【Project Euler】【Problem 2】Even Fibonacci numbers

本文探讨了在斐波那契数列中找到所有不超过四百万的偶数元素,并计算它们的总和。

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.


Answer:
4613732

偶数 斐波拉契数列?

Problem 2

斐波拉契数列中的每个元素都是根据已知的钱两个数列生成的。以1和2开始,前10个元素是:

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

思考元素值不超过4000000的斐波拉契数列,找出其中偶数元素的和


答案:
4613732
解法一:
#include <stdio.h>

int main(int argc, char *argv[])
{
    int fb1 = 1;
    int fb2 = 2;
    int sum = 0;

    int maxNum = 4000000;
    while (fb1<maxNum)
    {
	printf("%d \n",fb1);
	if(fb1 % 2==0)
            sum += fb1;

        fb2 = fb1+fb2;
	    fb1 = fb2-fb1;
    }
    printf("result is %d",sum);
    return 0;
}


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