今天是第一次作业，第一次接触Python，导致很多地方不太适应。第一题卡了我半个小时。没想到f可以代表方法名，真是一门灵活的语言。
第一题 ：两数之和
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)
这一题的目标就是不使用abs函数的情况下，获取a + |b|值；但是特别注意的是，return返回的是f(a,b)，所以我们这个f只能是方法名，如果我们写了具体实现，则返回值就有问题；
❌ 错误写法
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 = a - b # 这种方法f就代表一个具体值，但是我们需要考虑return的情况，所以不能写具体实现。
else:
f = a + b
return f(a, b)
☑️ 正确写法
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)
第二题 三数中最小的两个数的平方和；
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 _____
这题需要考虑的就是如何取得最大值，然后将最大值剔除就行，这个远比最小值要简单，因为最小值需要考虑的是负数的情况，需要穷举法才行；
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 * x + y * y + z * z - max(x, y, z) * max(x, y, z)
第三题 最大因子
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 ***"
最大因子这一题,就是需要求出除本数之外的最大因子,我一开始考虑的是穷举法,然后遍历取最大值,但是后面一想,a* b = n 当n是最大因子的时候,也就意味着b是除1之外的最小因子,所以我们只需要从1开始递增,找到最小因子,然后相除就可以得到最大因子的数了;
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 ***"
fa = 2
while (n % fa != 0):
if fa < n:
fa = fa + 1
if fa == n:
return 1
else:
return int(n / fa)
第四题 带条件的方法
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, 'equal', 'not equal')
'not equal'
>>> if_function(3>2, 'bigger', 'smaller')
'bigger'
"""
if condition:
return true_result
else:
return false_result
def with_if_statement():
"""
>>> result = with_if_statement()
61A
>>> print(result)
None
"""
if cond():
return true_func()
else:
return false_func()
def with_if_function():
"""
>>> result = with_if_function()
Welcome to
61A
>>> 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 ***"
这一题,需要我们知道方法调用的两种方式,with_if_statement这种调用方法,只会触发cond(),true_func(),false_func() 这里面的两个结果,因为code的结果只能是真/假,所以必然有一个不能执行.我们分下面两种情况讨论;
- code为True,当code为True的时候,**result = with_if_statement() **出现 61A的结果,那么true_func() 应该就是print('61a');但是我们从后面的题目中可以得知,还需要答应一个 Welcome to到终端,所以,code()必定为false;
- code为False的情况,那么,false_func()就必须打印'61a',那么true_func()打印的就是' Welcome to',这样顺序执行的时候,才不会导致61A出现在 Welcome to之前;
我们第二种调用方式,就是通过调用参数方法而不是直接调用参数,return if_function(cond(), true_func(), false_func()) 这就是这种情况,这种调用方式,首先会加载形参.而加载形参首先就需要对形参进行初始化,所以,cond(), true_func(), false_func()三个都会首先被加载一遍;
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 ***"
""" time = 1
while (n != 1):
print(n)
if n % 2 == 0:
n = int(n / 2)
else:
n = n * 3 + 1
time += 1
print(n)
return time """
step = 1
while True:
print(n)
if n == 1:
break
if n % 2 == 0:
n = n // 2 # 这里还学到一点,就是我们可以通过//语法向下取整,而避免了int方法转换的操作.
else:
n = n * 3 + 1
step += 1
return step
最后,附上通过截图.第一次作业提交完毕
