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

songyazh

songyazh 发表于 2023-7-8 18:24
import time
start = time.time()
NUM = 0

看了结果，我这错的没边儿了

摘星少年

马上大二，打卡打卡
感谢小甲鱼

Ewan-Ahiouy

/*
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。
*/

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

long long ans; // 记录答案
long long a = 1, b = 2, c = a + b, tmp;

int main() {
    while (c <= 4000000) {
        if (c % 2 == 0) ans += c;
        c = a + b; a = b; b = c;
    }
    cout << ans << endl;

    return 0;
}

输出为4613730

liangxixin

111

Ian_Li

public static void main(String[] args) {
        int res = getSum(4000000);
        System.out.println(res);
    }

    private static int getSum(int n) {
        int sum = 0;
        if (n < 2) return sum;
        int i = 1, j = 2;
        while (j <= n) {
            if (j % 2 == 0) sum += j;
            j += i;
            i = j - i;
        }
        return sum;
    }

xw20010211

学习

风眠

a=1
b=1
answer=0
while a<=4000000:
    if not a%2:
        answer+=a
    a,b = b,a+b
else:
    print(answer)

lU553178681

学习学习！

Mr.roushan

def fib(n):
    fibseq = [1,2]
    result = 0
    while fibseq[-1]<n:
        if fibseq[-1] % 2 == 0:
            result += fibseq[-1]
        fibseq.append(fibseq[-1]+fibseq[-2])
    return result
        
        
print(fib(4000000))

salt_eto

每三位出现一个偶数，每次算三位就不需要检查偶数了

ulong test2 (ulong n)
{
    ulong res = 0;
    ulong first = 1, second = 1, Xn = 0;
    while ((Xn = first + second) < n)
    {
        res += Xn;
        first = second+Xn;
        second = first+Xn;
    }
    return res;
}
int main (int argc, char *argv[])
{
  auto res = test2 (4000000);
  std::cout << res << std::endl;
}

1433391058

test

hejiage

开眼界了，还有这汇总项目

1Asdusdhjssd

学习了

Zchyu

sum = 0
a = 1
b = 2
while b <= 4000000:
        if b % 2 == 0:
                sum += b
        a, b = b, a + b
print(sum)