使用一次递归调用实现递归 [英] Implement recursion using one recursive call

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)

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