say withou itoa yaar. On Fri, Aug 14, 2009 at 7:35 PM, Yogesh Aggarwal < [email protected]> wrote:
> u can use the itoa function for that... > > > On Fri, Aug 14, 2009 at 7:31 PM, sharad kumar <[email protected]>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 < >> [email protected]> 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 < >>> [email protected]> 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 < >>>> [email protected]> 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 < >>>>> [email protected]> 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 <[email protected]> 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 >>>>>>> <[email protected]>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: [email protected] >>>>>> #9990956582 >>>>>> >>>>>> >>>>>> >>>>> >>>>> >>>>> >>>> >>>> >>>> -- >>>> Best Wishes & Regards >>>> Thank You >>>> Yogesh Aggarwal >>>> B.Tech(IT), >>>> University School of Information Technology >>>> GGS Indraprastha University >>>> Delhi >>>> mailto: [email protected] >>>> #9990956582 >>>> >>> >>> >>> >>> -- >>> Best Wishes & Regards >>> Thank You >>> Yogesh Aggarwal >>> B.Tech(IT), >>> University School of Information Technology >>> GGS Indraprastha University >>> Delhi >>> mailto: [email protected] >>> #9990956582 >>> >>> >>> >> >> >> > > > -- > Best Wishes & Regards > Thank You > Yogesh Aggarwal > B.Tech(IT), > University School of Information Technology > GGS Indraprastha University > Delhi > mailto: [email protected] > #9990956582 > > > > --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. 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