Sorry I cann tink of a 好的所有权。
我正在着手开展与信通技术相关的工作,并在此过程中:
计算alg a(n) and alg b(n) for n =1,2,3,4 and 5
(a)
alg_a(n):result
if n > 1 then
return(alg_a(n−1)+alg_a(n−1))
else return(1)
(b)
alg_b(n):result
if n > 1 then
return(2 · alg_b(n−1))
else return(1)
首先,第1行的代码是什么(alg_a(n):result)?
A: 问题请我计算出(n) al,这样,我就插入1,if n > 1-> no->tur 1。 但当我插入N=2时,情况如何。
Any help is appreciated,
thanks!
algorithm alg_a(n) calculates 2^(n-1) and alg_b(n) does the same thing.
Theese are recursive functions. For example for 4 alg_a returns:
alg_a(4)=
alg_a(3) + alg_a(3) =
alg_a(2) + alg_a(2) + alg_a(2) + alg_a(2) =
alg_a(1) + alg_a(1) + alg_a(1) + alg_a(1) + alg_a(1) + alg_a(1) + alg_a(1) + alg_a(1) = 8
该代码为某种形式的假编码。 “结果”工作只是意味着职能的结果。 因此,alg_a(1)给您1,而alg_a(2)。 继续得到你的其他答复。则赋予你<代码>(alg_a(1) + alg_a(1)),即
本案的问题没有要求比数字回答更加复杂。
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