Fibonacci序列中的每个新术语都是通过添加前两个术语生成的。从1和2开始,前10个术语将是:
1,2,3,5,8,13,21,34,55,89,......
通过考虑Fibonacci序列中的值不超过四百万的项,找到偶数项的总和。
我确定您以前在Project Euler上看到过有关此问题的问题,但我不确定为什么我的解决方案无法正常工作,所以我希望您可以帮忙!
public class Problem2Fibonacci {
public static void main(String[] args) {
/* 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. */
int sum = 0; // This is the running sum
int num1 = 1; // This is the first number to add
int num2 = 2; // This is the second number
int even1 = 0;
int even2 = 0;
int evensum = 0;
while (num2 <= 4000000){ // If i check num2 for the 4000000 cap, num will always be lower
sum = num1 + num2; // Add the first 2 numbers to get the 3rd
if(num2 % 2 == 0){ // if num2 is even
even1 = num2; // make even1 equal to num2
}
if(sum % 2 == 0){ // if sum is even
even2 = sum; // make even2 equal to sum
}
if (even1 != 0 && even2 != 0){ // If even1 and even2 both have values
evensum = even1 + even2; // add them together to make the current evensum
even2 = evensum;
}
num1 = num2;
num2 = sum;
System.out.println("The current sum is: " + sum);
System.out.println("The current Even sum is: " + evensum);
}
}
}
所以我的两个问题是, 1.为什么我的计划无法使偶数总和正常工作? 和 2.我的循环最后一次运行时,它使用的num2是&gt; 4000000.为什么?
谢谢!
答案 0 :(得分:0)
这可以帮到你:
int first = 0;
int second = 1;
int nextInSeq =first+second;
int sum =0;
while(nextInSeq < 4000000) {
first = second;
second = nextInSeq;
nextInSeq = first + second;
if(nextInSeq % 2 ==0)
sum = sum + nextInSeq;
System.out.println("Current Sum = " + sum);
}
System.out.println("Sum = " + sum);
对于您的代码:even1
和even2
不是必需的,并且在您继续操作时它们具有从之前的迭代中保留的值。