【Python】Codility in Python : Lesson 13 - Fibonacci Numbers【Ladder】

markdown Fibonacci Numbers 第一題：【Ladder】
Count the number of different ways of climbing to the top of a ladder.

---

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 zero-indexed 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.

Assume that:
　　L is an integer within the range [1..30,000];
　　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].

---

方法一：參考 Code Says 的做法
Correctness：100%、Performance：100%
```python def solution(A, B): limit = len(A) result = [0] * len(A) B = [(1 << item) - 1 for item in B] # 取得費氏數列 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(limit): result[i] = fib[A[i] + 1] & B[i] return result ```

完整練習題 source code 請參閱：github