def problem25():
for n, f in enumerate(fib()):
if (floor(log10(f if f != 0 else 1)) + 1) == 1000:
return n
# Problem 2 # 19 October 2001 # Each new term in the Fibonacci sequence is generated by adding the previous two terms. # By starting with 1 and 2, the first 10 terms will be: # 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... # By considering the terms in the Fibonacci sequence whose values do not exceed four million, # find the sum of the even-valued terms. from common import fib from itertools import takewhile print(sum(takewhile(lambda x: x <= 4000000, (x for x in fib() if x % 2 == 0))))
def foo():
for i, f in enumerate(fib(10**1000)):
if len(str(f)) >= 1000:
return i, f
def answer():
i = 0
for n in fib():
if len(str(n)) == 1000: return i
i += 1
# 导入模块
import sys
import common
# 一个模块只会被导入一次，不管你执行了多少次import。这样可以防止导入模块被一遍又一遍地执行。
# 现在可以调用模块里包含的函数了
common.print_func("Runoob")
common.fib(1000)
print(common.fib2(1000))
# 内置的函数 dir() 可以找到模块内定义的所有名称。以一个字符串列表的形式返回:
# 如果没有给定参数，那么 dir() 函数会罗列出当前定义的所有名称:
print(dir(common))
#! /usr/bin/env python
from common import fib
for n in xrange(4000, 5001):
fibn = fib(n)
#print n, fibn, len(str(fibn))
if (len(str(fibn)) == 1000):
print n, fibn, len(str(fibn))
#!/usr/bin/env python
from common import fib
from itertools import count
MAX = 1000
for a in count(1):
f = fib(a)
l = len(str(f))
if l == 1000:
print(a)
break