在 python 中使用递归函数和记忆化处理系列

发布于 2025-01-10 19:18:23 字数 561 浏览 5 评论 0原文

我正在处理像斐波那契数列这样的数列...（在斐波那契数列中，第 n 项只是 n-1 和 n-2 的总和。）但就我而言，我想要第 n 项是前一项一半的总和。例如：

n=5
Output should be: [0, 1, 1, 2, 3]

n=12
Output should be: [0, 1, 1, 2, 3, 6, 11, 22, 42, 84, 165, 330]

def my_function(n):
    
    list1=[0,1]
    for i in range(0,n-2):
        if(i>2):
            value=sum(list1[i//2+1:])
        else:
            value=list1[i]+list1[i+1]
        list1.append(value)
    return list1
    
print(my_function(5))

我已经在没有递归函数的情况下完成了此操作，但我正在考虑如何使用递归和记忆来完成此操作。你能帮我解决这个问题吗..

原文

I am working with series like Fibonacci series... (In Fibonacci series the nth term is the sum of n-1 and n-2 only.)
But in my case I want to nth term is the sum of half of the previous terms..
for example:

n=5
Output should be: [0, 1, 1, 2, 3]

n=12
Output should be: [0, 1, 1, 2, 3, 6, 11, 22, 42, 84, 165, 330]

def my_function(n):
    
    list1=[0,1]
    for i in range(0,n-2):
        if(i>2):
            value=sum(list1[i//2+1:])
        else:
            value=list1[i]+list1[i+1]
        list1.append(value)
    return list1
    
print(my_function(5))

I have done this without recursive function but I am thinking how to do it using recursively and memoization. Can you help me to solve this..

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太阳男子 2025-01-17 19:18:23

您在寻找这样的东西吗？

# Python has a built-in decorator for memoizing any function automagically.

from functools import lru_cache

@lru_cache(maxsize=None)
def my_func_rec(n):
    if n == 1:
        return [0]
    elif n == 2:
        return [0,1]
    else:
        prev = my_func_rec(n-1)
        prev.append(sum(prev[(n-1)//2:]))
        return prev

print(my_func_rec(12)) # [0, 1, 1, 2, 3, 6, 11, 22, 42, 84, 165, 330]

Are you looking for something like this?

# Python has a built-in decorator for memoizing any function automagically.

from functools import lru_cache

@lru_cache(maxsize=None)
def my_func_rec(n):
    if n == 1:
        return [0]
    elif n == 2:
        return [0,1]
    else:
        prev = my_func_rec(n-1)
        prev.append(sum(prev[(n-1)//2:]))
        return prev

print(my_func_rec(12)) # [0, 1, 1, 2, 3, 6, 11, 22, 42, 84, 165, 330]

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