.
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