Even Fibonacci numbers

Problem 2

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.

斐波那契数列中的每一项被定义为前两项之和。从1和2开始，斐波那契数列的前十项为：

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

考虑斐波那契数列中数值不超过4百万的项，找出这些项中值为偶数的项之和。

 

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算法

  该题目采用迭代算法思想，主要是考虑到算法的复杂度。当然也可以用递归，不过似乎会比较低效。

代码

#include<stdio.h>

int main(void)
{
	long n1=1, n2=2, sum=0;
	while(n1<4000000 && n2<4000000)
	{
		if(n1%2==0)
			sum+=n1;
		if(n2%2==0)
			sum+=n2;
		n1+=n2;
		n2+=n1;
	}
	printf("%ld\n",sum);
	n1=getchar();
	return 0;
}