关于python:cs61a-Fall-2020-hw01

A Plus Abs B

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 = sub
    else:
        f = add
    return f(a, b)

Two of Three

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 x**2+y**2+z**2 - max(x, y, z)**2

Largest Factor

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 ***"
    factor = n - 1
    while n > 0:
        if n % factor == 0:
            return factor
        factor -= 1

If Function vs Statement

这道题的关键在于，我不满足条件了，函数是否还会被执行？
能够看到，with_if_dunction中应用的是 call expression，即使cond 返回为False，第二个函数·true_func()·还是会被执行

def cond():
    "*** YOUR CODE HERE ***"
    return False

def true_func():
    "*** YOUR CODE HERE ***"
    print(42)
    
def false_func():
    "*** YOUR CODE HERE ***"
    print(47)

Hailstone

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 ***"
    count = 0
    while n != 1:
        count += 1
        print(n)
        if n % 2 == 1:
            n = n * 3 + 1
        else:
            n //= 2
    return count