POWFIB - Fibo and non fibo
The problem is simple.
Find (a^b) % MOD
a = Nth non-fibonacci number
b = (Nth fibonacci number)%MOD
MOD = 10^9+7
Consider fibonacci series as 1,1,2,3,....
Note : It is guaranteed that Nth non-fibonacci number will always be less than MOD value for every value of N used.
First line contains T , the number of test cases.
Each next T lines contains a number N.
Print T lines of output where each line corresponds to the required answer.
Announcement: Constraints are updated. Sorry for inconvenience occurred.
3 3 2 1
49 6 4
For N=3 : 3rd non fibo number =7, 3rd fibo number=2. ans= (7^2) %MOD =49
For N=2 : 2nd non fibo number =6, 2nd fibo number=1. ans=(6^1) %MOD=6
For N=1 : 1st non fibo number =4, 1st fibo number=1. ans= (4^1) %MOD =4
Note: Test cases have been updated and costraints are changed. Those who get TLE or WA are suggested to resubmit. GOOD LUCK there.
@Mohib Your solution is no more faster, see the updated list. Best is .17s
@Lakshman awesome speed!! Congratulations...!!Last edit: 2015-07-19 20:50:18
@All problem is ready http://www.spoj.com/problems/POWFIB2/
@wisfaq I mean with hard constraints we can get more elegant solution.
As this seems to easy we can create another version of this with higher constraints with n <=1e16 ~ 1e18 what other thinks.
@ivar.raknahs : Can you plz check my submission, I think my algorithm is correct. :-(