[欧拉计划]Problem 2.Even Fibonacci numbers

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.

首先最容易想到的方法:

int limit = 4000000, sum = 0, a = 1, b = 1;
while(b < limit)
{
if(b % 2 == 0)
sum += b;
int h = a + b;
a = b;
b = h;
}
cout << sum << endl;

这样做一定是把4000000以内的fibnacci数列遍历了一遍,那么有没有效率更高的算法呢?

我们先写出一段fibnacci数列:

F	1	1	2	3	5	8	13	21	34	55	89	144
	a	b	c	a	b	c	a	b	c	a	b	c
			*			*			*			*

不难证明每3项都是偶数,因此程序也可以写成这样:

int limit = 4000000, sum = 0, a = 1, b = 1, c = a + b;
while(c < limit)
{
sum += c;
a = b + c;
b = c + a;
c = a + b;
}
cout << sum << endl;

但是效率并没有提高,依然是将数列遍历了一遍.

我们把偶数项单独写出来,会发现一个优美的结构

2,8,34,144…

好像存在这种关系:

E(n)=4⋅E(n−1)+E(n−2)

如果我们能证明:

F(n)=4⋅F(n−3)+F(n−6)

便也就证明了上式.

证明如下:

F(n)

=F(n−1)+F(n−2)

=F(n−2)+F(n−3)+F(n−2)

=2F(n−2)+F(n−3)

=2(F(n−3)+F(n−4))+F(n−3)

=3F(n−3)+2F(n−4)

=3F(n−3)+F(n−4)+F(n−5)+F(n−6)

=4F(n−3)+F(n−6)

证毕.

这种形式可以避免奇偶判断.