u can use the itoa function for that... On Fri, Aug 14, 2009 at 7:31 PM, sharad kumar <aryansmit3...@gmail.com>wrote:
> brother how do u get the digits of number ???u use % and / rite?? > > > On Fri, Aug 14, 2009 at 7:28 PM, Yogesh Aggarwal < > yogesh.aggarwa...@gmail.com> wrote: > >> (CORRECTED ALGO.) >> We can do like dis... >> - add all d digits of the no. >> - if the result is MORE than 10, add all the digits of the result again. >> - continue step2 if the result is still MORE than 10 >> - if the result is either 0, 3 , 6 or 9 den the no. is divisible by 3. >> >> example 1 : >> num = 12345 >> sum1 = 15 (sum > 10) >> sum2 = 6 >> since sum < 10, we stop here and since final sum = 6 ....so the given no. >> is divisible by 3 >> >> >> On Fri, Aug 14, 2009 at 7:25 PM, Yogesh Aggarwal < >> yogesh.aggarwa...@gmail.com> wrote: >> >>> (CORRECTED ALGO.) >>> We can do like dis... >>> - add all d digits of the no. >>> - if the result is MORE than 10, add all the digits of the result again. >>> - continue step2 if the result is still less than 10 >>> - if the result is either 0, 3 , 6 or 9 den the no. is divisible by 3. >>> >>> example 1 : >>> num = 12345 >>> sum1 = 15 (sum > 10) >>> sum2 = 6 >>> since sum < 10, we stop here and since final sum = 6 ....so the given no. >>> is divisible by 3 >>> >>> On Fri, Aug 14, 2009 at 7:20 PM, santhosh venkat < >>> santhoshvenkat1...@gmail.com> wrote: >>> >>>> given a number n >>>> u can get the quotient when it is divided by 4 using right shift 2 >>>> times >>>> like n >> 2 this ll give u quotient(q) >>>> u can get the remainder by subtracting 4 * q from n which will give the >>>> remainder when divided by 4 >>>> >>>> by doing this u ll express n as n = 4q + r = 3q + (q+r) >>>> in this already 3q which is divisible by 3 .. u can apply the same logic >>>> recursively to q+r and return the remainder obtained for q+r.. >>>> >>>> >>>> On Fri, Aug 14, 2009 at 7:11 PM, Yogesh Aggarwal < >>>> yogesh.aggarwa...@gmail.com> wrote: >>>> >>>>> @arun : we are not supposed to use / operator. but in ur algo u r using >>>>> / or % has to be used to check wether the diff is divisible by 3. >>>>> We can do like dis... >>>>> - add all d digits of the no. >>>>> - if the result is less than 10, add all the digits of the result >>>>> again. >>>>> - continue step2 if the result is still less than 10 >>>>> - if the result is either 0, 3 , 6 or 9 den the no. is divisible by 3. >>>>> >>>>> example 1 : >>>>> num = 12345 >>>>> sum1 = 15 (sum > 10) >>>>> sum2 = 6 >>>>> since sum < 10, we stop here and since final sum = 6 ....so the given >>>>> no. is divisible by 3 >>>>> >>>>> >>>>> On Fri, Aug 14, 2009 at 3:09 PM, Arun N <arunn3...@gmail.com> wrote: >>>>> >>>>>> take an number find its binary >>>>>> add all odd bits and even bits seperately >>>>>> now check if the difference is divisible by 3 >>>>>> if yes it is >>>>>> say 6 110 -----> 1+0 - 1 =0 >>>>>> 9 1001 -----> 1+0 - 0+1 = 0 >>>>>> 12 1100 ------> 1+0 - 1+0 = 0 >>>>>> Arun, >>>>>> >>>>>> On Fri, Aug 14, 2009 at 1:15 PM, richa gupta >>>>>> <richa.cs...@gmail.com>wrote: >>>>>> >>>>>>> >>>>>>> can we check the divisibility of a given number by 3 withoutusing >>>>>>> operators like '/' or '%'. >>>>>>> I want the efficient solution to this problem .. >>>>>>> >>>>>>> can someone help ?? >>>>>>> -- >>>>>>> Richa Gupta >>>>>>> (IT-BHU,India) >>>>>>> >>>>>>> >>>>>>> >>>>>> >>>>>> >>>>>> -- >>>>>> Potential is not what U have, its what U think U have!!! >>>>>> It is better to worn out than rust. >>>>>> >>>>>> >>>>>> >>>>>> >>>>>> >>>>> >>>>> >>>>> -- >>>>> Best Wishes & Regards >>>>> Thank You >>>>> Yogesh Aggarwal >>>>> B.Tech(IT), >>>>> University School of Information Technology >>>>> GGS Indraprastha University >>>>> Delhi >>>>> mailto: yogesh.aggarwa...@gmail.com >>>>> #9990956582 >>>>> >>>>> >>>>> >>>> >>>> >>>> >>> >>> >>> -- >>> Best Wishes & Regards >>> Thank You >>> Yogesh Aggarwal >>> B.Tech(IT), >>> University School of Information Technology >>> GGS Indraprastha University >>> Delhi >>> mailto: yogesh.aggarwa...@gmail.com >>> #9990956582 >>> >> >> >> >> -- >> Best Wishes & Regards >> Thank You >> Yogesh Aggarwal >> B.Tech(IT), >> University School of Information Technology >> GGS Indraprastha University >> Delhi >> mailto: yogesh.aggarwa...@gmail.com >> #9990956582 >> >> >> > > > > -- Best Wishes & Regards Thank You Yogesh Aggarwal B.Tech(IT), University School of Information Technology GGS Indraprastha University Delhi mailto: yogesh.aggarwa...@gmail.com #9990956582 --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
