'''Task description
You have to climb up a ladder. The ladder has exactly N rungs, numbered from 1 to N. With each step, you can ascend by one or two rungs. More precisely:

with your first step you can stand on rung 1 or 2,
if you are on rung K, you can move to rungs K + 1 or K + 2,
finally you have to stand on rung N.
Your task is to count the number of different ways of climbing to the top of the ladder.

For example, given N = 4, you have five different ways of climbing, ascending by:

1, 1, 1 and 1 rung,
1, 1 and 2 rungs,
1, 2 and 1 rung,
2, 1 and 1 rungs, and
2 and 2 rungs.
Given N = 5, you have eight different ways of climbing, ascending by:

1, 1, 1, 1 and 1 rung,
1, 1, 1 and 2 rungs,
1, 1, 2 and 1 rung,
1, 2, 1 and 1 rung,
1, 2 and 2 rungs,
2, 1, 1 and 1 rungs,
2, 1 and 2 rungs, and
2, 2 and 1 rung.
The number of different ways can be very large, so it is sufficient to return the result modulo 2P, for a given integer P.

Write a function:

def solution(A, B)

that, given two non-empty arrays A and B of L integers, returns an array consisting of L integers specifying the consecutive answers; position I should contain the number of different ways of climbing the ladder with A[I] rungs modulo 2B[I].

For example, given L = 5 and:

    A[0] = 4   B[0] = 3
    A[1] = 4   B[1] = 2
    A[2] = 5   B[2] = 4
    A[3] = 5   B[3] = 3
    A[4] = 1   B[4] = 1
the function should return the sequence [5, 1, 8, 0, 1], as explained above.

Write an efficient algorithm for the following assumptions:

L is an integer within the range [1..50,000];
each element of array A is an integer within the range [1..L];
each element of array B is an integer within the range [1..30].'''

# O(L) complexity
# https://app.codility.com/demo/results/trainingBQVK3D-HXZ/
def solution(A, B):
    limit = max(A)         
    result = [0] * len(A)
    fib = [0] * (limit+2)
    fib[1] = 1
    for i in range(2, limit + 2):
        fib[i] = (fib[i - 1] + fib[i - 2])

    for i in range(len(A)):

        result[i] = fib[A[i]+1] & ((1<<B[i])-1)
    return result
