[Project Euler] 2. Even Fibonacci numbers
问题描述:
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.
解题思路:
这道题我是用暴力破解方法解的,即计算出fabonacci的每一个数字然后判断是否满足条件然后加入最后的和。
还有一种解法:
“I estimate that I had written about 3 million lines of assembler code in my whole life. Now, code only when strictly necessary.
Phi (golden ratio) is the approximate ratio between two consecutive terms in a Fibonacci sequence.
The ratio between consecutive even terms approaches phi^3 (4.236068) because each 3rd term is even.
Use a calculator and round the results to the nearest integer when calculating the next terms: 2,8,34,.. multiplying by 4.236068 each time: 144,610, 2584,10946,46368,196418 & 832040 The sum is 1089154
My codeless regards,
Rudy.”
这里说的是:黄金比例phi(1.618)是fabnacci数列两个相邻数字之间的比率,因为每隔两个数字有一个偶数,可以用一个偶数乘以 phi^3
不过我写代码计算了一下之后发现误差非常大。但是也是新思路和新知识!
代码:
void calFibb(int i, int j, int &sum){ int num = i + j; if(num < 4000000){ if(!(num & 1)){ sum += num; } calFibb(j, num, sum); } }