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

1. /*
   
2. Even Fibonacci numbers
   
3. 
   
4. 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:
   
5. 
   
6. 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
   
7. 
   
8. By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
   
9. 
   
10. 
    
11. 题目翻译：
    
12. 
    
13. 斐波那契数列中的每一项被定义为前两项之和。
    
14. 
    
15. 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, ...
    
16. 
    
17. 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。
    
18. */
    
19. 
    
20. #include <bits/stdc++.h>
    
21. using namespace std;
    
22. 
    
23. long long ans; // 记录答案
    
24. long long a = 1, b = 2, c = a + b, tmp;
    
25. 
    
26. int main() {
    
27.     while (c <= 4000000) {
    
28.         if (c % 2 == 0) ans += c;
    
29.         c = a + b; a = b; b = c;
    
30.     }
    
31.     cout << ans << endl;
    
32. 
    
33.     return 0;
    
34. }

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输出为4613730

liangxixin

111

Ian_Li

1. public static void main(String[] args) {
   
2.         int res = getSum(4000000);
   
3.         System.out.println(res);
   
4.     }
   
5. 
   
6.     private static int getSum(int n) {
   
7.         int sum = 0;
   
8.         if (n < 2) return sum;
   
9.         int i = 1, j = 2;
   
10.         while (j <= n) {
    
11.             if (j % 2 == 0) sum += j;
    
12.             j += i;
    
13.             i = j - i;
    
14.         }
    
15.         return sum;
    
16.     }

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xw20010211

学习

风眠

1. a=1
   
2. b=1
   
3. answer=0
   
4. while a<=4000000:
   
5.     if not a%2:
   
6.         answer+=a
   
7.     a,b = b,a+b
   
8. else:
   
9.     print(answer)

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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

1. ulong test2 (ulong n)
   
2. {
   
3.     ulong res = 0;
   
4.     ulong first = 1, second = 1, Xn = 0;
   
5.     while ((Xn = first + second) < n)
   
6.     {
   
7.         res += Xn;
   
8.         first = second+Xn;
   
9.         second = first+Xn;
   
10.     }
    
11.     return res;
    
12. }
    
13. int main (int argc, char *argv[])
    
14. {
    
15.   auto res = test2 (4000000);
    
16.   std::cout << res << std::endl;
    
17. }

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1433391058

test

hejiage

开眼界了，还有这汇总项目

1Asdusdhjssd

学习了

Zchyu

1. sum = 0
   
2. a = 1
   
3. b = 2
   
4. while b <= 4000000:
   
5.         if b % 2 == 0:
   
6.                 sum += b
   
7.         a, b = b, a + b
   
8. print(sum)

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HuangBin2025

1. sum=0
   
2. f1,f2=1,2
   
3. while f2<4000000:
   
4.     temp=f1+f2
   
5.     if temp%2==0:
   
6.         sum+=temp
   
7.     f1=f2
   
8.     f2=temp
   
9. print(sum)

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