SICP习题1.21 解答
题目:使用smallest-divisor过程找出下面各数的最小因子:199,1999,19999.
稍对原过程作出修改:
(define (divisor-iter n a)
(cond ((>= (square a) n) "prime")
((= (remainder n a) 0) a)
(else (divisor-iter n (+ a 1)))))
(define (smallest-divisor n)
(divisor-iter n 2))
(define (square x)
(* x x))
稍对原过程作出修改:
(define (divisor-iter n a)
(cond ((>= (square a) n) "prime")
((= (remainder n a) 0) a)
(else (divisor-iter n (+ a 1)))))
(define (smallest-divisor n)
(divisor-iter n 2))
(define (square x)
(* x x))
