在保证提供专业高质量的CS代写服务的同时，我们将为同学们争取最经济实惠的代写价格。由于CS作业的类型很多，CS作业代做、CS程序代写、CS代码代做、CS测验等，所以我们很难指定CS代写价格。在CS专家查看具体要求，评估难度，并结合deadline，才能提出价格。欢迎联系我们的微信客服获取报价和获取优惠。
CS代考价格说明
时长：要说影响代考价格的首要因素，就是代考时长。在大部分代考服务机构，基本上是按照小时收费的，所以代考时间越长，题量越多价格也越高啦。一般来说，日常的test都是小型考试，在一个小时左右，价格相交与final exam会便宜一些。
难度：第二个会影响价格的因素就是难度，高中、本科、以及硕士研究生等不同学术阶段的难度是不一样的，这很容易理解，代考价格也不同。
个人需求：在EssayOne帮助过的客户中，不同的客户有不同的要求。有些同学只需要pass，有些同学则需要保证拿A，那么代考价格就会在基础报价上有所增加。
一般来说，Java代考的价格在600元/小时~1500元/小时不等，具体价格还是根据每个人不同的具体需求确定。
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TopMask CS代写王牌导师团队介绍
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CS Final Exam 代考案例
Q1 – Python to MIPS translation
func:
# save the $fp and $ra into the stack
addi $sp, $sp -8 # make space in the stack for the two registers
sw $ra, 4($sp) # save $ra onto stack
sw $fp, 0($sp) # save $fp onto stack
addi $fp, $sp, 0 # copy $sp into $fp
addi $sp, $sp, -4 # make space for local var (result)
# if n< = 0
lw $t0, 8($fp) # load argument n
slt $t0, $0, $t0 # if 0 < n then $t0 = 1
bne $t0, $0, else # if $t0 = 1 (i.e., n > 0) go to else
sw $0, -4($fp) # result = 0
j endif # jump over else branch
else:
# compute n-1 and store it in $t0
lw $t0, 8($fp)
addi $t0, $t0, -1
# save the argument (n-1) in the stack
addi $sp, $sp, -4
sw $t0, 0($sp)
jal func # call func with n-1 as argument
addi $sp, $sp, 4 # remove argument
# result = 4*n + func(n-1)
lw $t0, 8($fp) # load n into $t0
sll $t0, $t0, 2 # 4*n shifting by 2
addi $t0, $t0, $v0, # 4*n + func(n-1)
sw $t0, -4($fp) # store it in result
endif:
lw $v0, -4($fp) # put result in $v0
addi $sp, $sp, 4 # remove local
lw $fp, 0($sp) # restore $fp and $ra
lw $ra 4($sp)
addi $sp, $sp, 8
jr $ra #go back to the callee
Part 1.f
The iterative version will require exactly the same number of bytes (N) as the recursive
version, since the number of dynamic objects created during their executions (which are the
only ones stored by functions in the Heap) will not change.
For the MIPS code provided, N is 0, as nothing is created in the Heap.
Note that, in practice, Python would create objects for integers and will indeed use the Heap.
Part 1.h
The Stack for the iterative version of func(n) will contain the argument n (4 bytes), the saved
$ra and $fp (4+4 bytes), and the local variable result (4 bytes). This means a total of N = 16
bytes for the iterative version.
In the recursive version, the callee will call func(n) which will then call itself n} times. And
each time it will take N (16 as we shown above) bytes. That means a total of (n+1)*N bytes.
Q2 – CS saves the world
Write the output of the function mystery for the input values:
1
1
2
3
1
4
What does the function mystery compute?
It computes the sum of the digits of x in base 2.
What is the time complexity of mystery, using the O() notation? Prove
your answer.
O(log x). See solutions of Exercise 2 of tute 5.
Write the output of the function enigma for the input …
1
1
3
6
1
5
The subquestions below this line have been removed the exam (11 out of 20 marks). It was of the same difficulty of the digital root question (Exercise 6 of tute 5), which is a non-starred exercise. This was hard but doable, as the histogram shows. However, since there were other hard questions, this one was removed.
What does the function enigma compute?
It computes mystery(x) + mystery(mystery(x)) + …
What is the time complexity of enigma, using the O() …
The output of mystery(x) has size log(x).
Since enigma computes mystery(x) + mystery(mystery(x)) + …, it requires T(x) = log(x) +
log( log(x)) + … operations.
Since log(x) <=x/2, we have \log\log(x) <= log (x/2)<= x/4$.
Hence T(x) <= x/2 + x/4 + … = x.
Computing enigma(x) is thus in O(x).
See solutions of Exercise 6 of tute 5.
What does enigma(4095) return? Justify your answer.
Observe that 4095 = 4096 – 1 = 2^{12} -1 = 2^{11} + 2^{10} + … + 2^0, hence mystery(4095)
returns 12.
Therefore enigma(4095) returns 12 + enigma(12).
mystery(12) returns 2, and mystery(2) returns 1.
Hence enigma(4095) returns 15
