##### Short Problem Definition:

Count the number of different ways of climbing to the top of a ladder.

##### Link

##### Complexity:

expected worst-case time complexity is `O(L)`

expected worst-case space complexity is `O(L)`

##### Execution:

We first compute the Fibonacci sequence for the first L+2 numbers. The first two numbers are used only as fillers, so we have to index the sequence as A[idx]+1 instead of A[idx]-1. The second step is to replace the modulo operation by removing all but the n lowest bits. A discussion can be found on Stack Overflow.

##### Solution:

```
def solution(A, B):
L = max(A)
P_max = max(B)
fib = [0] * (L+2)
fib[1] = 1
for i in xrange(2, L + 2):
fib[i] = (fib[i-1] + fib[i-2]) & ((1 << P_max) - 1)
return_arr = [0] * len(A)
for idx in xrange(len(A)):
return_arr[idx] = fib[A[idx]+1] & ((1 << B[idx]) - 1)
return return_arr
```