实验4 函数与异常处理编程

实验任务一

task1：

程序源代码：

 1 print(sum)
 2 sum = 42
 3 print(sum)
 4 
 5 def inc(n):
 6     sum = n+1
 7     print(sum)
 8     return sum
 9 
10 sum = inc(7) + inc(7)
11 print(sum)

运行结果：

实验任务二

task2-1

源代码：

 1 def func1(a,b,c,d,e,f):
 2     return [a,b,c,d,e,f]
 3 
 4 def func2(a,b,c,*,d,e,f):
 5     return [a,b,c,d,e,f]
 6 
 7 def func3(a,b,c,/,d,e,f):
 8     return [a,b,c,d,e,f]
 9 
10 print(func1(1,9,2,0,5,3))
11 print(func1(a=1,b=9,c=2,d=0,e=5,f=3))
12 print(func1(1,9,2,f=3,d=0,e=5))
13 
14 print(func2(11,99,22,d=0,e=55,f=33))
15 print(func2(a=11,b=99,c=22,d=0,e=55,f=33))
16 
17 print(func3(111,999,222,0,555,333))
18 print(func3(111,999,222,d=0,e=555,f=333))

 

运行结果：

 

line16 增加  print(func2(11,99,22,0,55,33))：

line18 增加 print(func3(a=111, b=999, c=222, 0, 555, 333) )

 

 

task2-2

源代码：

1 list1 = [1,9,8,4]
2 
3 print(sorted(list1))
4 print(sorted(list1,reverse = True))
5 print(sorted(list1,True))

 

运行结果：

问题回答：sorted内置函数中，reverse必须按照关键字传递。

task2-3

源代码：

1 def func(a,b,c,/,*,d,e,f):
2     return([a,b,c,d,e,f])
3 
4 print(func(1,2,3,d=4,f=5,e=6))

 

运行结果：

 

 

 

实验任务三

task3

源代码：

 1 def solve(a,b,c):
 2     '''
 3     求解一元二次方程, 返回方程的两个根
 4 
 5     :para: a,b,c: float 方程系数
 6     :return: tuple
 7     '''
 8     delta = b*b - 4*a*c
 9     delta_sqrt = abs(delta)**0.5
10     p1 = -b/2/a
11     p2 = delta_sqrt/2/a
12 
13     if delta > 0:
14         root1 = p1+p2
15         root2 = p1- p2
16 
17     else:
18         root1 = complex(p1,p2)
19         root2 = complex(p1,-p2)
20 
21     return root1,root2
22 
23 while True:
24     try:
25         t = input('输入一元二次方程系数a b c, 或者，输入#结束:')
26         if t == '#':
27             print('结束计算，退出')
28             break
29         a,b,c = map(float,t.split())
30         if a == 0:
31             raise ValueError('a = 0, 不是一元二次方程')
32     except ValueError as e:
33         print(repr(e))
34         print()
35     except:
36         print('有其它错误发生\n')
37     else:
38         root1,root2 = solve(a,b,c)
39         print(f'root1 = {root1:.2f},root2 = {root2:.2f}')
40         print()

 

 

运行结果：

line23增加 print(solve.__doc__):

 

 

实验任务四

task4

源代码：

 1 def list_genetator(q,z,b = 1):
 2     lst = []
 3     while q <= z:
 4         if q <= z+1:
 5             lst.append(q)
 6             q = q+b
 7         else:
 8             break
 9 
10     return lst
11 
12 list1 = list_genetator(-5,5)
13 print(list1)
14 
15 list2 = list_genetator(-5,5,2)
16 print(list2)
17 
18 list3 = list_genetator(1,5,0.5)
19 print(list3)

运行结果：

 

 

实验任务五

task5

源代码：

 1 def is_prime(n):
 2     if n > 1:
 3         if n%2 > 0:
 4             return True
 5         elif n == 2:
 6             return True
 7         else:
 8             return False
 9     else:
10         return False
11 
12 s = [i for i in range(1,21)]
13 s_1 = []
14 s_2 = []
15 s_3 = {}
16 for j in s:
17     if is_prime(j) == True:
18         s_1.append(j)
19 for y in s:
20     if y > 2 and y%2 == 0:
21         s_2.append(y)
22 for i in range(len(s_2)):
23     for z in s_1:
24         for q in s_1:
25             if z + q == s_2[i]:
26                 s_3.setdefault(s_2[i],(z,q))
27 for key,value in s_3.items():
28     a = value[0]
29     b = value[1]
30     print(f'{key} = {a}+{b}')

 

运行结果：

 

 

实验任务六

task6

源代码：

 1 def encoder(s):
 2     al = [chr(i) for i in range(97,123)]*2
 3     al_m = [chr(i) for i in range(65,91)]*2
 4     s_1 = list(s)
 5     for j in s_1:
 6         if j in al:
 7             a = al.index(j)
 8             a1 = s_1.index(j)
 9             t = al[a+5]
10             s_1[a1] = t
11         elif j in al_m:
12             b = al_m.index(j)
13             b1 = s_1.index(j)
14             t = al_m[b+5]
15             s_1[b1] = t
16         else:
17             pass
18     s_s = ''.join(s_1)
19     return s_s
20 
21 def decoder(text):
22     al = [chr(i) for i in range(97,123)]*2
23     al_m = [chr(i) for i in range(65,91)]*2
24     text_1 = list(text)
25     for j in text_1:
26         if j in al:
27             a = al.index(j)
28             a1 = text_1.index(j)
29             t = al[a-5]
30             text_1[a1] = t
31         elif j in al_m:
32             b = al_m.index(j)
33             b1 = text_1.index(j)
34             t = al_m[b-5]
35             text_1[b1] = t
36         else:
37             pass
38     text_s = ''.join(text_1)
39     return text_s
40 
41 
42 
43 
44 text = input('请输入英文文本：')
45 
46 encoded_text = encoder(text)
47 print('编码后的文本：',encoded_text)
48 
49 decoded_text = decoder(encoded_text)
50 print('对编码后的文本解码：',decoded_text)

 

运行结果：

说明：这个有点无能为力，知道为什么第二次测试结果不对的原因，因为替换后使用index函数只会获得列表中第一个出现该字母所对应的索引，但是暂时没想到解决办法。

 

实验任务七

task7

源代码：

 1 def collatz(n):
 2     s = []
 3     for i in range(n**2):
 4         s.append(n)
 5         if n == 1:
 6             break
 7         else:
 8             if n%2 == 0:
 9                 n = n/2
10             else:
11                 n = n*3+1
12     return s
13 
14 
15 try:
16     n = eval(input('Enter an interger:'))
17     ans = collatz(n)
18     if n<=0:
19         raise ValueError('Error: n must be an interger')
20 except ValueError as e:
21     print(repr(e))
22 except:
23     print('Error: n must be an interger')
24 else:
25     print(ans)

 

运行结果：

 

 

实验任务八

task8

源代码：

 1 def func(n):
 2     if n == 0:
 3         return 0
 4     else:
 5         return func(n-1)*2+1
 6 
 7 while True:
 8     x = input()
 9     if x == '#':
10         print('计算结束')
11         break
12     n = int(x)
13     ans = func(n)
14     print(f'n = {n},ans = {ans}')

 

运行结果：