题目2：在斐波那契数列中，找出所有值不大于4,000,000，并且为偶数的项之和

欧拉计划

题目2：在斐波那契数列中，找出所有值不大于4,000,000，并且为偶数的项之和

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.

题目翻译：

斐波那契数列中的每一项被定义为前两项之和。

从 1 和 2 开始，斐波那契数列的前十项为：1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

考虑斐波那契数列中数值不超过 4 百万的项，找出这些项中值为偶数的项之和。


视频讲解：




思路解析及源码参考（C & Python）：

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zhangjinxuan

#include <bits/stdc++.h>
using namespace std;

long long f[51], ans = 0;

int main() {
        f[1] = 1;
        f[2] = 2;
        for (int i = 3; i <= 50 && f[i - 1] <= 4000000; ++i) {
                f[i] = f[i - 1] + f[i - 2];
                if (f[i - 1] % 2 == 0) ans += f[i - 1];
        }
        printf("%lld\n", ans);
    return 0;
}

Kazimierz

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int N = 55,INF = 4000000;
int f[N];
LL ans=2;

int main()
{
    f[1] = 1;
    f[2] = 2;
    for(int i=3;f[i-1]<=INF;i++)
    {
        f[i] = f[i-1] + f[i-2];
        if(f[i]<=INF&&f[i]%2==0) ans += f[i];
    }
    cout<<ans<<endl;
    return 0;
}

dabaiea

1

白于玉

list1 = [1,2]
while list1[-1] <= 4000000 * (pow(5,1/2)-1)/2:
    list1.append(list1[-2]+list1[-1])

print(sum([each for each in list1 if each % 2 == 0]))

小甲鱼的二师兄

学习

猫熊同学

def fib():
    fib1, fib2 = 1, 1
    while True:
        yield fib1
        fib1, fib2 = fib2, fib1 + fib2

sum = 0
f = fib()
for x in f:
    if x > 4000000:
        break
    else:
        if x % 2 == 0:
            sum += x
print(f"值为偶数的项之和：{sum}")

a8634944

开看

auend

学习学习

KentChen

111

N781791878

twices

2

大饼弟弟

6

Afragrant

Rust代码

fn fibonacci_sum(n: i64) -> i64 {
    let mut v: Vec<i64> = [1, 2].to_vec();

    let mut i = 2;

    while v[i - 1] <= n {
        v.push(v[i - 1] + v[i - 2]);
        i += 1;
    }

    v.iter().filter(|&&i| i % 2 == 0).sum()
}

巅峰老甲鱼1

a = []
i = 1
j= 2
sum = 0
while i <= 4000000:
    a.append(i)
    if i % 2 == 0:
        sum += i
    i, j = j, i+j
print("偶数项的和为：%d" % sum)
print(a)

偶数项的和为：4613732
[1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578]

白山莫念

fn main() {

    let mut fibonacci: Vec<u32> = vec![1, 2];
    let mut i = 2;
    while fibonacci[i - 2] + fibonacci[i - 1] <= 4000000 {
        fibonacci.push(fibonacci[i - 2] + fibonacci[i - 1]);
        i += 1;
    }
    let sum: u32 = fibonacci.iter().filter(|x| *x % 2 == 0).sum();
    println!("数列为：{:?}", fibonacci);
    println!("偶数项的和为：{}", sum);
}

xiaqiyi

#include <stdio.h>

int main(void) {
    int a = 1,b=2,sum = 2;
    int c = a + b;
    while(c < 4000000){
        if (c % 2 == 0){
            sum += c;
        }
        a = b;
        b = c;
        c = a + b;
    }
    printf("%d\n", sum);
    return 0;
}

上官秀

111111111111

pixie99

打卡

songyazh

import time
start = time.time()
NUM = 0
xulie = [1, 4000000]
for i in range(len(xulie)):
    xin = xulie[i-1] + xulie[i]
    for k in range(1, xin):
        NUM = NUM + k
print(NUM)
end = time.time()
print(end-start)

这么写的，运行了有结果，但不知道对不对，特来验证