求斐波那契数列的第101项

方法一：

a = 1
b = 1

count = 2
while True:
    c = a + b
    count += 1
    if count >= 101:
        print(c)
        break
    a = b
    b = c

 

a = 0
b = 1

count = 2
print(0)
print(1)
while True:
    c = a + b
    count += 1
    print(c)
    if count >= 101:
        break
    a = b
    b = c
print(c)

 

import functools
import datetime

start = datetime.datetime.now()


@functools.lru_cache()
def fib(n):
    return 1 if n < 3 else fib(n - 1) + fib(n - 2)


delta = (datetime.datetime.now() - start).total_seconds()

print(fib(101), delta)