i think itoa wont work for long long type and as far i know it is designed to work only for integers only... so u cant find remainder for longer numbers with the logic u said above On Fri, Aug 14, 2009 at 7:36 PM, sharad kumar <aryansmit3...@gmail.com>wrote:
> say withou itoa yaar. > > > On Fri, Aug 14, 2009 at 7:35 PM, Yogesh Aggarwal < > yogesh.aggarwa...@gmail.com> wrote: > >> 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. 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