[CS61A-Fall-2020]学习记录一 Homework1 题解思路分享
前言
观前提示,笔者写的代码答案放在github仓库中,此处仅记录过程与心得
正文
首先来看下hw的第一道题
Q2: A Plus Abs B
Fill in the blanks in the following function for adding a to the absolute value of b, without calling abs. You may not modify any of the provided code other than the two blanks.
from operator import add, sub
def a_plus_abs_b(a, b):
"""Return a+abs(b), but without calling abs.
>>> a_plus_abs_b(2, 3)
5
>>> a_plus_abs_b(2, -3)
5
>>> # a check that you didn't change the return statement!
>>> import inspect, re
>>> re.findall(r'^\s*(return .*)', inspect.getsource(a_plus_abs_b), re.M)
['return f(a, b)']
"""
if b < 0:
f = _____
else:
f = _____
return f(a, b)
题目大意,赋予f一个函数满足在不使用abs函数的情况下,实现a+abs(b)的效果
解题过程,刚开始没注意到是赋予函数,于是分别填了a - b和a + b的错误答案,看了报错类型才反应过来,然后想起python可以函数赋予,于是将sub和add分别赋予f即可
Q3: Two of Three
Write a function that takes three positive numbers as arguments and returns the sum of the squares of the two smallest numbers. Use only a single line for the body of the function.
def two_of_three(x, y, z):
"""Return a*a + b*b, where a and b are the two smallest members of the
positive numbers x, y, and z.
>>> two_of_three(1, 2, 3)
5
>>> two_of_three(5, 3, 1)
10
>>> two_of_three(10, 2, 8)
68
>>> two_of_three(5, 5, 5)
50
>>> # check that your code consists of nothing but an expression (this docstring)
>>> # a return statement
>>> import inspect, ast
>>> [type(x).__name__ for x in ast.parse(inspect.getsource(two_of_three)).body[0].body]
['Expr', 'Return']
"""
return _____
题目大意,返回三个参数中最小两个的平方和
解题过程,刚开始想了会如何在一行里选出三个参数中最小两个,然后马上想到全部一起平方加起来再减去最大的平方不就行了吗
Q4: Largest Factor
Write a function that takes an integer n that is greater than 1 and returns the largest integer that is smaller than n and evenly divides n.
def largest_factor(n):
"""Return the largest factor of n that is smaller than n.
>>> largest_factor(15) # factors are 1, 3, 5
5
>>> largest_factor(80) # factors are 1, 2, 4, 5, 8, 10, 16, 20, 40
40
>>> largest_factor(13) # factor is 1 since 13 is prime
1
"""
"*** YOUR CODE HERE ***"
题目大意,找到比整数n小的最大的能整除n的数字
解题过程,直接循环,从1开始穷举,可以对循环条件除2或取根号2来减少循环次数
Q5: If Function vs Statement
Let's try to write a function that does the same thing as an if statement.
def if_function(condition, true_result, false_result):
"""Return true_result if condition is a true value, and
false_result otherwise.
>>> if_function(True, 2, 3)
2
>>> if_function(False, 2, 3)
3
>>> if_function(3==2, 3+2, 3-2)
1
>>> if_function(3>2, 3+2, 3-2)
5
"""
if condition:
return true_result
else:
return false_result
Despite the doctests above, this function actually does not do the same thing as an if statement in all cases. To prove this fact, write functions cond, true_func, and false_func such that with_if_statement prints the number 47, but with_if_function prints both 42 and 47.
def with_if_statement():
"""
>>> result = with_if_statement()
47
>>> print(result)
None
"""
if cond():
return true_func()
else:
return false_func()
def with_if_function():
"""
>>> result = with_if_function()
42
47
>>> print(result)
None
"""
return if_function(cond(), true_func(), false_func())
def cond():
"*** YOUR CODE HERE ***"
def true_func():
"*** YOUR CODE HERE ***"
def false_func():
"*** YOUR CODE HERE ***"
题目大意,通过编辑三个函数cond(),true_func(),false_func(),使得with_if_statement和with_if_fuction()产生预想的结果
解题过程,关键之关键在于了解到python中,将函数作为参数时,会先将函数运行,再将其结果作为参数返回,因此填上对应的数字,并调好条件cond就能成功解题
Q6: Hailstone
Douglas Hofstadter's Pulitzer-prize-winning book, Gödel, Escher, Bach, poses the following mathematical puzzle.
Pick a positive integer n as the start.
If n is even, divide it by 2.
If n is odd, multiply it by 3 and add 1.
Continue this process until n is 1.
The number n will travel up and down but eventually end at 1 (at least for all numbers that have ever been tried -- nobody has ever proved that the sequence will terminate). Analogously, a hailstone travels up and down in the atmosphere before eventually landing on earth.
Breaking News (or at least the closest thing to that in math). There was a recent development in the hailstone conjecture last year that shows that almost all numbers will eventually get to 1 if you repeat this process. This isn't a complete proof but a major breakthrough.
This sequence of values of n is often called a Hailstone sequence. Write a function that takes a single argument with formal parameter name n, prints out the hailstone sequence starting at n, and returns the number of steps in the sequence:
def hailstone(n):
"""Print the hailstone sequence starting at n and return its
length.
>>> a = hailstone(10)
10
5
16
8
4
2
1
>>> a
7
"""
"*** YOUR CODE HERE ***"
题目大意,有一种数学规律(尚未完全证明),对一个正整数,若其为偶数,则除2,若其为奇数,则乘3并加1,重复上述步骤,此数定终为1。现有n,输出n经上述规律到1的过程,并返回计算次数
解题过程,写循环,循环条件是数为1,每次循环输出该数,再设一个变量每次循环++即可
小结
虽然题目不难,但仍发现了自己一些问题,比如审题不清,有一定的原因是题目是英语而非中文,不过都要适应嘛