Question

If anyone can help understand how i solve this

Batmanacci

The Fibonacci sequence can be defined as follows:

fib1fib2fibn=1=1=fibn−2+fibn−1fib1=1fib2=1fibn=fibn−2+fibn−1

We get the sequence 1,1,2,3,5,8,13,21,…1,1,2,3,5,8,13,21,…. But there are many generalizations of the Fibonacci sequence. One of them is to start with other numbers, like:

f1f2fn=5=4=fn−2+fn−1f1=5f2=4fn=fn−2+fn−1

And we get the sequence 5,4,9,13,22,35,57,…5,4,9,13,22,35,57,…. But what if we start with something other than numbers? Let us define the Batmanacci sequence in the following manner:

s1s2sn=N=A=sn−2+sn−1s1=Ns2=Asn=sn−2+sn−1

where ++ is string concatenation. Now we get the sequence N, A, NA, ANA, NAANA, ….

Given NN and KK, what is the KK-th letter in the NN-th string in the Batmanacci sequence?

Input

Input consists of a single line containing two integers NN (1≤N≤1051≤N≤105) and KK (1≤K≤10181≤K≤1018). It is guaranteed that KK is at most the length of the NN-th string in the Batmanacci sequence.

Output

Output the KK-th letter in the NN-th string in the Batmanacci sequence.

Sample Input 1	Sample Output 1
7 7	N

Answer 1

Code:-

from collections import defaultdict

fibs = defaultdict(lambda: -1)
fibs[1] = 1
fibs[2] = 1

def length(n):
if n > 87:
return -1
if n not in fibs:
fibs[n] = length(n - 2) + length(n-1)
return fibs[n]

def bat(n, k):
if n == 1:
return 'N'
if n == 2:
return 'A'
l = length(n-2)
while l == -1:
n -= 1
l = length(n-2)
if l < k:
return bat(n-1, k-l)
return bat(n-2, k)

a, b = tuple(map(int, input().split()))
print(bat(a, b))

Code Screenshot:-

Output Screenshots:-

Note: Could you please consider my effort on this work and give up vote. Thank you :)