@Navin: No problem. Just print a 1 instead of a quotient digit. That makes the code even simpler, like this: int n=<the number that ends with 3>; int divisor=1; printf("1"); while( divisor != 0 ) { printf("1"); divisor = 10 * (divisor % n) + 1; } printf("\n"); Dave
On Friday, September 21, 2012 2:05:55 AM UTC-5, Navin Kumar wrote: > @Dave sir: Thanx for reply. your solution gives the exact multiple like > 37 for 3, 8547 for 13. In the question i think we have to print the number > which is 13x8547 ....which will be very large (out of integer range).In > that case we have to store result in string. > > > On Fri, Sep 21, 2012 at 12:09 PM, Dave <dave_an...@juno.com > <javascript:>>wrote: > >> @Navin: It means that given a positive integer n whose decimal >> representation ends in 3, find a multiple, m*n, which is written solely >> with the digit 1. E.g., 3: 37 * 3 = 111; 13: 8547 * 13 = 111,111. >> >> Dave >> >> On Thursday, September 20, 2012 11:56:08 PM UTC-5, Navin Kumar wrote: >> >>> @all: Please explain question number 8. I am not getting the question >>> exactly what it says ? >>> >>> On Fri, Sep 21, 2012 at 9:30 AM, Dave <dave_an...@juno.com> wrote: >>> >>>> @Bharat: Simulate long division, dividing a number 1111...1 by the >>>> number. You can do this one digit at a time, printing the quotient digit >>>> by >>>> digit until you "bring down" a zero. It could look something like this: >>>> >>>> int n=<the number that ends with 3>; >>>> int divisor=1; >>>> while( divisor < n ) >>>> divisor = 10 * divisor + 1; >>>> while( divisor != 0 ) >>>> { >>>> printf("%d",divisor / n); >>>> divisor = 10 * (divisor % n) + 1; >>>> } >>>> printf("\n"); >>>> >>>> Dave >>>> >>>> On Thursday, September 20, 2012 9:45:55 PM UTC-5, bharat wrote: >>>> >>>>> what is the solution(not brute force) for 8th question ? >>>>> >>>>> On Fri, Sep 14, 2012 at 5:19 PM, Bhupendra Dubey >>>>> <bhupen...@gmail.com>wrote: >>>>> >>>>>> Which edition of barron? >>>>>> >>>>>> >>>>>> On Wed, Sep 28, 2011 at 6:05 PM, VIHARRI <vihar...@gmail.com> wrote: >>>>>> >>>>>>> 1. Java uses stack for byte code in JVM - each instruction is of one >>>>>>> byte, so how many such instructions are possible in an operating >>>>>>> system. >>>>>>> >>>>>>> 2. Three processes p1, p2, p3, p4 - each have sizes 1GB, 1.2GB, 2GB, >>>>>>> 1GB. And each processes is executed as a time sharing fashion. Will >>>>>>> they be executed on an operating system. >>>>>>> >>>>>>> 3. write a recursive program for reversing the linked list. >>>>>>> >>>>>>> 4. write a program for checking the given number is a palindrome. >>>>>>> ( dont use string / array for converting number ). >>>>>>> >>>>>>> 5. write a recursive program for multiplying two numbers a and b, >>>>>>> with >>>>>>> additions. The program should take care of doing min # additions as >>>>>>> that of which ever number is lower between a and b. >>>>>>> >>>>>>> 6. There are two sets A and B with n integers, write a program to >>>>>>> check the whether there exists two numbers a in A and b in B such >>>>>>> that >>>>>>> a+b = val ( val is given ); >>>>>>> >>>>>>> 7. write a program to return the row number which has max no of one's >>>>>>> in an array of NxN matrix where all 1's occur before any 0's starts. >>>>>>> >>>>>>> 8. For every number that has 3 in its units place has one multiple >>>>>>> which has all one's i.e. 111 is such multiple and 13 has a multiple >>>>>>> 111111. Write a program to find such multiple for any number that has >>>>>>> 3 at its units place. >>>>>>> >>>>>>> 9. what are the maximum no of edges that can be connected in a graph >>>>>>> of n vertices and 0 edges such that after adding edges given graph is >>>>>>> still disconnected. >>>>>>> >>>>>>> 10. One Question on critical section. >>>>>>> >>>>>>> For Analytical Test - Prepare the Questions in the barrons book of >>>>>>> sample paper - 2 ( they have give two passages ) >>>>>>> >>>>>>> -- >>>>>>> You received this message because you are subscribed to the Google >>>>>>> Groups "Algorithm Geeks" group. >>>>>>> To post to this group, send email to algo...@googlegroups.com. >>>>>>> To unsubscribe from this group, send email to algogeeks+...@** >>>>>>> googlegroups.com**. >>>>>>> >>>>>>> For more options, visit this group at http://groups.google.com/** >>>>>>> group**/algogeeks?hl=en<http://groups.google.com/group/algogeeks?hl=en> >>>>>>> . >>>>>>> >>>>>>> >>>>>> >>>>>> >>>>>> -- >>>>>> Thanks & regards >>>>>> Bhupendra >>>>>> >>>>>> >>>>>> >>>>>> -- >>>>>> You received this message because you are subscribed to the Google >>>>>> Groups "Algorithm Geeks" group. >>>>>> To post to this group, send email to algo...@googlegroups.com. >>>>>> To unsubscribe from this group, send email to algogeeks+...@** >>>>>> googlegroups.com**. >>>>>> >>>>>> For more options, visit this group at http://groups.google.com/** >>>>>> group**/algogeeks?hl=en<http://groups.google.com/group/algogeeks?hl=en> >>>>>> . >>>>>> >>>>> >>>>> -- >>>> You received this message because you are subscribed to the Google >>>> Groups "Algorithm Geeks" group. >>>> To view this discussion on the web visit https://groups.google.com/d/** >>>> msg/algogeeks/-/pDBPxDR3R1oJ<https://groups.google.com/d/msg/algogeeks/-/pDBPxDR3R1oJ> >>>> . >>>> >>>> To post to this group, send email to algo...@googlegroups.com. >>>> To unsubscribe from this group, send email to algogeeks+...@** >>>> googlegroups.com. >>>> For more options, visit this group at http://groups.google.com/** >>>> group/algogeeks?hl=en <http://groups.google.com/group/algogeeks?hl=en>. >>>> >>> >>> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To view this discussion on the web visit >> https://groups.google.com/d/msg/algogeeks/-/LpHDrQKDb90J. >> >> To post to this group, send email to algo...@googlegroups.com<javascript:> >> . >> To unsubscribe from this group, send email to >> algogeeks+...@googlegroups.com <javascript:>. >> 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. 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