#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
#include<string.h>
#include<math.h>
int multiply(int a,int b)
{
int i;
int temp=a;
printf("\na=%d b=%d\n",a,b);
for(i=1;i<b;i++)
a+=temp;
printf("\nfinal a = %d",a);
return(a);
}
void main ()
{
int x,rem,quo=0,i,j;
char p[20];
clrscr();
scanf("%d",&x);
itoa(x,p,3);
rem=(int)p[strlen(p)-1]-48;
printf("%dRemainder\n",rem);
for(i=strlen(p)-2,j=0;i>=0;i--,j++)
{
printf("p[j] = %d",p[j]-48);
// quo+=(p[j]-48)*pow(3,i);
quo+=multiply(p[j]-48,pow(3,i));
printf("\nquo=%d",quo);
}
printf("\nQuotient=%d",quo);
printf("\n%s",p);
getch();
}
taking base 3 will convert the number into the base 3 form ..
for example let x=100 the number to be divided by three.
so 100 in base 10 = 10201 in base 3 .
we get base 3 by consecutive divisions by 3 so the last number will always
be the remainder ie 1 .
rest 1 0 2 0 will be the quotient.
| | | |
index: 3 2 1 0
so p[j]-48 will convert the char to integer and mutiply it wil 3 ^ power (
index )
the answer will be the quotient.
On Sat, Jul 30, 2011 at 12:15 AM, aditi garg <aditi.garg.6...@gmail.com>wrote:
> @Samm : Im not able to understand ur logic...im not getting the correct
> ans...can u explain the working taking n as 7?
>
>
> On Sat, Jul 30, 2011 at 12:25 PM, SAMM <somnath.nit...@gmail.com> wrote:
>
>> Single bit shift...
>>
>> int divide(int n)
>> {
>> n-=1;
>> n>>=1;
>> return n;
>> }
>>
>> On 7/30/11, tech rascal <techrascal...@gmail.com> wrote:
>> > hw will u get the ans on repeated subtraction from the sum of the
>> digits??
>> > I mean ....if I hv 2 divide 27 by 3 thn first I'll find sum of the
>> digits
>> > i.e, 2+7=9
>> > then I'll apply repeated subtraction on 9, so hw will i reach to the
>> ans??
>> >
>> >
>> > On Sat, Jul 30, 2011 at 10:05 AM, nivedita arora <
>> > vivaciousnived...@gmail.com> wrote:
>> >
>> >>
>> >> @brijesh-
>> >> itoa basically converts integer to string ..we are using the fact tht
>> >> a number is multiple of 3 if its sum is multiple of 3
>> >> . we have int as string and we can traverse it ..for each character
>> >> apply
>> >> int sum+=*c-'0' (ankur missed the star :P)
>> >> then on sum we use repeated subtraction...i hope its clear .
>> >>
>> >> we are using all this just coz we have to use itoa ..otherwise there
>> >> are more methods (check my frst post )
>> >>
>> >>
>> >> On Jul 30, 4:34 am, brijesh <brijeshupadhyay...@gmail.com> wrote:
>> >> > @ankur I didnt get this... could u or anyone please elaborate!
>> >> >
>> >> > On Jul 30, 12:43 am, Ankur Khurana <ankur.kkhur...@gmail.com> wrote:
>> >> >
>> >> >
>> >> >
>> >> >
>> >> >
>> >> >
>> >> >
>> >> > > when you use itoa , what you get is a string. get the sum of all
>> the
>> >> digits
>> >> > > , using c-'0' and then use repeated subtraction . . .
>> >> >
>> >> > > On Sat, Jul 30, 2011 at 1:01 AM, sukhmeet singh <
>> >> sukhmeet2...@gmail.com>wrote:
>> >> >
>> >> > > > repeated subtraction !!
>> >> >
>> >> > > > On Sat, Jul 30, 2011 at 12:52 AM, nivedita arora <
>> >> > > > vivaciousnived...@gmail.com> wrote:
>> >> >
>> >> > > >> Without using /,% and * operators. write a function to divide a
>> >> number
>> >> > > >> by 3. itoa() function is available.
>> >> >
>> >> > > >> all i cn thnk of is to use shift operator and addition ,
>> >> x/3=e^(logx-
>> >> > > >> log3) or repetitive subtraction
>> >> >
>> >> > > >> but none of them uses itoa() ..ne idea how its done?
>> >> > > >> thnks !
>> >> >
>> >> > > >> --
>> >> > > >> You received this message because you are subscribed to the
>> Google
>> >> Groups
>> >> > > >> "Algorithm Geeks" group.
>> >> > > >> To post to this group, send email to algogeeks@googlegroups.com
>> .
>> >> > > >> To unsubscribe from this group, send email to
>> >> > > >> algogeeks+unsubscr...@googlegroups.com.
>> >> > > >> For more options, visit this group at
>> >> > > >>http://groups.google.com/group/algogeeks?hl=en.
>> >> >
>> >> > > > --
>> >> > > > You received this message because you are subscribed to the
>> Google
>> >> Groups
>> >> > > > "Algorithm Geeks" group.
>> >> > > > To post to this group, send email to algogeeks@googlegroups.com.
>> >> > > > To unsubscribe from this group, send email to
>> >> > > > algogeeks+unsubscr...@googlegroups.com.
>> >> > > > For more options, visit this group at
>> >> > > >http://groups.google.com/group/algogeeks?hl=en.
>> >> >
>> >> > > --
>> >> > > Ankur Khurana
>> >> > > Computer Science
>> >> > > Netaji Subhas Institute Of Technology
>> >> > > Delhi.
>> >>
>> >> --
>> >> You received this message because you are subscribed to the Google
>> Groups
>> >> "Algorithm Geeks" group.
>> >> To post to this group, send email to algogeeks@googlegroups.com.
>> >> To unsubscribe from this group, send email to
>> >> algogeeks+unsubscr...@googlegroups.com.
>> >> For more options, visit this group at
>> >> http://groups.google.com/group/algogeeks?hl=en.
>> >>
>> >>
>> >
>> > --
>> > You received this message because you are subscribed to the Google
>> Groups
>> > "Algorithm Geeks" group.
>> > To post to this group, send email to algogeeks@googlegroups.com.
>> > To unsubscribe from this group, send email to
>> > algogeeks+unsubscr...@googlegroups.com.
>> > For more options, visit this group at
>> > http://groups.google.com/group/algogeeks?hl=en.
>> >
>> >
>>
>>
>> --
>> Somnath Singh
>>
>> --
>> You received this message because you are subscribed to the Google Groups
>> "Algorithm Geeks" group.
>> To post to this group, send email to algogeeks@googlegroups.com.
>> To unsubscribe from this group, send email to
>> algogeeks+unsubscr...@googlegroups.com.
>> For more options, visit this group at
>> http://groups.google.com/group/algogeeks?hl=en.
>>
>>
>
>
> --
> Aditi Garg
> Undergraduate Student
> Electronics & Communication Divison
> NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
> Sector 3, Dwarka
> New Delhi
>
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To post to this group, send email to algogeeks@googlegroups.com.
> To unsubscribe from this group, send email to
> algogeeks+unsubscr...@googlegroups.com.
> For more options, visit this group at
> http://groups.google.com/group/algogeeks?hl=en.
>
--
The more you sweat in the field, the less you bleed in war."
Ankit Minglani
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
