python做斐波那契数列通项公式_python实现斐波那契数列

斐波那契数列:第0项是0,第1项是第一个1。这个数列从第3项开始,每一项都等于前两项之和。如下:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368.....

实现1:

if __name__ == '__main__':

f = [0,1]

print(type(f))

for n in range(2,20):

# f[n]=f[n-1]+f[n-2] # 会报错IndexError: list assignment index out of range;

f.append(f[n - 1] + f[n - 2])

# print(f)

n = n + 1

print(f)

结果如下:

[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181]

Process finished with exit code 0

实现2:

# Fibonacci series: 斐波纳契数列

# 两个元素的总和确定了下一个数

a, b= 0, 1

whileb< 1000:

print(b, end=',')  # 关键字end可以用于将结果输出到同一行,或者在输出的末尾添加不同的字符

a, b= b, a+b

结果如下:

1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,

Process finished with exit code 0

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