使用一次递归调用实现递归

本文介绍了使用一次递归调用实现递归的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

给定如下函数:f(n) = f(n-1) + f(n-3) + f(n-4)

Given a function as follow : f(n) = f(n-1) + f(n-3) + f(n-4)

f(0) = 1 f(1) = 2 f(2) = 3 f(3) = 4

我知道在一个函数内使用三个递归调用来实现它.但我只想在函数内部进行一次递归调用.怎么做?

I know to implement it using recursion with three recursive calls inside one function. But I want to do it with only one recursion call inside the function. How it can be done ?

要使用 3 个递归调用来实现，这是我的代码:

To implement using 3 recursive calls here is my code :

def recur(n): if n == 0: return 1 elif n == 1: return 2 elif n == 2: return 3 elif n == 3: return 4 else: return recur(n-1) + recur(n-3) + recur(n-4) #this breaks the rule because there are 3 calls to recur

推荐答案

您的尝试方向正确，但需要稍作改动:

Your attempt is in the right direction but it needs a slight change:

def main(): while True: n = input("Enter number : ") recur(1,2,3,4,1,int(n)) def recur(firstNum,secondNum,thirdNum,fourthNum,counter,n): if counter==n: print (firstNum) return elif counter < n: recur (secondNum,thirdNum,fourthNum,firstNum+secondNum+fourthNum,counter+1,n)