1.23
secje
posted @ 2010年5月12日 05:24
in sicp
, 874 阅读
这题没什么难度，just modify a little bit from the previos procedure (prime? n)
(define (runtime) (current-milliseconds))
(define (square x) (* x x))
(define (next-divisor n)
(if (= n 2)
3
(+ n 2)))
(define (smallest-divisor n)
(find-divisor n 2))
(define (find-divisor n test-divisor)
(cond ((> (square test-divisor) n) n)
((divides? test-divisor n) test-divisor)
(else (find-divisor n (next-divisor test-divisor)))))
(define (divides? a b)
(= (remainder b a) 0))
(define (prime? n)
(= n (smallest-divisor n)))
(define (timed-prime-test n)
(start-prime-test n (runtime)))
(define (start-prime-test n start-time)
(if (prime? n)
(- (runtime) start-time)))
(timed-prime-test 10000000019)
(timed-prime-test 10000000033)
(timed-prime-test 10000000061)
然后输出
594
523
600
previous version
(define (runtime) (current-milliseconds))
(define (square x) (* x x))
(define (smallest-divisor n)
(find-divisor n 2))
(define (find-divisor n test-divisor)
(cond ((> (square test-divisor) n) n)
((divides? test-divisor n) test-divisor)
(else (find-divisor n (+ test-divisor 1)))))
(define (divides? a b)
(= (remainder b a) 0))
(define (prime? n)
(= n (smallest-divisor n)))
(define (timed-prime-test n)
(start-prime-test n (runtime)))
(define (start-prime-test n start-time)
(if (prime? n)
(- (runtime) start-time)))
(timed-prime-test 10000000019)
(timed-prime-test 10000000033)
(timed-prime-test 10000000061)
输出
1507
1029
1203
两倍的关系还是比较相近的
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