Python中的斐波那契-简单的解决方案 [英] Fibonacci in Python - Simple solution

问题描述

n1 = 1
n2 = 1
n3 = n1 + n2
for i in range(10):
   n1 + n2
   print(n3)
   n1 = n2
   n2 = n3

据我所知，这应该是输出系列的前10位数字的最简单方法，但是，它会打印 2 10次.我不明白为什么 n1 没有设置为 n2 ，为什么 n2 没有设置为 n3 n3 之后的code>.

According to what I know, this should be the simplest way of outputting the first 10 digits of the series, however, it prints 2 10 times. I don't understand why n1 doesn't get set to n2, and n2 doesn't get set to n3 after n3 has been printed.

推荐答案

您的代码有很多问题.而且，您首先应该自己学习和尝试.我也是一个初学者，所以我知道您在想什么.为了使其快速生效，可以进行一些

There are many issues with your code. And you should first learn and try as much as you can on your own. I am also a beginner so I know what you are thinking. For some quick edits to make it workable:

n1 = 0
n2 = 1
n3 = 0
for i in range(10):
   n3 = n1 + n3
   print(n3)
   n1 = n2
   n2 = n3

1. 该系列以0开头，您以1进行了初始化.
2. 更新语句 n3 = n1 + n2 在循环之外，它将如何更新?此处发生的是 n3 = 1 + 1 = 2 ，您的代码保持不变，并且没有变化.

1. The series starts with 0, you initialized it with 1.
2. The update statement n3=n1+n2 is outside the loop, how will it update? What is happening here is n3 = 1 + 1 = 2 in your code stays the same and it doesn't change.

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