400 万以下的斐波那契数列 [英] Fibonacci under 4 millions

问题描述

可能的重复:
查找斐波那契数列的 Python 程序.更多 Pythonic 方式.

嘿，我正在尝试编写一个脚本，将斐波那契数列"中的所有偶数项汇总到 400 万以下.

Hey, i was trying to write a script which sums all the even terms in "Fibonacci Sequence" under 4 millions.

Fibonacci1 = 1
Fibonacci2 = 2
a = 2
i = 4
for i in range(1,4000000):
 Fibonacci1 = Fibonacci1 + Fibonacci2
 if Fibonacci1 % 2 == 0:
  a = a + Fibonacci1
 Fibonacci2 = Fibonacci1 + Fibonacci2
 if Fibonacci2 % 2 == 0:
  a = a + Fibonacci2
print a
raw_input()

应该不到一分钟的时间，但花了整整一夜的时间都没有解决！

it should takes less than a minute, but it took all night and it wasn't solved !

抱歉各位，我误解了这个问题.我虽然这意味着我必须将所有偶数项相加到 400 万！但解决方案是将所有偶数项相加，直到 400 万.

Sorry guys, i misunderstood the problem. I though it means that i have to sum all the even terms UP TO 4 million ! But the solution was to sum all the even terms UNTIL 4 million.

工作代码(不到一秒完成):

Working code (finished in less than one second):

Fibonacci1 = 1
Fibonacci2 = 2
a = 2
while a < 4000000:
 Fibonacci1 = Fibonacci1 + Fibonacci2
 if Fibonacci1 % 2 == 0:
  a = a + Fibonacci1
 Fibonacci2 = Fibonacci1 + Fibonacci2
 if Fibonacci2 % 2 == 0:
  a = a + Fibonacci2
print a
raw_input()

推荐答案

循环条件不对，应该是这样的:

The loop condition is wrong, it should be something like this:

while True:
    Fibonacci1 = Fibonacci1 + Fibonacci2
    if Fibonacci1 % 2 == 0:
        if a + Fibonacci1 > 4000000:
            break
        a = a + Fibonacci1
    Fibonacci2 = Fibonacci1 + Fibonacci2
    if Fibonacci2 % 2 == 0:
        if a + Fibonacci2 > 4000000:
            break
        a = a + Fibonacci2

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