E1.2.5you shouldn't be that's way of the programmor in this

gyumni | Aug. 22, 2019, 5:33 a.m.

Alyssa P. Hacker迷之思,世界上的hacker who is istp

如果我想use exp log I will don't do that way,because is pus the reminder ,is the featur not belong to exp (log)

I will do redinmder plus exp (log)


(define (expmod base exp m)
  (remainder (fast-expt base exp) m))


(define (expmod base exp m)
  (remainder (fast-expt base exp) m))

but if use this why don't use fast-expt linterly

because I can't think it reduce steps,when you use a fuction,to complate the feaure,of reminder only is has the fuction enothgh for this ,why write a procedure again?

if you want do a procedure let to do two futures,should let them use in same time,or don't put thgethor

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brown1730and...命令a去执行b,a要知道b在哪,的唯一方法是,一直看b,如果,b不在a里面,就要,一直看着b实现,用余数,如果是放在执行,那么a来执行b,a不需要用额外的hide去监督b,如果a等待执行b,其他人等待执行b,too,a如果返回给其他人一个执行结果,或是一个执行程序b,prime?问得是a的value 还是 b的值,但是a告诉?,去找b,a,只负责执行结果的一小部分,话说,我把a的一小部分和b的执行结果放在prime?会怎么样,首先是一个迭代算法,然而,在已经define的情况,是这样的

exp被实现以迭代的乘以自身,那么如果需要处理reminder最好用自身来处理,这样步骤只是1,如果被别人来处理一个迭代自身的结果,需要一种,keep b 的信息。over2019-08-222:28:21 PM

(define (expmod base exp m)
  (cond ((= exp 0) 1)
        ((even? exp)
         (remainder (* (expmod base (/ exp 2) m)
                       (expmod base (/ exp 2) m))
         (remainder (* base (expmod base (- exp 1) m))

这不符合b b 2n的公式,如果

,it’s(*(* b b b b)(* b b b b))Exercise 1.262:35:43 PM46 The reduction steps in the cases where the exponent e is greater than 1 are based on the fact that, for any integers xy, and m, we can find the remainder of x times y modulom by computing separately the remainders of x modulo m and y modulo m, multiplying these, and then taking the remainder of the result modulo m. For instance, in the case where e is even, we compute the remainder of be/2 modulo m, square this, and take the remainder modulo m. This technique is useful because it means we can perform our computation without ever having to deal with numbers much larger than m. (Compare exercise 1.25.)

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