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22.sicp每日一题[2.20]
23.sicp每日一题[2.21]24.sicp每日一题[2.22-2.23]25.sicp每日一题[2.24-2.27]26.sicp每日一题[2.28]27.sicp每日一题[2.29]28.sicp每日一题[2.30]29.sicp每日一题[2.31]30.sicp每日一题[2.32]31.sicp每日一题[2.33]32.sicp每日一题[2.34]33.sicp每日一题[2.35]34.sicp每日一题[2.36-2.37]35.sicp每日一题[2.38-2.39]36.sicp每日一题[2.40]37.sicp每日一题[2.41]38.sicp每日一题[2.42]39.sicp每日一题[2.43]40.sicp每日一题[2.44]41.sicp每日一题[2.45]42.sicp每日一题[2.46]43.sicp每日一题[2.47]44.sicp每日一题[2.48]45.sicp每日一题[2.49]46.sicp每日一题[2.50]47.sicp每日一题[2.51]48.sicp每日一题[2.52]49.sicp每日一题[2.53-2.55]50.sicp每日一题[2.56]51.sicp每日一题[2.57]52.sicp每日一题[2.58]53.sicp每日一题[2.59]54.sicp每日一题[2.60]55.sicp每日一题[2.61]56.sicp每日一题[2.62]57.sicp每日一题[2.63-2.64]58.sicp每日一题[2.65]59.sicp每日一题[2.66]60.sicp每日一题[2.67-2.68]61.sicp每日一题[2.69]62.sicp每日一题[2.70]63.sicp每日一题[2.71]64.sicp每日一题[2.72]65.sicp每日一题[2.73]66.sicp每日一题[2.74]67.sicp每日一题[2.75]68.sicp每日一题[2.76]69.sicp每日一题[2.77]70.sicp每日一题[2.78]71.sicp每日一题[2.79]72.sicp每日一题[2.80]73.sicp每日一题[2.81]74.sicp每日一题[2.82]75.sicp每日一题[2.83]76.sicp每日一题[2.84]77.sicp每日一题[2.85-2.86]
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Exercise 2.20

The procedures +, *, and list take arbitrary numbers of arguments. One way to define such procedures is to use define with dotted-tail notation. In a procedure definition, a parameter list that has a dot before the last parameter name indicates that, when the procedure is called, the initial parameters (if any) will have as values the initial arguments, as usual, but the final parameter’s value will be a list of any remaining arguments. For instance, given the definition
(define (f x y . z) ⟨body⟩)
the procedure f can be called with two or more arguments. If we evaluate
(f 1 2 3 4 5 6)
then in the body of f, x will be 1, y will be 2, and z will be the list (3 4 5 6). Given the definition
(define (g . w) ⟨body⟩)
the procedure g can be called with zero or more arguments. If we evaluate
(g 1 2 3 4 5 6)
then in the body of g, w will be the list (1 2 3 4 5 6).
Use this notation to write a procedure same-parity that takes one or more integers and returns a list of all the arguments that have the same even-odd parity as the first argument. For example,

(same-parity 1 2 3 4 5 6 7)
(1 3 5 7)
(same-parity 2 3 4 5 6 7)
(2 4 6)

这道题也是看着唬人，题干挺长，但是主要在介绍如何定义参数个数不定的函数，真正的题目就最后一段，就是要根据第一个参数的奇偶性，把后面参数中跟第一个参数奇偶性相同的找出来。

(define (same-parity x . y)
  (define (iter items r result)
    (cond ((null? items) result)
          ((= r (remainder (car items) 2)) (iter (cdr items) r (cons (car items) result)))
          (else (iter (cdr items) r result))))
  (let ((r (remainder x 2)))
    (reverse (iter y r (cons x nil)))))


(same-parity 1 2 3 4 5 6 7)
(same-parity 2 3 4 5 6 7)
(same-parity 2)

; 执行结果
'(1 3 5 7)
'(2 4 6)
'(2)
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