[Project Euler] 2. Even Fibonacci numbers

问题描述：

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.

解题思路：

这道题我是用暴力破解方法解的，即计算出fabonacci的每一个数字然后判断是否满足条件然后加入最后的和。

还有一种解法：

“I estimate that I had written about 3 million lines of assembler code in my whole life. Now, code only when strictly necessary.

Phi (golden ratio) is the approximate ratio between two consecutive terms in a Fibonacci sequence.

The ratio between consecutive even terms approaches phi^3 (4.236068) because each 3rd term is even.

Use a calculator and round the results to the nearest integer when calculating the next terms: 2,8,34,.. multiplying by 4.236068 each time: 144,610, 2584,10946,46368,196418 & 832040 The sum is 1089154

My codeless regards,

Rudy.”

这里说的是：黄金比例phi(1.618)是fabnacci数列两个相邻数字之间的比率，因为每隔两个数字有一个偶数，可以用一个偶数乘以 phi^3

不过我写代码计算了一下之后发现误差非常大。但是也是新思路和新知识！

代码：

void calFibb(int i, int j, int &sum){
    int num = i + j;
    if(num < 4000000){
        if(!(num & 1)){
            sum += num;
        }
        calFibb(j, num, sum);
    }
}