The number which has 5 as last digit can be written as
n = 4*a + b where b = 1 or 3
so if 4*a + 1 == n or 4*a + 3 == n; then n has 5 as a last digit.
So code looks like:
void main()
{
int n, m;
scanf("%d", &n);
m >>= 2;
m <<= 2;
if(add(m, 1) == n || add(m,3) == n)
puts("yes");
else
puts("no");
}
int add(int x, int y) {
int a, b;
do {
a = x & y; b = x ^ y; x = a << 1; y = b;
}while(b);
return b;
}
On Fri, Oct 12, 2012 at 12:46 AM, Dave <dave_and_da...@juno.com> wrote:
> @Jaspreet: The % operator is implemented using division, which is
> considered an arithmetic operation, not a bitwise operation.
>
> On primitive chips, as may be used in specialized hardware, division may
> not be implemented, in which case a non-division based algorithm may be
> desired.
>
> The question may be based on that idea, or just on how to do it faster
> than using division, which typically is a very slow instruction (compared
> to other machine instructions).
>
> Dave
>
> On Thursday, October 11, 2012 4:14:17 AM UTC-5, Jaspreet Singh wrote:
>
>> Nice solution Dave sir .. but if you know can you please tell us what is
>> the internal structure of "%" operator .. i mean it has also to be done by
>> bitwise any way .. so can't we implement that in HHLs.
>>
>> Thanks
>>
>> On Thu, Oct 11, 2012 at 1:25 AM, Dave <dave_an...@juno.com> wrote:
>>
>>> @Mohit: The decimal representation of a number ends in 5 if its low
>>> order bit is 1 and it is divisibile by 5.
>>>
>>> An algorithm using bitwise operations to check for divisibility by 5 is
>>> given at https://groups.google.com/d/**msg/algogeeks/I5HWmwKW_ks/**
>>> n38FWJSd0l8J<https://groups.google.com/d/msg/algogeeks/I5HWmwKW_ks/n38FWJSd0l8J>.
>>>
>>>
>>> It probably is not as fast as (n & 1) && (n % 5 == 0), though.
>>>
>>> Dave
>>>
>>> On Wednesday, October 10, 2012 4:06:28 AM UTC-5, mohit mishra wrote:
>>>
>>>>
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