欧拉计划第2题题解
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, …
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
偶斐波那契数
斐波那契数列中的每一项都是前两项的和。由1和2开始生成的斐波那契数列前10项为:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
考虑该斐波那契数列中不超过四百万的项,求其中为偶数的项之和。
解题思路
暂时没有想到更优的解法。
目前的想法是枚举出所有不超过 \(4 \times 10^6\) 的斐波那契数列中的元素,然后把他们中的所有偶数都加到总和中。
实现代码如下:
#include <bits/stdc++.h>
using namespace std;
long long f[1000] = {1, 2}, sum = 2;
int main() {
for (int i = 3; ; i ++) {
f[i] = f[i-2] + f[i-1];
if (f[i] > 4000000) break;
if (f[i] % 2 == 0) sum += f[i];
}
cout << sum << endl;
return 0;
}
得到答案为 \(7049156\)。