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 h(a, b)']
"""
if b >= 0:
h = add
else:
h = sub
return h(a, b)
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
def largest_factor(x):
"""Return the largest factor of x that is smaller than x.
>>> 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 ***"
for i in range(x//2, 0, -1):
if x % i == 0:
return i
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
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 ***"
return False
def true_func():
"*** YOUR CODE HERE ***"
print(42)
def false_func():
"*** YOUR CODE HERE ***"
print(47)
def hailstone(x):
"""Print the hailstone sequence starting at x and return its
length.
>>> a = hailstone(10)
10
5
16
8
4
2
1
>>> a
7
"""
"*** YOUR CODE HERE ***"
step = 1
while x != 1:
print(x)
if x % 2 == 0:
x //= 2
else:
x = 3 * x + 1
step += 1
print(1)
return step
最后修改:2022 年 06 月 24 日
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