尤拉計劃002--偶斐波那契數

首先看一下題目

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:

\[1,2,3,5,8,13,21,34,55,89,\cdots \]

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

偶斐波那契數

斐波那契數列中的每一項都是前兩項的和。由1和2開始生成的斐波那契數列的前\(10\)項為：

\[1,2,3,5,8,13,21,34,55,89,\cdots \]

考慮該斐波那契數列中不超過四百萬的項，求其中為偶數的項之和。

解題思路：

其實這一道題，就是單純的遍歷一次，並沒有取巧的方法。退出迴圈的條件條件就是Fibonacci[i] > 4000000。判斷是否為偶數的方法就是如果i%2==0成立，那麼i就為偶數。

#include <iostream>
using namespace std;
long long Fibonacci[1000] = {1, 2};
long long sum = 2;
int main() {
    for (int i = 3; ; i ++) {
        Fibonacci[i] = Fibonacci[i-2] + Fibonacci[i-1];
        if (Fibonacci[i] > 4000000) 
            break;
        if (Fibonacci[i] % 2 == 0) 
            sum += Fibonacci[i];
    }
    cout << sum << endl;
    return 0;
}
 

最後算出來的結果為7049156。