I know that modular exponentiation is enough to solve this problem but I came across a formula on Quora by Divanshu Garg ( http://qr.ae/R753zZ ). Using his formula the problem simply boils down to calculate and print pow(a%10,b%4)%10 but this approach is giving me Runtime error. Here is my code. http://ideone.com/Y5VQLJ Problem link : http://www.spoj.com/problems/LASTDIG/ Thanks for reading! :) asked 17 Jul '15, 03:19

include <stdio.h>include <math.h>int main() {
} Try this code, as you said, a%10, it will contain 1 digit only and b%4, it will also contain one digit only. So, you can use pow() function only. I think the problem is with your modexpo() function. It gives runtime error because of the bitwise operator used. I changed b>>=1 to b/=2 and it gave Wrong answer on spoj but not Runtime error. So, change your modexpo() function. I better suggest you to use the code given above. answered 17 Jul '15, 04:01
Simply pow(a%10,b%4) gives WA. Tried a simple implementation here, http://ideone.com/jpkR2K
(17 Jul '15, 11:45)

@h1ashdr@gon... i think you have not yet understood the concept (as mentioned by you on quora) well. For example let a = 2 and b=4, then ans is last digit of 2^4=16 i.e 6. But your code prints out 1 which is why it is WA. Understand that the answer for any a repeats after a max of 4 iterations. For eg for a=3 it repeats as 3, 9, 7 and 1 while for a=6, it repeats as 6,6,6,6 so after taking b%4 which gives answer in the range [0,3] doesn't actually give the correct answer when b is a multiple of 4. So, as i wrote my code above consider the code which checks explicitly for the case when b%4=0, just add this condition and i think your code will run fine now... Happy coding. answered 18 Jul '15, 08:15
