使用循环和递归的斐波那契数列

发布于 2024-11-16 14:34:40 字数 445 浏览 2 评论 0原文

我正在 Clojure 中进行 Project Euler 挑战，我想找到斐波那契数列中直到某个数字的所有偶数之和。

执行此操作的函数的代码如下。我知道有更快更简单的方法来做到这一点，我只是尝试使用循环和递归进行递归。然而，代码似乎不起作用，它永远不会返回答案。

(defn fib-even-sum [upto]
  (loop [previous 1 nxt 1 sum 0]
    (if (or (<= upto 1) (>= nxt upto))
     sum)
    (if (= (mod nxt 2) 0)
       (recur nxt (+ previous nxt) (+ sum nxt))
       (recur nxt (+ previous nxt) sum))))

我不确定是否可以在同一个循环中重复两次。我不确定这是否导致了问题？

原文

I am doing the Project Euler challenge in Clojure and I want to find the sum of all the even numbers in a fibonacci sequence up to a certain number.

The code for a function that does this is below. I know there are quicker and easier ways of doing this, I am just experimenting with recursion using loop and recur. However the code doesn't seem to work it never returns an answer.

(defn fib-even-sum [upto]
  (loop [previous 1 nxt 1 sum 0]
    (if (or (<= upto 1) (>= nxt upto))
     sum)
    (if (= (mod nxt 2) 0)
       (recur nxt (+ previous nxt) (+ sum nxt))
       (recur nxt (+ previous nxt) sum))))

I was not sure if I could do recur twice in the same loop or not. I'm not sure if this is causing the problem?

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回心转意 2024-11-23 14:34:40

您在第一个 IF 中放错了右括号（在 sum 之后）...

(defn fib-even-sum [upto]
  (loop [previous 1 nxt 1 sum 0]
    (if (or (<= upto 1) (>= nxt upto))
        sum
        (if (= (mod nxt 2) 0)
           (recur nxt (+ previous nxt) (+ sum nxt))
           (recur nxt (+ previous nxt) sum)))))

现在它可以快速工作了

You have a misplaced closing paren in the first IF (after sum)...

(defn fib-even-sum [upto]
  (loop [previous 1 nxt 1 sum 0]
    (if (or (<= upto 1) (>= nxt upto))
        sum
        (if (= (mod nxt 2) 0)
           (recur nxt (+ previous nxt) (+ sum nxt))
           (recur nxt (+ previous nxt) sum)))))

Now it works and fast

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