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

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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）：

游客，如果您要查看本帖隐藏内容请回复

zhangjinxuan

1. #include <bits/stdc++.h>
   
2. using namespace std;
   
3. 
   
4. long long f[51], ans = 0;
   
5. 
   
6. int main() {
   
7.         f[1] = 1;
   
8.         f[2] = 2;
   
9.         for (int i = 3; i <= 50 && f[i - 1] <= 4000000; ++i) {
   
10.                 f[i] = f[i - 1] + f[i - 2];
    
11.                 if (f[i - 1] % 2 == 0) ans += f[i - 1];
    
12.         }
    
13.         printf("%lld\n", ans);
    
14.     return 0;
    
15. }

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Kazimierz

1. #include <bits/stdc++.h>
   
2. using namespace std;
   
3. 
   
4. typedef long long LL;
   
5. const int N = 55,INF = 4000000;
   
6. int f[N];
   
7. LL ans=2;
   
8. 
   
9. int main()
   
10. {
    
11.     f[1] = 1;
    
12.     f[2] = 2;
    
13.     for(int i=3;f[i-1]<=INF;i++)
    
14.     {
    
15.         f[i] = f[i-1] + f[i-2];
    
16.         if(f[i]<=INF&&f[i]%2==0) ans += f[i];
    
17.     }
    
18.     cout<<ans<<endl;
    
19.     return 0;
    
20. }

复制代码

dabaiea

1

白于玉

1. list1 = [1,2]
   
2. while list1[-1] <= 4000000 * (pow(5,1/2)-1)/2:
   
3.     list1.append(list1[-2]+list1[-1])
   
4. 
   
5. print(sum([each for each in list1 if each % 2 == 0]))

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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

1. a = []
   
2. i = 1
   
3. j= 2
   
4. sum = 0
   
5. while i <= 4000000:
   
6.     a.append(i)
   
7.     if i % 2 == 0:
   
8.         sum += i
   
9.     i, j = j, i+j
   
10. print("偶数项的和为：%d" % sum)
    
11. print(a)

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1. 偶数项的和为：4613732
   
2. [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]

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白山莫念

1. fn main() {
   
2. 
   
3.     let mut fibonacci: Vec<u32> = vec![1, 2];
   
4.     let mut i = 2;
   
5.     while fibonacci[i - 2] + fibonacci[i - 1] <= 4000000 {
   
6.         fibonacci.push(fibonacci[i - 2] + fibonacci[i - 1]);
   
7.         i += 1;
   
8.     }
   
9.     let sum: u32 = fibonacci.iter().filter(|x| *x % 2 == 0).sum();
   
10.     println!("数列为：{:?}", fibonacci);
    
11.     println!("偶数项的和为：{}", sum);
    
12. }

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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