projecteuler.net 2--Even Fibonacci numbers

原创于 2017-10-21 09:31:41 发布 · 183 阅读

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本文介绍了一个关于斐波那契数列的问题：计算不超过四百万的所有偶数值的斐波那契数之和，并提供了一段C++代码实现。通过迭代的方式生成斐波那契数列并累加符合条件的项。

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.

#include <iostream>
#include <cmath>
int main()
{
    int a,b,c,target,sum;
    sum=2;
    a=1;
    b=2;
    c=2;
    target=4000000;
    while(c<target)
    {
        c=a+b;
        if(c<target)
        {
            a=b;
            b=c;
            if (c % 2 == 0)
            {
                sum = sum + c;
            }
            c=a+b;
            if(c<target)
            {
                a = b;
                b = c;
                if (c % 2 == 0)
                {
                    sum = sum + c;
                }
            }
        }
    }
    std::cout<<sum<<std::endl;
}