N! overflows... Try to write a program to find the value of 30! You don't have a variable that is large enough to store such a big number...
2009/7/31 sharad kumar <[email protected]> > check this out > > Let x and y be the missing number, > > Now equation 1 is : x + y = [n(n+1)/2] - S > equation 2 is: x * y = N! /P > solve both we get elements > > On Fri, Jul 31, 2009 at 8:27 PM, Devi G <[email protected]> wrote: > >> The logic is actually simple. Tot if we mark in some way an element when >> it's scanned, we can find the missing numbers in the second scannin. >> >> 3,5,1,2,9,10,8,6 >> >> When for loop sees '3' it knows elt 3 is there. So multiplies the number >> at 3rd position by some arbitrary number. (* I've taken the arbitrary >> number to be n here but CORRECT ONE IS n+3 cos n will fail in some cases* >> ) >> >> so, when it sees '5' multiplies the number at 5th position by n+3. >> It skips when the numbr is greater than n. >> >> n+3 = 11 here. >> >> So,after first loop, >> 33, 55, 11, 2 , 99, 110, 8, 66. >> >> So now, in the second scan, the indices of all elts that are divisible by >> n+3 are present in the array. >> elts at 4th and 7th positions are not divisible. hence missing numbers are >> 4 and 7. >> >> >> >> >> > > > > -- "Reduce, Reuse and Recycle" Regards, Vivek.S --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
